Showing posts with label 12th. Show all posts
Showing posts with label 12th. Show all posts

Class 12 exams postponed, cancels Class 10 exams

 


CBSE Class 12 exams postponed, cancels Class 10 exams

The Ministry of Education has decided to postpone the class 12 exams and cancel the class 10 board exams. For CBSE class 12 boards, the situation will be reviewed on June 1 and revised dates will be announced thereafter.

For CBSE class 10 students, result will be prepared on the basis of an “objective criterion” to be developed by the Board. Any candidate who is not satisfied with the marks allocated to him/her on this basis will be given an opportunity to sit in an exam as and when the conditions are conducive to hold the exams. Last year too CBSE had devised a new scheme for class 10 students.

 



 

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New CBSE Syllabus 2021-22 for 9th, 10th, 11th, 12th Released: Applicable for CBSE Academic Session 2021-22







Subject-wise New CBSE Syllabus 2021-22 (PDF) for 9th, 10th, 11th & 12th has been officially released online at cbseacademic.nic.in. The new CBSE Syllabus is applicable for CBSE Academic Session 2021-22. The board has already instructed CBSE Schools to start new CBSE Academic Session 2021-22 from April and the board has released the new syllabus before April 2021. 

Also Check: CBSE Date Sheet 2021 (Revised) - 10th & 12th Board Exams 2021: Download PDF Now!

CBSE Syllabus 2021-22 for Class 9 & CBSE Syllabus 2021-22 for Class 10 

⇒ Download CBSE Syllabus 2021-22 for Maths: Class 9 and Class 10

 

⇒ Download CBSE Syllabus 2021-22 for Science: Class 9 and Class 10

⇒ Download CBSE Syllabus 2021-22 for Social Science: Class 9 and Class 10

⇒ Download CBSE Syllabus 2021-22 for Computer Applications: Class 9 and Class 10

⇒ Download CBSE Syllabus 2021-22 for Hindi Course-B: Class 9 and Class 10

⇒ Download CBSE Syllabus 2021-22 for Hindi Course-A: Class 9 and Class 10

With the links given above, students of CBSE Class 9th & Class 10th can check & download the latest CBSE Syllabus 2021-22 (PDF). 

CBSE Syllabus 2021-22 for Class 11 & CBSE Syllabus 2021-22 for Class 12

⇒ Download CBSE Syllabus 2021-22 for Physical Education: Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for (English Core): Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for (English Elective): Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for Physics: Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for Chemistry: Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for Biology: Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for Mathematics: Class 11 and Class 12

 Download CBSE Syllabus 2021-22 for Applied Mathematics: Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for Biotechnology: Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for Physics: Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for Information Prac: Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for Computer Science: Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for Accountancy: Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for Business Studies: Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for Psychology: Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for History: Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for Political Science: Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for Hindi Core: Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for Hindi Elective: Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for Geography: Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for Home Science: Class 11 and Class 12

⇒ Download CBSE Syllabus 2021-22 for Sociology: Class 11 and Class 12

With the links given above, students of CBSE Class 9th & Class 10th can check & download the latest CBSE Syllabus 2021-22 (PDF). 

CBSE Syllabus 2021-22 is the most important resource for the students & teachers as well. New CBSE Syllabus provides a roadmap for the academic year. It also provides details about chapters, topics, practical to be studied in the entire academic year. It also provides information about the format of the question paper & important details about practical exams.

Students are also advised to check the YouTube channel of Jagran Josh where students can also access important videos with tips & strategies.

Students preparing for CBSE 10th & 12th board exams 2021 can also check important questions, previous year papers, tips & strategies for exam preparation.

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Class 12 Chapter 11 (Three Diamensional Geometry) Class Notes Part II


  Three Dimensional Geometry 


1. CENTRAL IDEA OF 3D  
There are infinite number of points in space. We want to identify each and every point of space with the help of three mutually perpendicular coordinate axes OX, OY and OZ. 
2. AXES 
Three mutually perpendicular lines OX, OY, OZ are considered as three axes. 
3. COORDINATE PLANES 
Planes formed with the help of x and y axes is known as x-y plane similarly y and z axes y – z plane and with z and x axis z - x plane. 
4. COORDINATE OF A POINT 
Consider any point P on the space drop a perpendicular form that point to x - y plane then the algebraic length of this perpendicular is considered as z-coordinate and from foot of the perpendicular drop perpendiculars to x and y axes these algebraic length of perpendiculars are considered as y and x coordinates respectively. 
5. VECTOR REPRESENTATION OF A POINT IN SPACE 
If coordinate of a point P in space is (x, y, z) then the position vector of the point P with respect to the same origin is \(x\hat i + y\hat j + z\hat k\).
6. DISTANCE FORMULA
Distance between any two points (x1, y1, z1) and (x2, y2, z2) is given as \(\sqrt {{{({x_1} - {x_2})}^2} + {{({y_1} - {y_2})}^2} + {{({z_1} - {z_2})}^2}} \)
Vector Method
 We know that if position vector of two points A and B are given as \(\overrightarrow {OA} \) and \(\overrightarrow {OB} \) then
\[ \Rightarrow |\overrightarrow {AB} | = |\overrightarrow {AB}  - \overrightarrow {OA} |\] \[ \Rightarrow |\overrightarrow {AB} | = |({x_2}\hat i + {y_2}\hat j + {z_2}\hat k) - ({x_1}\hat i + {y_1}\hat j + {z_1}\hat k)|\] \[ \Rightarrow |\overrightarrow {AB} | = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} \]

7. DISTANCE OF A POINT P FROM COORDINATE AXES

Let PA, PB and PC are distances of the point P(x, y, z) from the coordinate axes OX, OY and OZ respectively, then \(PA = \sqrt {{y^2} + {z^2}} ,\;\;PB = \sqrt {{z^2} + {x^2}} ,\;\;PC = \sqrt {{x^2} + {z^2}} \)

8. SECTION FORMULA

(i) Internal Division : 

If point P divides the distance between the points A (x₁, y₁, z₁,) and B (x₂, y₂, Z₂,) in the ratio of m: n (internally). The coordinate of P is given as

12 class Maths Notes Chapter 11 Three Dimensional Geometry free PDF| Quick revision Three Dimensional Geometry Notes class 12 maths
12 class Maths Notes Chapter 11 Three Dimensional Geometry free PDF| Quick revision Three Dimensional Geometry Notes class 12 maths


Note:

All these formulae are very much similar to two dimension coordinate geometry.

9. CENTROID OF A TRIANGLE

12 class Maths Notes Chapter 11 Three Dimensional Geometry free PDF| Quick revision Three Dimensional Geometry Notes class 12 maths

10. INCENTRE OF TRIANGLE ABC

12 class Maths Notes Chapter 11 Three Dimensional Geometry free PDF| Quick revision Three Dimensional Geometry Notes class 12 maths

11. CENTROID OF A TETRAHEDRON

12 class Maths Notes Chapter 11 Three Dimensional Geometry free PDF| Quick revision Three Dimensional Geometry Notes class 12 maths

12. RELATION BETWEEN TWO LINES 

Two lines in the space may be coplanar and may be none coplanar. Non coplanar lines are called skew lines if they never intersect each other. Two parallel lines are also non intersecting lines but they are coplanar. Two lines whether intersecting or non intersecting, the angle between them can be obtained.

13. DIRECTION COSINES AND DIRECTION RATIOS 

Direction cosines : Let a, ß, be the angles which directed line makes with the positive directions of the  axes of x, y and z respectively, the \(\cos \alpha ,\cos \beta ,\cos \gamma \) are called the direction cosines of the line. The direction cosine denoted (l, m, n).

12 class Maths Notes Chapter 11 Three Dimensional Geometry free PDF| Quick revision Three Dimensional Geometry Notes class 12 maths


(ii) If l, m, n, be the direction cosines of a lines, then + m² + n² = 1

(iii) Direction ratios: Let a, b, c be proportional to the direction cosines, l, m, n, then a, b, c are called the direction ratios.

If a, b, c are the direction ratio of any line L the \(a\hat i + b\hat j + c\hat k\) will be a vector parallel to the line L.

If l, m, n are direction cosine of line L then \(l\hat i + m\hat j + n\hat k\) is a unit vector parallel to the line L.

(iv) If l, m, n be the direction cosines and a, b, c be the direction ratios of a vector, then

12 class Maths Notes Chapter 11 Three Dimensional Geometry free PDF| Quick revision Three Dimensional Geometry Notes class 12 maths

(v) If OP = r, when O is the origin and the direction cosines of OP are l, m, n then the coordinates of P are (lr, mr, nr).

If direction cosine of the line AB are l, m, n, | AB |= r, and the coordinate of A is (x₁, y₁, z₁,) then the coordinate of B is given as (x₁, + rl, y₁, + rm, z₁ + rn)

(vi) If the coordinates P and Q are (x₁, y₁, z₁) and (x₂, y₂, z₂) then the direction ratios of line PQ are, a = x₂ − x₁, b = y₂ - y₁ and c = z₂ - z₁, and the direction cosines of line PQ are l = \(\dfrac{{{x_2} - {x_1}}}{{|\overrightarrow {PQ} |}}\), m = \(\dfrac{{{y_2} - {y_1}}}{{|\overrightarrow {PQ} |}}\) and n = \(\dfrac{{{z_2} - {z_1}}}{{|\overrightarrow {PQ} |}}\)


(vii) Direction cosines of axes : Since the positive x-axis makes angles 0°, 90°, 90° with axes of x, y and z respectively. Therefore

Direction cosines of x-axis are (1,0, 0)
Direction cosines of y-axis are (0, 1, 0)
Direction cosines of z-axis are (0, 0, 1)

 
14. ANGLE BETWEEN TWO LINE SEGMENTS

If two lines having direction ratios a₁, b₁, c₁, and a₂, b₂, c₂, respectively then we can consider two vector parallel to the lines as \({a_1}\hat i + {b_1}\hat j + {c_1}\hat k\) and \({a_2}\hat i + {b_2}\hat j + {c_2}\hat k\) and angle between them can be given as: \[\cos \theta  = \dfrac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {a_1^2 + b_1^2 + c_1^2}  \cdot \sqrt {a_2^2 + b_2^2 + c_2^2} }}\]
(i) The line will be perpendicular if a1a2 + b1b2 + c1c2 = 0
(ii) The lines will be parallel, if \(\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{{{b_1}}}{{{b_2}}} = \dfrac{{{c_1}}}{{{c_2}}}\).
(iii) Two parallel lines have same direction cosines i.e. l1 = l2, m1 = m2 and n1 = n2.



15. PROJECTION OF A LINE SEGMENT ON A LINE 

(i) If the coordinates P and Q are (x₁, Y₁, Z₁,) and (X₂, Y₂, Z₂) then the projection of the line segments PQ on a line having direction cosines l, m, n is |l(x₂ - x₁) + m(y₂ - y₁) + n(z₂ - z₁)|
12 class Maths Notes Chapter 11 Three Dimensional Geometry free PDF| Quick revision Three Dimensional Geometry Notes class 12 maths

 

A PLANE 

If line joining any two points on a surface lies completely on it then the surface is a plane.
OR
If line joining any two points on a surface is perpendicular to some fixed straight line. Then this surface is called a plane. This fixed line is called the normal to the plane.

16. EQUATION OF A PLANE 

(i) Normal form of the equation of a plane is lx + my + nz = p, where, l, m, n are the direction cosines of the normal to the plane and p is the distance of the plane from the origin.

(ii) General form : ax + by + cz + d = 0 is the equation of a plane, where a, b, c are the direction ratios of the normal to the plane.

(iii) The equation of a plane passing through the point (x₁, y₁, z₁) is given by \[a({x_1} - {x_2}) + b({y_1} - {y_2}) + c({z_1} - {z_2}) = 0\] where a, b, c are the direction ratios of the normal to the plane.

(iv) Plane through three points : The equation of the plane through three non-collinear points (x1, Y1, z1), (x2, y2, z2) and (x3, y3, z3) is:   \[\left| {\begin{array}{*{20}{c}}x&y&z&1\\{{x_1}}&{{y_1}}&{{z_l}}&1\\{{x_2}}&{{y_2}}&{{z_2}}&1\\{{x_3}}&{{y_3}}&{{z_3}}&1\end{array}} \right| = 0\]
 
(v) Intercept Form: The equation of a plane cutting intercept a, b, c on the axes is \(\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1\)

(vi) Vector Form : The equation of a plane passing through a point having position vector \({\vec a}\) and normal to vector \({\vec n}\) is \((\vec r - \vec a) \cdot \vec n\) or \(\vec r \cdot \vec n = \vec a \cdot \vec n\).

Note: 
(a) Vector equation of a plane normal to unit vector \({\hat n}\) and at a distance d from the origin is: \(\vec r \cdot \hat n = d\)

(b) Planes parallel to the coordinate planes 

(i) Equation of yz - plane is x = 0
(ii) Equation of xz - plane is y = 0
(iii) Equation of xy - plane is z = 0

(c) Planes parallel to the axes : 

If a = 0, the plane is parallel to x-axis i.e. equation of the plane parallel to the x-axis is by + cz + d = 0. Similarly, equation of planes parallel to y-axis and parallel to z-axis are ax + cz + d = 0 and ax + by + d = 0 respectively.

(d) Plane through origin : Equation of plane passing through origin is ax + by + cz = 0.

(e) Transformation of the equation of a plane to the normal form: To reduce any equation ax + by + cz - d = 0 to the normal form, first write the constant term on the right hand side and make it positive, then divided each term by \(\sqrt {{a^2} + {b^2} + {c^2}} \), where a, b, c are coefficients of x, y and z respectively \[\dfrac{{ax}}{{ \pm \sqrt {{a^2} + {b^2} + {c^2}} }} + \dfrac{{by}}{{ \pm \sqrt {{a^2} + {b^2} + {c^2}} }} + \dfrac{{cz}}{{ \pm \sqrt {{a^2} + {b^2} + {c^2}} }} = \dfrac{d}{{ \pm \sqrt {{a^2} + {b^2} + {c^2}} }}\]
Where (+) sign is to be taken if d > 0 an (-) sign is to be taken if d < 0.

(f) Any plane parallel to the given plane ax + by + cz + d = 0 is ax + by + cz + λ = 0 distance between two parallel planes ax + by + cz + d1 = 0 and ax + dy + xz + d2 = 0 is given as: \[\frac{{|{d_1} - {d_2}|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]
 
(g) Equation of a plane passing through a given point and parallel to the given vectors: The equation of a plane passing through a point and having position vector \({\vec a}\) and parallel to \({\vec b}\) and \({\vec c}\) is \[\vec r = \vec a + \lambda \vec b + \mu \vec c\] parametric form (where \(\lambda \) and \(\mu \) are scalers).
OR
\(\vec r \cdot (\vec b \times \vec c) = \vec a \cdot (\vec b \times \vec c)\) (non parametric form)

(h) A plane ax + by + cz + d = 0 divides the line segment joining (x1, y1, z1) and (x2, y2, z2) in the ratio: \(\left( { - \dfrac{{a{x_1} + b{y_1} + c{z_1} + d}}{{a{x_2} + b{y_2} + c{z_2} + d}}} \right)\)
 
(i) The xy-plane divides the line segment joining the point (x1, y1, z1) and (x2, y2, z2) in the ratio \( - \dfrac{{{z_1}}}{{{z_2}}}\). Similarly yz - plane divides in \( - \dfrac{{{x_1}}}{{{x_2}}}\) and zx - plane divides in \( - \dfrac{{{y_1}}}{{{y_2}}}\).


17. ANGLE BETWEEN TWO PLANES

(i) Consider two planes ax + by + cz + d = 0 and a'x + b'y + c'z+ d' = 0. Angle between these planes is the angle between their normal. Since direction ratios of their normal are (a, b, c) and (a', b', c') respectively, hence \(\theta \) the angle between them is given by: \[\cos \theta  = \frac{{aa' + bb' + cc'}}{{\sqrt {{a^2} + {b^2} + {c^2}}  \cdot \sqrt {a{'^2} + b{'^2} + c{'^2}} }}\]
Planes are perpendicular if aa' + bb' + c'' =0 and planes are parallel, if \(\dfrac{a}{{a'}} = \dfrac{b}{{b'}} = \dfrac{c}{{c'}}\).

(ii) The angle \(\theta \) between the plane \(\vec r \cdot {{\vec n}_1} = {d_1}\) and \(\vec r \cdot {{\vec n}_1} = {d_1}\) is given by \[\cos \theta  = \dfrac{{{{\vec n}_1} \cdot {{\vec n}_2}}}{{|{{\vec n}_1}||{{\vec n}_2}|}}\]
Planes are perpendicular if \({{{\vec n}_1} \cdot {{\vec n}_2} = 0}\) and planes are parallel if \({{{\vec n}_1} = \lambda {{\vec n}_2}}\) .


18. A PLANE AND A POINT 

(i) Distance of the point (x', y', z') from the plane ax + by + cz + d = 0 is given by \[\dfrac{{ax' + by' + cz' + d}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]
(ii) The length of the perpendicular from a point having position vector \({\vec n}\) to the plane \(\vec r \cdot {\vec n} = {d}\) is given by \[p = \dfrac{{|\vec a \cdot \vec n - d|}}{{|\vec n|}}\]


19. ANGLE BISECTORS

(i) The equations of the planes bisecting the angle between two given planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 are\[\dfrac{{{a_1}x + {b_1}y + {c_1}z + {d_1}}}{{\sqrt {a_1^2 + b_1^2 + c_1^2} }} =  \pm \dfrac{{{a_2}x + {b_2}y + {c_2}z + {d_2}}}{{\sqrt {a_2^2 + b_2^2 + c_2^2} }}\]
 
(ii) Equation of bisector of the angle containing origin: First make both the constant terms positive. then the positive sign in \(\dfrac{{{a_1}x + {b_1}y + {c_1}z + {d_1}}}{{\sqrt {a_1^2 + b_1^2 + c_1^2} }} =  \pm \dfrac{{{a_2}x + {b_2}y + {c_2}z + {d_2}}}{{\sqrt {a_2^2 + b_2^2 + c_2^2} }}\) gives the bisector of the angle which contains the origin.

(iii) Bisector of acute/obtuse angle : First make both the constant terms positive. Then
     a1a2 + b1b2 + c1c2 > 0 ⇒ origin lies obtuse angle
     a1a2 + b1b2 + c1c2 < 0 ⇒ origin lies in acute angle


20. FAMILY OF PLANES

(i) Any plane passing through the line of intersection of non- parallel planes OR equation of the plane through the given line in non symmetrical form. a1x + b1y + c1z + d1 =0 and a2x + b2y + c2z + d2 = 0 is  a1x + b1y + c1z+ d1 + λ(a2x + b2y + c2z + d2) = 0
 
(ii) The equation of plane passing through the intersection of the planes \(\vec r \cdot {{\vec n}_1} = {d_1}\) and \(\vec r \cdot {{\vec n}_2} = {d_2}\) is \(\vec r \cdot ({{\vec n}_1} + \lambda {{\vec n}_2}) = {d_1} + \lambda {d_2}\).

(iii) Plane through a given line : Equation of any plane through the line in symmetrical form \(\dfrac{{x - {x_1}}}{l} = \dfrac{{y - {y_1}}}{m} = \dfrac{{z - {z_1}}}{n}\) is A(x - x1) + B(y - y1) + c(z - z1) = 0, where Al + Bm + Cn = 0.


21. AREA OF A TRIANGLE
Let A(x1, y1, z1), B(x2, y2, z2), C(x3, y3, z3) be the vertices of a triangle then Area of triangle is \(\Delta  = \sqrt {\Delta _x^2 + \Delta _y^2 + \Delta _z^2} \), where \({\Delta _x} = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}{{y_1}}&{{z_1}}&1\\{{y_2}}&{{z_2}}&1\\{{y_3}}&{{z_3}}&1\end{array}} \right|\), \({\Delta _y} = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}{{z_1}}&{{x_1}}&1\\{{z_2}}&{{x_2}}&1\\{{z_3}}&{{x_3}}&1\end{array}} \right|\) and \({\Delta _z} = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}{{x_1}}&{{y_1}}&1\\{{x_2}}&{{y_2}}&1\\{{x_3}}&{{y_3}}&1\end{array}} \right|\)
 
Vector Method: From Two vectors \(\overrightarrow {AB} \)  and \(\overrightarrow {AC} \). then area is given by \[\dfrac{1}{2}|\overrightarrow {AB}  \times \overrightarrow {AC} | = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}{\hat i}&{\hat j}&{\hat k}\\{{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}}\\{{x_3} - {x_1}}&{{y_3} - {y_1}}&{{z_3} - {z_1}}\end{array}} \right|\]
 


22. VOLUME OF A TETRAHEDRON 

Volume of a tetrahedron with vertices A(x1, y1, z1), B(x2, y2, z2), C(x3, y3, y3) and D(x4, y4, y4,) is given by: \[V = \frac{1}{6}\left| {\begin{array}{*{20}{c}}{{x_1}}&{{y_1}}&{{z_1}}&1\\{{x_2}}&{{y_2}}&{{z_2}}&1\\{{x_3}}&{{y_3}}&{{z_3}}&1\\{{x_4}}&{{y_4}}&{{z_4}}&1\end{array}} \right|\]
 
 
23. EQUATION OF A LINE 

(i) A straight line in space is characterized by the intersection of two planes which are not parallel and therefore, the equation of a straight line is a solution of the system constituted by the equations of the two planes, a1x + b1y + c1z +d1 = 0 and a2x + b2y + c2z + d2 = 0. This form is also known as non-symmetrical form.
(ii) The equation of a line passing through the point (x1, y1, z1,) and having direction ratios a, b, c is \[\frac{{x - {x_1}}}{a} = \frac{{y - {y_1}}}{b} = \frac{{z - {z_1}}}{c} = r\] This is called the symmetric form. A general point on the line is given by (x1 + ar, y1 + br, z1 + cr).
(iii) Vector Equation: Vector equation of a straight line passing through a fixed point with position vector \({\vec a}\) and parallel to a given vector \({\vec b}\) is \(\vec r = \vec a + \lambda \vec b\) where \(\lambda \) is scalar.
 
(iv) The equation of the line passing through the points (x1, y1, z1) and (x2, y2, z2) is: \[\dfrac{{x - {x_1}}}{{{x_2} - {x_1}}} = \dfrac{{y - {y_1}}}{{{y_2} - {y_1}}} = \dfrac{{z - {z_1}}}{{{z_2} - {z_1}}}\]
(v) Vector equation of a straight line passing through two points with position vectors \({\vec a}\) and \({\vec b}\) is \(\vec r = \vec a + \lambda (\vec b - \vec a)\).
(vi) Reduction of cartesion form of equation of a line to vector form and vice versa \[\frac{{x - {x_1}}}{a} = \frac{{y - {y_1}}}{b} = \frac{{z - {z_1}}}{c} \Leftrightarrow \vec r = ({x_1}\hat i + {y_1}\hat j + {z_1}\hat k) + \lambda (a\hat i + b\hat j + c\hat k)\]
 

NOTE:
Straight lines parallel to coordinate axes: 
 Straight Lines
Equation
(i)Through Origin
y = mx, z = nx
(ii) x - axis
y = 0, z = 0
(iii) y - axis
x = 0, z = 0
(iv) z - axis
x = 0, y = 0
(v) Parallel to x - axis
y = q, z = r
(vi) Parallel to y - axis
x = p, z = r
(vii) Parallel to z - axis
x = p, y = q

 
24. ANGLE BETWEEN A PLANE AND A LINE

12 class Maths Notes Chapter 11 Three Dimensional Geometry free PDF| Quick revision Three Dimensional Geometry Notes class 12 maths

 
25. CONDITION FOR A LINE TO LIE IN A PLANE
 
(i) Cartesian form: Line \(\dfrac{{x - {x_1}}}{l} = \dfrac{{y - {y_1}}}{m} = \dfrac{{z - {z_1}}}{n}\) would lie in a plane ax + by + cz + d = 0, if ax1 + by1 + cz1 + d = 0 and al + bm + cn = 0.

(ii) Vector form: Line \(\vec r = \vec a + \lambda \vec b\) would lie in the plane \(\vec r \cdot \vec n = d\) if \(\vec b \cdot \vec n = 0\) and \(\vec a \cdot \vec n = d\).
 
 
26. COPLANER LINES

12 class Maths Notes Chapter 11 Three Dimensional Geometry free PDF| Quick revision Three Dimensional Geometry Notes class 12 maths
12 class Maths Notes Chapter 11 Three Dimensional Geometry free PDF| Quick revision Three Dimensional Geometry Notes class 12 maths
27. SKEW LINES 

(i) The straight lines which are not parallel and non-coplanar i.e. non-intersecting are called skew lines.
If \[\Delta  = \left| {\begin{array}{*{20}{c}}{\alpha ' - \alpha }&{\beta ' - \beta }&{\gamma ' - \gamma }\\l&m&n\\{l'}&{m'}&{n'}\end{array}} \right| \ne 0\]
then the lines are skew.
 
(ii) Vector Form: For lines \({{\vec a}_1} + \lambda {{\vec b}_1}\) and \({{\vec a}_2} + \lambda {{\vec b}_2}\) to be skew, when \(({{\vec b}_1} \times {{\vec b}_2})({{\vec a}_2} - {{\vec a}_1}) \ne 0\) or \([{{\vec b}_1}{{\vec b}_2}({{\vec a}_2} - {{\vec a}_1})] \ne 0\) 
 
(iii) Shortest distance between the two parallel lines \(\vec r = {{\vec a}_1} + \lambda \vec b\) and \(\vec r = {{\vec a}_2} + \mu \vec b\) is \[d = \left| {\frac{{({{\vec a}_2} - {{\vec a}_1}) \times \vec b}}{{|\vec b|}}} \right|\]


28. COPLANARITY OF FOUR POINTS 

The points A(x1, y1, z1), B(x2, y2, z2), C(x3, y3, z3) and D(x4, y4, z4) are coplaner, then \[\begin{array}{l}\left| {\begin{array}{*{20}{c}}{{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}}\\{{x_3} - {x_1}}&{{y_3} - {y_1}}&{{z_3} - {z_1}}\\{{x_4} - {x_1}}&{{y_4} - {y_1}}&{{z_4} - {z_1}}\end{array}} \right| = 0\\\end{array}\]


29. SIDES OF. PLANE 

A plane divides the three dimensional space two equal parts. Two points A (x1,y1, z1,) and B (x2, y2, z2) are on the same side of the plane ax + by + cz + d = 0 if ax1 + by1 + cz1 + d and ax2 + by2 + cz2 + d and both positive or both negative and are opposite side of plane if both of these values are in opposite sign.

30. LINE PASSING THROUGH THE GIVEN POINT (x1 y1 z1) AND INTERSECTING BOTH THE LINES    (P1 = 0, P2 = 0)    AND     (P3 = 0, P4 = 0) 
Get a plane through (x1, y1, z1) and containing the line (P1 = 0, P2 = 0) as P5 = 0
Also get a plane through (x1, y1, z1) and containing the line P3 = 0, P4 = 0 as P6 = 0
Equation of the required line is (P5 = 0, P6 = 0)

31. TO FIND IMAGE OF A POINT W.R.T. A LINE

Let is a given line. Let (x', y', z') is the image of the point P(x1, y1, z1) with respect to the line L.
then,
(i) a(x1 - x') + b(y1 - y') +c(z1 - z') = 0
(ii) \(\dfrac{{\dfrac{{x + {x_1}}}{2} - {x_2}}}{a} = \dfrac{{\dfrac{{{y_1} - y'}}{2} - {y_2}}}{b} = \dfrac{{\dfrac{{{z_1} - z'}}{2} - {z_2}}}{c} = \lambda \)
from (ii) get the value of x', y', z' in terms of λ as x' = 2aλ + 2x2 - x1, y' = 2bλ + 2y2 - y1 and z' = 2cλ + 2z2 - z1. Now, put the values of x', y', z' in (i) get λ and resubstitute the value of λ to get (x', y', z').

 

32. TO FIND IMAGE OF A POINT W.R.T. A PLANE

Let P(x1, y1, z1) is a given point and ax + by + cz + d = 0 is given plane. Let (x', y', z') is the image point, then
  1. x' - x1 = λa, y' - y1 = λb, z' - z1 = λc, \( \Rightarrow \) x' = λa + x1, y' = λb + y1, z' = λc + z1,
  2.  \(a\left( {\dfrac{{x' + {x_1}}}{2}} \right) + b\left( {\dfrac{{y' + {y_1}}}{2}} \right) + c\left( {\dfrac{{z' + {z_1}}}{2}} \right) + d = 0\)

 from (i) put the values of x', y', z' in (ii) and get the values of λ and re-substitute in (i) to get (x', y', z').



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Class 12 Chapter 10 (Vector Algebra) Class Notes Part II

 

Product of Two Vectors

Considering examples:-

Suppose we have a wooden stick and in order to move the stick we will apply some force onto it.

The stick will move this shows some work is done in moving the stick.

Work done = Force × Displacement

Where Force and Displacement both are vector quantities and the work done is a scalar quantity.

 

Consider the case when the wooden stick is fixed on the wall with the help of a nail.

Again force is applied the wooden stick will start rotating.

Torque = Force × Distance 

Where Force and Distance both are vector quantities and the torque is also a vector quantity.

 Whenever cross product is considered then the resultant will be a vector quantity.

 Whenever dot product is considered then the resultant is a scalar quantity.


Scalar (or Dot Product) of two vectors

The scalar  product of  two nonzero vectors \({\overrightarrow a }\) and \({\overrightarrow b }\) is denoted by \(\overrightarrow a  \cdot \overrightarrow b \). It is defined as:  \(\overrightarrow a  \cdot \overrightarrow b  = |\overrightarrow a ||\overrightarrow b |\cos \theta \), Where θ = angle between \(\overrightarrow a \) and \(\overrightarrow b \), \(|\overrightarrow a |\) = magnitude of \(\overrightarrow a \) and \(|\overrightarrow a |\) = magnitude of \(\overrightarrow b \).

 

Derivation with the help of example:-

Consider a log of wood and if a  Force F  is  applied  on it, the log  of wood moves. From the  figure  we  can  see  that the θ  is the angle  between displacement and force.

Force can  be  written into 2 components i.e. Fcosθ and Fsinθ.

Work done = (Fx Displacement + Fy Displacement), Where FX = Force in x direction and FY = Force in y direction.

Work done = \(F\cos \theta \overrightarrow D  + F\sin \theta \), Where \(\overrightarrow D \) = displacement in x direction and displacement in y direction is 0.

Therefore Work done = \(F\cos \theta \overrightarrow D  \Rightarrow W = |\overrightarrow F ||\overrightarrow D |\cos \theta \) 

 

Observations:-

\(\overrightarrow a  \cdot \overrightarrow b \) is a real number.

Let \(\overrightarrow a \) and \(\overrightarrow b \) be two  nonzero  vectors, then \(\overrightarrow a  \cdot \overrightarrow b  = 0\), if and only if  \(\overrightarrow a \) and  \(\overrightarrow b \) are perpendicular to each other i.e. \(\overrightarrow a  \cdot \overrightarrow b  = 0\) iff \(\overrightarrow a \) ⊥ \(\overrightarrow b \).

( Î¸ = 900 then cos900 = 0 therefore  \(\overrightarrow a  \cdot \overrightarrow b  = 0\)).

If θ = 0, then \(\overrightarrow a  \cdot \overrightarrow b  = |\overrightarrow a ||\overrightarrow b |\)

 

If θ = Ï€, then \(\overrightarrow a  \cdot \overrightarrow b  =  - |\overrightarrow a ||\overrightarrow b |\)

 

Therefore \(\left( {\widehat i \cdot \widehat i = \widehat j \cdot \widehat j = \widehat k \cdot \widehat k = 1} \right)\) as  they  are in same  direction which shows they are  parallel to each other and \(\left( {\widehat i \cdot \widehat k = \widehat j \cdot \widehat k = \widehat k \cdot \widehat i = 0} \right)\) as they are perpendicular to each other. The scalar product is commutative i.e. \(\overrightarrow a  \cdot \overrightarrow b  = \overrightarrow b  \cdot \overrightarrow a \).

 

Angle between Vectors: Scalar (or Dot product)

To Prove: \(\cos \theta  = \dfrac{{\overrightarrow a  \cdot \overrightarrow b }}{{|\overrightarrow a ||\overrightarrow b |}}\) or \(\theta  = {\cos ^{ - 1}}\left( {\dfrac{{\overrightarrow a  \cdot \overrightarrow b }}{{|\overrightarrow a ||\overrightarrow b |}}} \right)\).

Consider \({\overrightarrow a  \cdot \overrightarrow b  = |\overrightarrow a ||\overrightarrow b |\cos \theta }\)   (equation 1)

Dividing both the sides of equation(1)   by \({|\overrightarrow a ||\overrightarrow b |}\) we get,

\(\cos \theta  = \dfrac{{\overrightarrow a  \cdot \overrightarrow b }}{{|\overrightarrow a ||\overrightarrow b |}}\) or \(\theta  = {\cos ^{ - 1}}\left( {\dfrac{{\overrightarrow a  \cdot \overrightarrow b }}{{|\overrightarrow a ||\overrightarrow b |}}} \right)\).


Properties of scalar product

Let \(\overrightarrow a \) and \(\overrightarrow b \) be any two vectors and λ be any scalar. Then \[(\lambda \overrightarrow a ) \cdot \overrightarrow b  = \lambda (\overrightarrow a  \cdot \overrightarrow b ) = \overrightarrow a  \cdot (\lambda \overrightarrow b )\]

Derivation of scalar product:-

To derive :- (\overrightarrow a  \cdot \overrightarrow b ) = (a1 b1) +( a2 b2) + (a3 b3)

Consider 2 vectors such that \(\overrightarrow a  = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k\)  and \(\overrightarrow b  = {b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k\).

\(\overrightarrow a  \cdot \overrightarrow b  = ({a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k) \cdot ({b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k)\)

\( = ({a_1}.{b_1})(\widehat i.\widehat i) + ({a_1}.{b_2})(\widehat i.\widehat j) + ({a_1}.{b_3})(\widehat i.\widehat k) + ({a_2}.{b_1})(\widehat j.\widehat i) + ({a_2}.{b_2})(\widehat j.\widehat j) + ({a_2}.{b_3})(\widehat j.\widehat k) + ({a_3}.{b_1})(\widehat k.\widehat i) + ({a_3}.{b_2})(\widehat k.\widehat j) + ({a_3}.{b_3})(\widehat k.\widehat k)\) equation (1)

Using \(\left( {\widehat i \cdot \widehat i = \widehat j \cdot \widehat j = \widehat k \cdot \widehat k = 1} \right)\)  and \(\left( {\widehat i \cdot \widehat k = \widehat j \cdot \widehat k = \widehat k \cdot \widehat i = 0} \right)\)  equation(1) becomes

Therefore \(\overrightarrow a  = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k\)  and \(\overrightarrow b  = {b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k\).  Hence proved.


Projection of a vector on a line

Suppose a vector \(\overrightarrow {AB} \) makes  an angle θ with a given directed line l (say). Then the projection of \(\overrightarrow {AB} \) on l is a  vector \(\overrightarrow {p} \) (say) with magnitude | AB| cosθ, and the direction of \(\overrightarrow {p} \) being the same (or opposite) to that of the line l, depending upon whether cos θ is positive or negative.

The vector \(\overrightarrow {p} \) is called the projection vector and its magnitude |\(\overrightarrow {p} \)|  is called as the projection of the vector \(\overrightarrow {AB} \) on line l

Problem:-

Find the angle between two vectors \(\overrightarrow {a} \) and \(\overrightarrow {b} \) with magnitudes \(\sqrt 3 \) and 2, respectively having  \(\overrightarrow a  \cdot \overrightarrow b  = \sqrt 6 \).

Answer: It is given that, \({|\overrightarrow a | = \sqrt 3 ,|\overrightarrow b | = 2}\) and \({\overrightarrow a  \cdot \overrightarrow b  = \sqrt 6 }\).

\(\overrightarrow a  \cdot \overrightarrow b  = |\overrightarrow a ||\overrightarrow b |\cos \theta \)

Therefore, \(\sqrt 6  = \sqrt 3  \times 2 \times \cos \theta \)

\( \Rightarrow \cos \theta  = \dfrac{1}{{\sqrt 2 }}\)

\( \Rightarrow \theta  = \dfrac{\pi }{4}\)

Hence, the angle between the given vectors \(\overrightarrow {a} \) and  \(\overrightarrow {b} \) is \(\dfrac{\pi }{4}\).

 

Problem:  Find the angle between the vectors \(\widehat i - 2\widehat j + 3\widehat k\) and \(3\widehat i - 2\widehat j + \widehat k\)?

Answer:  The given vectors are \(\overrightarrow a  = \widehat i - 2\widehat j + 3\widehat k\) and \(\overrightarrow b  = 3\widehat i - 2\widehat j + \widehat k\).

\(|\overrightarrow a | = \sqrt {{1^2} + {{( - 2)}^2} + {3^2}}  = \sqrt {14} \) and \(|\overrightarrow b | = \sqrt {{3^2} + {{( - 2)}^2} + {1^2}}  = \sqrt {14} \).

Now \(\overrightarrow a  \cdot \overrightarrow b  = (\widehat i - 2\widehat j + 3\widehat k) \cdot (3\widehat i - 2\widehat j + \widehat k)\) = 1.3 + (-2) (-2) + 3.1 = (3 + 4 + 3) = 10.

Also, we know that \(\overrightarrow a  \cdot \overrightarrow b  = |\overrightarrow a ||\overrightarrow b |\cos \theta \)

Therefore, \(10 = \sqrt {14} \sqrt {14} \cos \theta  \Rightarrow \cos \theta  = \left( {\dfrac{5}{7}} \right)\)

\( \Rightarrow \theta  = {\cos ^{ - 1}}\left( {\dfrac{5}{7}} \right)\)

 

Problem:  Find the projection of the vector î – ĵ  on the vector î + Äµ.

Answer:  Let \(\overrightarrow a \) = ( î – ĵ) and \(\overrightarrow b \) = ( î + Äµ)

Now, the projection of vector \(\overrightarrow {a} \) on \(\overrightarrow {b} \) is given by,

\({\dfrac{{\overrightarrow a  \cdot \overrightarrow b }}{{|\overrightarrow b |}} = \dfrac{{1 - 1}}{{\sqrt {{1^2} + {1^2}} }} = 0}\)

Hence, the projection of vector \({\overrightarrow a }\) on \({\overrightarrow b }\) is  0.

Problem:-

Find the projection of the vector (î +3 Äµ +7 k̂) on the vector (7î - Äµ +8 k̂).

Answer:-

Let \(\overrightarrow a  = \widehat i + 3\widehat j + 7\widehat k\) and \(\overrightarrow b  = 7\widehat i - \widehat j + 8\widehat k\)

 Now, projection of vector \(\overrightarrow a \) on \(\overrightarrow b \) is given by,

\(\dfrac{{\vec a \cdot \vec b}}{{|\vec b|}} = \dfrac{{7 - 3 + 36}}{{\sqrt {{1^2} + {{( - 1)}^2} + {8^2}} }} = \dfrac{{60}}{{\sqrt {114} }}\) 
Problem:-

Show that each of the given three vectors is a unit vector: \(\dfrac{{2\hat i + 3\hat j + 6\hat k}}{7},\;\dfrac{{3\hat i - 6\hat j + 2\hat k}}{7},\;\dfrac{{6\hat i + 2\hat j - 3\hat k}}{7}\). Also, show that they are mutually perpendicular to each other.

Answer:-

Let \(\vec a = \dfrac{{2\hat i + 3\hat j + 6\hat k}}{7} = \dfrac{2}{7}\hat i + \dfrac{3}{7}\hat j + \dfrac{6}{7}\hat j,\),   \(\vec b = \;\dfrac{{3\hat i - 6\hat j + 2\hat k}}{7} = \dfrac{3}{7}\hat i - \dfrac{6}{7}\hat j + \dfrac{2}{7}\hat j,\),  \(\vec c = \;\dfrac{{6\hat i + 2\hat j - 3\hat k}}{7} = \dfrac{6}{7}\hat i + \dfrac{2}{7}\hat j - \dfrac{3}{7}\hat j\)

\(|\vec a| = \sqrt {{{\left( {\dfrac{2}{7}} \right)}^2} + {{\left( {\dfrac{3}{7}} \right)}^2} + {{\left( {\dfrac{6}{7}} \right)}^2}}  = \sqrt {\dfrac{{4 + 9 + 36}}{{49}}}  = 1\),  \(|\vec b| = \sqrt {{{\left( {\dfrac{3}{7}} \right)}^2} + {{\left( {\dfrac{{ - 6}}{7}} \right)}^2} + {{\left( {\dfrac{2}{7}} \right)}^2}}  = \sqrt {\dfrac{{9 + 36 + 4}}{{49}}}  = 1\),  \(|\vec c| = \sqrt {{{\left( {\dfrac{3}{7}} \right)}^2} + {{\left( {\dfrac{2}{7}} \right)}^2} + {{\left( {\dfrac{{ - 3}}{7}} \right)}^2}}  = \sqrt {\dfrac{{36 + 4 + 9}}{{49}}}  = 1\).

Thus, each of the given three vectors is a unit vector.

\(\vec a \cdot \vec b = \dfrac{2}{7} \times \dfrac{3}{7} + \dfrac{3}{7} \times \dfrac{{ - 6}}{7} + \dfrac{6}{7} \times \dfrac{2}{7} = \dfrac{6}{{49}} - \dfrac{{18}}{{49}} + \dfrac{{12}}{{49}} = 0\).

\(\vec b \cdot \vec c = \dfrac{3}{7} \times \dfrac{6}{7} + \dfrac{{ - 6}}{7} \times \dfrac{2}{7} + \dfrac{2}{7} \times \dfrac{{ - 3}}{7} = \dfrac{{18}}{{49}} - \dfrac{{12}}{{49}} - \dfrac{6}{{49}} = 0\).

\(\vec c \cdot \vec a = \dfrac{6}{7} \times \dfrac{2}{7} + \dfrac{2}{7} \times \dfrac{3}{7} + \dfrac{{ - 3}}{7} \times \dfrac{6}{7} = \dfrac{{12}}{{49}} + \dfrac{6}{{49}} - \dfrac{{18}}{{49}} = 0\).

Hence, the given three vectors are mutually perpendicular to each other.

 

Problem:-

 Find \(|\vec a|\)  and \(|\vec b|\), if \((\vec a + \vec b) \cdot (\vec a - \vec b) = 8\) and \(|\vec a|\) = 8 \(|\vec b|\).

Answer:-

\( \Rightarrow (\vec a \cdot \vec a - \vec a \cdot \vec b + \vec b \cdot \vec a - \vec b \cdot \vec b) = 8\)

\( \Rightarrow |\vec a{|^2} - |\vec b{|^2} = 8\) \( \Rightarrow {(8|\vec b|)^2} - |\vec b{|^2} = 8\)  using \(|\vec a|\) = 8 \(|\vec b|\).

\( \Rightarrow 64|\vec b{|^2} - |\vec b{|^2} = 8\) \( \Rightarrow 63|\vec b{|^2} = 8\)

\( \Rightarrow |\vec b| = \sqrt {\dfrac{8}{{63}}}  = \dfrac{{2\sqrt 2 }}{{3\sqrt 7 }}\)

\( \Rightarrow |\vec a| = 8|\vec b| = 8 \times \dfrac{{2\sqrt 2 }}{{3\sqrt 7 }} = \dfrac{{16\sqrt 2 }}{{3\sqrt 7 }}\)

 

Problem:-

Evaluate the product (3 a ⃗ - 5 b ⃗). (2 a ⃗ + 7 b ⃗).

Answer:-

(3 a ⃗ - 5 b ⃗). (2 a ⃗ + 7 b ⃗)

= (3 a ⃗. 2 a ⃗ + 3 a ⃗. 7 b ⃗ - 5 b ⃗.2 a ⃗ - 5 b ⃗.7 b ⃗)

= (6 a ⃗. a ⃗+21 a ⃗. b ⃗-10 a ⃗ b ⃗-35 b ⃗. b ⃗)

=6 I a ⃗I2 + 11 a ⃗. b ⃗ - 35 I b ⃗I2

Problem:-

Find the magnitude of two vectors a ⃗ and b ⃗, having the

same magnitude and such  that  the angle between them is 60°  and

their scalar product is (1/2).

Answer:-

Let θ be the angle between the vectors a ⃗ and b ⃗.

It is given that I a ⃗I = I b ⃗ I, (a ⃗. b ⃗) =(1/2), and θ  = 60°.       (1)

We know that a ⃗. b ⃗ = I a ⃗I I b ⃗ I cos θ.

Therefor (1/2) = I a ⃗I I a ⃗ I cos 60° Using (1)

=> (1/2) = I a ⃗I2 x (1/2)

=> I a ⃗I2 = 1

=> I a ⃗I = I b ⃗I = 1

Problem:-

Find I x ⃗I,  if for a unit vector a ⃗, (x ⃗- a ⃗) (x ⃗+ a ⃗) =12?

Answer:-

(x ⃗- a ⃗) (x ⃗+ a ⃗) = 12

=> (x ⃗. x ⃗ + x ⃗. a ⃗ - a ⃗. x ⃗ - a ⃗.a ⃗)  = 12

=> I x ⃗I2 - I a ⃗I2  = 12

=> I x ⃗I2 – 1 = 12   (I a ⃗I = 1 as a ⃗ is a unit vector).

=> I x ⃗I2  = 13

Therefore I x ⃗I =  √ (13).

 

Problem:-

If a ⃗ = 2î +2 Äµ +3 k̂, b ⃗=-î +2 Äµ +k̂ and c ⃗ = 3î +ĵ  are such that a ⃗+ λ b ⃗ 

Is  perpendicular to c ⃗, then find the value of λ.

Answer:-

The given vectors are a ⃗ = 2î +2 Äµ +3 k̂,  b ⃗=-î +2 Äµ +k̂ and c ⃗ = 3î + ĵ

Now, a ⃗+ λ b ⃗ = (2î +2 Äµ +3 k̂) + λ (-î +2 Äµ +k̂)

= (2- λ) Ã® + (2+2 λ) Äµ + (3+ λ) k̂

If (a ⃗+ λ b ⃗) is perpendicular to c ⃗, then

(a ⃗+ λ b ⃗). c ⃗=0.

[(2- Î») î +(2+2 Î») Äµ+(3+ Î») k̂].(3 Ã®+ ĵ)=0

(2- Î») (3) + (2+2 Î») (1) + (3+ Î») (0) =0

=> 6- 3 λ + 2+ 2 λ =0

=>- λ+8=0

=> λ=8.

Hence, the required value of λ is 8.

 

Problem:-

Show that: (I a ⃗ I b ⃗) +( I b ⃗ I a ⃗) is perpendicular to

 (I a ⃗ I b ⃗) -( I b ⃗ I a ⃗), for any two nonzero vectors a ⃗ and b ⃗.

Answer:-

((I a ⃗ I b ⃗) + (I b ⃗ I a ⃗)). ((I a ⃗ I b ⃗) - (I b ⃗ I a ⃗))

= (I a ⃗I2 b ⃗. b ⃗) - (I a ⃗ I I b ⃗ I) (b ⃗. a ⃗) + (( I b ⃗ I I a ⃗ I a ⃗. b ⃗) – (I b ⃗I2 a ⃗. a ⃗))

= I a ⃗I2 I b ⃗I2 - I b ⃗I2 I a ⃗I2

=0

Hence (I a ⃗ I b ⃗) +( I b ⃗ I a ⃗)  and (I a ⃗ I b ⃗) -( I b ⃗ I a ⃗),

are perpendicular to each other.

 

Problem:-

If (a ⃗. a ⃗=0) and (a ⃗. b ⃗=0),  then what can be concluded about  the vector b ⃗?

Answer:-

It is given that (a ⃗. a ⃗=0) and (a ⃗. b ⃗=0).

Now,

(a ⃗. a ⃗)=0 => I a ⃗ I2 =0 => I a ⃗ I =0

Therefore a ⃗  is  a  zero vector.

Hence, vector b ⃗ satisfying  (a ⃗. b ⃗) can  be  any  vector.

 

 

Problem:-

If the vertices A, B, C of a triangle ABC are (1, 2, 3), (–1, 0, 0), (0, 1, 2), respectively, then find ∠ ABC. [∠ ABC is the angle between the vectors BA ⃗ and BC ⃗]

Answer:-

The vertices of ∆ABC are given as A (1, 2, 3), B (–1, 0, 0), and C (0, 1, 2).

Also, it is given that ∠ ABC is the angle between the vectors BA ⃗ and BC ⃗.

BA ⃗= {1-(-1)} Ã® + (2-0) ĵ + (3-0) k̂ = 2î +2 Äµ +3 k̂

BC ⃗ = {0-(-1)} Ã® + (1-0) ĵ + (2-0) k̂ =î + Äµ +2k̂

Therefore (BA ⃗. BC ⃗) = (2î +2 Äµ +3 k̂). (î + Äµ +2k̂)

=2x1+2x1+3x2 = 2+2+6 =10

| BA ⃗|=√ (2)2+ (2)2+ (3)2 = √4+4+9=√17

| BC ⃗|=√1+1+ (2)2 = √6.

Now, it is known that:

(BA ⃗. BC ⃗) = | BA ⃗|| BC ⃗| cos [∠ ABC]

Therefore, 10 =√17 x√6 cos [∠ ABC]

=> cos [∠ ABC] = (10)/ (√17 x√6)

=> ∠ ABC = cos-1(10/√102)

Problem:-

Show that the points A (1, 2, 7), B (2, 6, 3) and C (3, 10, –1) are collinear.

Answer:-

The given points are A (1, 2, 7), B (2, 6, 3), and C (3, 10, –1).

Therefore    AB ⃗ = (2-1) Ã® + (6-2) Äµ + (3-7) k̂ = î+ 4 ĵ -4 k̂

 BC ⃗ = (3-2) Ã® + (10-6) Äµ + (-1-3) k̂ = î+ 4 ĵ -4 k̂

AC ⃗ = (3-1) Ã® + (10-2) Äµ + (-1-7) k̂= 2î+ 8 ĵ -8 k̂

I AB ⃗I=√ ((1)2 + (4)2+ (-4)2) = √1+16+16= √33

I BC ⃗I=√ ((1)2 + (4)2+ (-4)2) = √1+16+16= √33

I AC ⃗I=√ ((2)2 + (8)2+ (8)2) =√4+64+64 =√132=2√33

Therefore I AC ⃗I = I AB ⃗I + I BC ⃗I

Hence, the given points A, B, and C are collinear.


Vector (Cross) Product

The vector product of two nonzero vectors  a-> and  b-> is  denoted  by a->  × b->.

Vector product is given as:- (a× b) =| a || b | sin θ nˆ

Where n ˆ = unit vector perpendicular to both  a-> and b->.  The value of n ˆ is  given by the right hand thumb rule.

 

Consider a wooden stick which is fixed in the wall with the help of a nail.

If we apply force as shown in the figure the stick will topple.

Torque produced = F ⃗x  r ⃗

θ = angle between  the  r ⃗ and  the  force 

The 2 components of force will be F cos θ (vertical) and F sin θ (horizontal).

Form the figure we can see as F cos θ  and  r ⃗  are both in  same    directions  so  Fcos θ  won’t produce any effect.

Therefore Torque =| F ⃗ || r ⃗ | sin θ.

Conclusion:-

Result of cross product of any 2 vectors is a vector.

Direction is perpendicular to both inputs (F and r).

 

Vector Product: Right handed rectangular coordinate systems

In order to  find  the  vector product of 2  vectors    a->  and    b->, then  the direction of   a->  × b-> will  be  given  by the thumb.

 

Observations:-

a->  × b-> is a vector.

Let a⃗and b⃗  be two nonzero vectors. Then a⃗ × b⃗ = 0  if and  only  if a⃗ and b⃗ are parallel (or collinear) to  each  other, i.e., a⃗ × b⃗ = 0 ⃗ ⇔ a⃗ || b ⃗.

If θ = (Ï€/2) then ( a-> x  b->) =|  a->  ||  b-> | nˆ.

îx î =  Äµ x ĵ =   k̂ x k̂ =0  and  î x Äµ = k̂ ,  ĵ x k̂ = Ã®  and  k̂ x Ã®  = ĵ

 

sin θ =(| a-> x  b->|)/(|  a->|| b->|)

It is always true that the vector product is not commutative, as

( a-> x  b->) = - ( b-> x  a->). The magnitudes are same but the directions are different.

 

Cross  Product  as  Area of triangle

If  \({\overrightarrow a }\) and  \({\overrightarrow b }\)  represent the adjacent sides  of a  triangle then its area is given as : \(\dfrac{1}{2}\left| {\overrightarrow a  \times \overrightarrow b } \right|\)

 

By the definition of area of triangle, from the figure:

Area of triangle ABC = \(\dfrac{1}{2}AB \times CD\)

But \(AB = |\vec b|\) (given) and \(CD = |\vec a|\sin \theta \).

Thus area of triangle ABC = \(\frac{1}{2}|\vec b||\vec a|\sin \theta  = \frac{1}{2}|\vec a \times \vec b|\).

Cross Product as Area of Parallelogram

If  a-> and  b-> represent the sides of a parallelogram, then area is given as \(|\vec a \times \vec b|\).

 

Area of parallelogram ABCD =  DE

But AB=| b->| (given)  and  DE = |  a->| sin θ.

Therefore Area of parallelogram ABCD = | b->||  a->| sin θ = |  a-> x  b->|.

Vector (Cross) Product: Distributive Property

If  a-> ,  b-> and c ⃗ are  any three vectors and λ be a scalar, then:- 

a× (b + c ) = a × b + a × c .

λ(a× b) = (λ a) x (b) = a x (λ b)

Vector(Cross) Product: General Vector:-

Let a ⃗and b⃗ be two vectors and in component form they are given as:

(a1î +a2 Äµ +a3 k̂) and (b1 Ã® +b2 Äµ +b3 k̂) respectively.

To Prove:-

 

Using ( a-> x b->) = (a1 Ã® +a2 Äµ +a3 k̂) x (b1 Ã® +b2 Äµ +b3 k̂)

= a1 b1 (î x î) + a1 b2 (î x ĵ) + a1 b3 (î x k̂) + a2 b1 (ĵ x î) + a2b2 (ĵ x ĵ)

+ a2 b3 (ĵ x k̂) + a3 b1 (k̂ x î) + a3 b2 (k̂ x ĵ) + a3 b3 (k̂ x k̂)

= a1 b2 (î x ĵ) - a1 b3 (k̂ x î) - a2 b1 (î x ĵ) + a2 b3 (ĵ x k̂) + a3 b1 (k̂ x î)

- a3 b2 (ĵ x k̂)

(Using  î x î = ĵ x ĵ  = k̂ x k̂ = 0  and  î x k̂ = - k̂ x î,   ĵ x î  = - î x ĵ 

and  k̂ x ĵ  = - ĵ x k̂)

= a1 b2 k̂ - a1 b3 Äµ - a2 b1 k̂ + a2 b3 Ã® + a3 b1 Äµ - a3 b2 Ã®

Problem: -

Find a unit vector perpendicular to each ( a-> +  b->) and ( a-> -  b->), where  a->  = î +ĵ + k̂ and  b->= î +2 Äµ +3 k̂?

Now | c ⃗| = √ (4+16+4) = √24 = 2√6.

Therefore, the required unit vector is (c ⃗)/ (| c ⃗|)

= (-1/√6) Ã® + (2/ √6) Äµ – (1/√6) k̂

 

Problem:-

If a unit vector a⃗ makes  an  angle (Ï€/3)  with î,  (Ï€/4)  with ĵ and an acute angle  θ  with k̂, then find θ and hence,  the components of a⃗.

Answer:-

Let unit vector have (a1, a2, a3) components. a⃗ = a1î + a2ĵ + a3k̂.

Since a⃗ is a unit vector, | a⃗|=1.

Also it is given that a⃗ makes angles (Ï€/3) with î and (Ï€/4) ĵ with, an

acute angle θ with k̂.

Then we have:

cos (Ï€/3) = (a1/|a|)

=> (1/2) = a1 (because |a|= 1)

cos (Ï€/4) = (a2/|a|)

=> (1/√2) = a2   (because |a|= 1)

Also, cos θ = (a3/|a|)

=> a3 = cos θ

Now, |a|= 1

=> √ ((a1)2+ (a2)2+ (a3)2) =1

=> (1/2)2 + (1/√2)2 + cos 2 θ=1

=> (1/4) + (1/2) + cos 2 θ=1

=> (3/4) cos 2 θ=1

=> cos 2 θ = 1 – (3/4) (1/4)

=>cos θ = (1/2) => θ = (π/3)

Therefore a3 = cos (Ï€/3) = (1/2)

Hence, θ = (Ï€/3) and the components of a⃗ = ((1/2), (1/√2), (1/2))

Problem:  Find the area of the triangle with vertices A (1, 1, 2), B (2, 3, 5) and C (1, 5, 5).

Answer:  The vertices of triangle ABC  are given as A (1, 1, 2), B (2, 3, 5)  and C (1, 5, 5).

The adjacent sides  \(\overrightarrow {AB} \) and \(\overrightarrow {BC} \) of ∆ABC  are given as:

\(\overrightarrow {AB}  = (2 - 1)\widehat i + (3 - 1)\widehat j + (5 - 2)\widehat k =   \widehat i + 2\widehat j + 3\widehat k \) and \(\overrightarrow {BC}  = (1 - 2)\widehat i + (5 - 3)\widehat j + (5 - 5)\widehat k =  - \widehat i + 2\widehat j\)

Area of ∆ABC = \(\dfrac{1}{2}\left| {\overrightarrow {AB}  \times \overrightarrow {BC} } \right|\).

\(\left| {\overrightarrow {AB}  \times \overrightarrow {BC} } \right| = \left| {\begin{array}{*{20}{c}}{\widehat i}&{\widehat j}&{\widehat k}\\1&2&3\\{ - 1}&2&0\end{array}} \right| = | - 6\widehat i - 3\widehat j + 4\widehat k|\)

Therefore \(\left| {\overrightarrow {AB}  \times \overrightarrow {BC} } \right| = \sqrt {{{( - 6)}^2} + {{( - 3)}^2} + {4^2}}  = \sqrt {36 + 9 + 16}  = \sqrt {61} \) 

Hence, the area of triangle ABC is \(\dfrac{{\sqrt {61} }}{2}\) square units.


Triple product

Vector product of two vectors can be made to undergo dot or cross product with any third vector.
(a) Scalar tripple product:-
  • For three vectors A, B, and C, their scalar triple product is defined as A . (B × C) = B . (C × A) = C . (A × B) obtained in cyclic permutation.
  • Switching the two vectors in the cross product negates the triple product, i.e.:   C . (A x B) = - C . (B × A)
  • If A = (Ax, Ay, Az) , B = (Bx, By, Bz) , and C = (Cx, Cy, Cz) then, A . (B × C) is the volume of a parallelepiped having A, B, and C as edges and can easily obtained by finding the determinant of the 3 x 3 matrix formed by AB, and C. \[A.(B \times C) = \left| {\begin{array}{*{20}{c}}{{A_x}}&{{A_y}}&{{A_z}}\\{{B_x}}&{{B_y}}&{{B_z}}\\{{C_x}}&{{C_y}}&{{C_z}}\end{array}} \right|\]
  • If the scalar triple product is equal to zero, then the three vectors a, b, and c are coplanar, since the "parallelepiped" defined by them would be flat and have no volume.
(b) Vector Triple Product:-
  • For vectors A, B, and C, we define the vector tiple product as A × (B × C) = B(A . C) - C(A - B)
  • Note that (A × B) ×C ≠ A× (B × C)



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Class 12 Chapter 13 (Probability) Class Notes)

Probability

Introduction:

Probability: probability measures how much chance is there for my expected
result to happen.

\[{\rm{Probability  =  }}\dfrac{{{\rm{Number \,\, of \,\, Favourable \,\, Outcomes}}}}{{{\rm{Total \,\, Number \,\, of \,\, Outcomes}}}}\]

Example: Let there be a basket with 3 balls: Red, Green, Blue. If I want to pick a Red ball,
we can calculate the probability of picking a red ball.

Total number of possibilities when we pick a ball from the basket = 3 ( The ball we pick can be Red, Green or blue)

Total number of favorable possibility i.e. possibility to get a red ball = 1

Therefore, Probability of getting a red ball, \(P(red) = \dfrac{1}{3}\)


Sample space: set of all possible outcomes of a random experiment is called Sample space. It is denoted by S.


Sample point: each element of the sample space is called the sample point.
Example: for Tossing a coin, the sample space is S = {HEAD, TAIL} and and the element in sample space namely head, or tail, each of them can be called as a sample point


Event: it is the set of favorable outcome.
Example: if you throw a coin and if you are looking for head, then head is the favorable outcome.


Mutually Exclusive events: two events are said to be mutually exclusive if
there is no common element between them.


Exhaustive events: the given events are Exhaustive if when I take the
elements in those events forms the given sample space.

 

Conditional Probability:

Given a condition and you have to find the probability

Example: when I throw a dice, what is the probability that the outcome is 3 given that the outcome is odd.


The answer is 1/3 as the odd numbers are 1, 3 and 5. So we need the probability of getting the number 3 out of the numbers 1,3,5. The chance is one out of three so The probability is 1/3.


∴ if number is odd,then the probability that it is 3 is = P(3|number is odd) = 1/3


Note:But P(3) = 1/6 If there is no condition. As the probability of getting a single number named 3, out of total 6 numbers namely 1,2,3,4,5,6 is 1/6

 

Conditional Probability Definition:

Probability of event E is called the conditional probability of F given that E has already occurred, and it's defined by P(F|E).

Formula for conditional probability
P(F|E) = P(F∩E) / P(E)

 

Conditional Probability Problems:

Example: In a survey in a class it was found that the probability of a student watching ABC videos is 0.8. and the probability that a student is both topper and also watches ABC videos is 0.792. what is the probability that a student is a topper if he watches ABC videos?

Solution:
Let E denotes the event that a person watches ABC videos 
Let F denotes the event that a person is topper .

Then, P(E) = The probability that a person watches ABC videos = 0.8
P(E ∩ F ) = probability that one is both topper and also watches ABC videos = 0.792
and P(F|E) = the probability that a person is a topper if he watches ABC videos = ?
then according to Formula for conditional probability: P(F|E) = P(F∩E)/ P(E)
= 0.792/0.8
= 0.99
∴ The probability that a person is a topper if he watches ABC videos is 0.99

  

Properties of conditional probability:

• P(S|F) = 1
• P(F|F) = 1
• P((A∪B)|F) = P(A|F) + P(B|F) – P((A∩B)|F)
• P(E'|F) = 1- P(E|F)

Example: Evaluate P(A ∪ B), if 2P(A) = P(B) = 5/13 and P(A|B) =2/5
Solution: ∵ 2P(A) = P(B) = 5/13
=> P(B) = 5/13 and P(A) = 1/2 x P(B) = 1/2  × 5/13 = 5/26


Now the formula of conditional probability is P(A|B) = P( A∩B)/ P(B)
∴ 2/ 5 = P( A∩B) /P(B)  [Since it was given in question that P(A|B)=2/5 ]

=> P(A∩B) = 2/5 X P(B)
= 2/5 × 5/13
= 2/13

So, P(A∩B)= 2/13 , but we need to find P(A ∪ B)
Now is there any formula relating P(A ∪ B) and P(A∩B) ?
Yes , the formula is P(A ∪ B)=P(A)+P(B)−P(A∩B)
Now you know values of P(A)= 5/ 26 , P(B)= 5/13 and P(A∩B)= 2/13.

So substitute these values in the formula, you will get value of P(A∪B).
That is P(A∪B)=P(A)+P(B)−P( A∩B)
=> P(A∪B)= 5/26 + 5/13 - 2/13
=> P(A∪B)= (5+10−4 )/26
=> P(A∪B)=11/26

∴ The value of P(A∪B) is 11/26

 

Example: Mother, father and son line up at random for a family picture
E : son on one end, F : father in middle , what is the probability that the son is on one end given that the father is in middle?

Solution:

let Mother be denoted with M, Father be denoted with F and Son be denoted with S. If the sample space for all the possible ways of arranging them is denoted by S.
Then the possible elements in S are, S = {MFS, MSF, FMS, FSM, SMF, SFM}
Now E denotes son on one end as given in question so A={MFS, FMS, SMF, SFM}
and F denoted Father in middle as given in question , so B={MFS, SFM}
Since the common elements between A and B are MFS and SFM ∴A∩B ={MFS, SFM}
Now, P(E∩F) = probability of getting two elements of A∩B I.e; MFS and SFM out of the total 6 elements MFS, MSF, FMS, FSM, SMF, SFM
= 2/6 = 1/3
and P(F)= probability of getting two elements of B i.e; MFS and SFM out of the total 6 elements MFS, MSF, FMS, FSM, SMF, SFM.
= 2/6 = 1/3
∴ P(E|F) = P(E∩F) /P(F)
= (1/3)  / (1/3)
= 1
∴The probability that the son is on one end given that the father is in middle is 1.


Example: If A and B are events such that P(A|B) = P(B|A), then
(A)A ⊂ B but A ≠ B

(B)A = B

(C)A ∩ B = Φ

(D)P(A) = P(B)

Solution:
Given in question that P(A|B) = P(B|A)
But we know P(A|B) = P( A∩B) /P(B)
and P(B|A) = P( A∩B)/ P(A)
since P(A|B) = P(B|A)
=> P( A∩B)/ P(B) = P( A∩B)/ P(A)
=> P(A) = P(B)
Thus the correct option is (D).

 

Multiplication Theorem on Probability:

Let E and F be two events associated with a sample space of an experiment.
Then 

P(E ∩ F) = P(E) P(F|E), P(E) ≠ 0
= P(F) P(E|F), P(F) ≠ 0

If E, F and G are three events associated with a sample space, then

P(E∩F∩G) = P(E) P(F|E) P(G|E∩F)


Example: In a survey in a class, the probability for a person to watch ABC videos is 0.8 and the probability for a person to be a topper, if given that he watches ABC videos is 0.99. find the probability for a person to be both topper and watches ABC videos.

Solution:
let Event E denotes the event that a person watches ABC videos
let Event F denotes the event that a person is topper
then P(E) = the probability that a person watches ABC videos =0.8
and P(F|E)= the probability that a person is a topper if he watches ABC videos=0.99
then P (E ∩ F)= the probability that a person is both topper and also watches ABC videos
then according to Multiplication Theorem on probability
P (E ∩ F) = P (E) P (F | E)
= 0.8 x 0.99
= 0.792
∴ The probability that a person is both topper and also watches ABC videos = 0.792

 

Independent Events:
Let E and F be two events associated with a sample space S. If the probability of occurrence of one of them is not affected by the occurrence of the other, then we say that the two events are independent. Thus, two events E and F will be independent, if

(a) P(F | E) = P(F), provided P (E) ≠ 0
(b) P(E | F) = P(E), provided P (F) ≠ 0

Using the multiplication theorem on probability, we have

(c) P(E ∩ F) = P(E) P(F)

Three events A, B and C are said to be mutually independent if all the
following conditions hold:

P(A ∩ B) = P(A) P(B)
P(A ∩ C) = P(A) P(C)
P(B ∩ C) = P(B) P(C)
and P(A ∩ B ∩ C) = P(A) P(B) P(C)


Example: If you throw two coins, then probability of getting head or tail in second coin is independent of probability of getting head or tail in first coin.


Example: Let A and B be independent events with P(A)=0.3 and P(B) = 0.4.
Find(i) P(A ∩ B) (ii) P(A ∪ B) (iii) P(A|B) (iv) P(B|A)
Solution:
(i) we have to find the value of P(A ∩ B) and they gave P(A) =0.3 and P(B) = 0.4.
given A and B are independent events.
We know that if A and B are independent events then P(A ∩ B) = P(A) P(B)
=>P(A ∩ B) = 0.3×0.4 = 0.12


(ii)we have to find the value of P(A ∪ B)
the formula for P(A ∪ B) is P (A ∪B)=P(A)+P(B)−P( A∩B)
=> P(A ∪ B)=0.3+0.4-0.12 = 0.58


(iii) P(A|B)=?
the formula for P(A|B) is P(A|B)= P( A∩B)/ P(B)
=> P(A|B)= 0.12/ 0.4 = 0.3


(iv) P(B|A)=? 
the formula for P(B|A) is P(B|A)= P( A∩B)/ P(A)
=> P(B|A)= 0.12/0.3 = 0.4


Example:If A and B are two events such that P(A)=1/4 ,P(B)=1/2 , ,P(A∩B)=1/8
find P(not A and not B).
Solution:
we know that P(not A and not B)=P(A'∩B')
But we also know that A'∩B'=(A∪B)' => P(A'∩B')=P((A∪B))'
∴P(not A and not B)=P(A'∩B')
=P((A∪B))'
=1-P(A∪B) ( ∵ P(A)' = 1-P(A) )
=1- (P(A)+P(B)−P( A∩B)) (∵ P(A∪B)=P(A)+P(B)−P( A∩B) )
=1- ( 1/4 + 1/2 - 1/8 )
=1- 5/8
=3/8

 

Bayes' Theorem:

If E1, E2,..., En are mutually exclusive and exhaustive events associated
with a sample space, and A is any event of non zero probability, then

 

Example: In a survey in a classroom it was found that chance of a person becoming a topper is 10% . if a person is a topper, there is a 99% chance that he watches ABC videos. One who is not a topper there is 1% chance of watching ABC videos. Then, what is the Probability for one to be a Topper , given if he watches ABC videos?

Solution:
Let “A person to be a Topper” is Event A and “To watch ABC videos” is Event B

Given P(A) = 0.1 = probability for a person to be a Topper.

So P(A')= 1-P(A)=1 - 0.1= 0.9 = probability for a person to be not a Topper

Given P(B|A) = 0.99 =  probability that a person watches ABC videos, if he is a topper, 

so P(B'|A) = 1 - P(B|A) = 1 - 0.99 = 0.01 = probability that a person not watches ABC videos, if he is a topper

Given P(B|A')=0.01=probability that a person watches ABC videos , if he is not a topper 
so P(B'|A')=1-0.01=0.99=probability that a person not watches ABC videos , if he is not a topper

Therefore according to Bayes' Theorem,Probability for one to be a Topper , if he watches ABC videos is

= (0.1x0.99)/ {( (0.1x0.99)+(0.01x0.9))}
= 0.9909

∴ if one watches ABC videos , for him there is 99.09% chance to become topper.

Example: A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.

Solution:
Let E1= the event of selecting first bag.
E2=the event of selecting second bag.
A = The event of getting red ball.
Since there is equal chance of selecting first bag or selecting second bag,
P(E1)=P(E2)= 1/2


now P(A|E1)=P(Drawing a red ball from first bag )= 4/8
and P(A|E2)=P(Drawing a red ball from second bag )= 2/8 =1/4


probability that ball is drawn from the first bag given that the ball drawn is red is P(E1|A)

 

Example:Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group.

Solution:
Let E1= event that the first group wins the competition
E2 = event that the second group wins the competition
A = event of introducing a new product.
Then,
P(E1) = Probability that the first group wins the competition = 0.6
P(E2) = Probability that the second group wins the competition = 0.4
P(A|E1)=Probability of introducing new product if the first group wins =0.7
P(A|E2)=Probability of introducing new product if the second group wins =0.3
The probability that the new product is introduced by the second group is given by

 

 

Random Variables and its probability distributions:



A random variable is a real valued function whose domain is the sample space of a random experiment.
The probability distribution of a random variable X is the system of numbers

where Pi > 0, i = 1 to n and P1 + P2 + P3 + …..... + Pn =1


Example: Suppose that a coin is tossed twice so that the sample space is S = {HH,HT,TH,TT}. Let X represent the number of heads that can come up. So with each sample point we can associate a number for X so for HH the value of X is 2 as in HH there are 2 heads, for HT the value of X is 1 as there is only one head in HT. similarly for TH, X = 1 and for TT value of X = 0 as there are no heads in TT.


So in the sample space S = {HH,HT,TH,TT} assuming that the coin is fair,
P(HH)= ¼ , P(HT)= ¼ , P(TH)= ¼ , P(TT)= ¼
therefore P(X=0) = P(TT) = ¼
P(X=1) = P(HT ∪ TH) = P(HT) + P(TH) = ¼ + ¼ = ½
P(X=2) = P(HH) = ¼
Thus the table formed is shown below:


Example: A coin is biased so that the head is 3 times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tails.

Solution:
Let the probability of getting a tail in the biased coin be x.
∴ P (T) = x
⇒ P (H) = 3x
For a biased coin, P (T) + P (H) = 1
⇒ x + 3x = 1
⇒ 4x = 1
⇒ x = \(\frac{1}{4}\)
∴ P(T) =  \(\frac{1}{4}\)   and   P(H) =  \(\frac{3}{4}\)


When the coin is tossed twice, the sample space is {HH, TT, HT, TH}.
Let X be the random variable representing the number of tails.
∴ P (X = 0) = P (no tail) = P (H) × P (H)= \(\frac{3}{4}\)  × \(\frac{3}{4}\)
=  \(\frac{9}{16}\)


P (X = 1) = P (one tail) = P (HT) + P (TH) = \(\frac{3}{4} \times \frac{1}{4} + \frac{1}{4} \times \frac{3}{4}\)
= \(\frac{3}{{16}} + \frac{3}{{16}}\)
= \(\frac{3}{8}\)


P (X = 2) = P (two tails) = P (TT)= \(\frac{1}{4} \times \frac{1}{4} = \frac{1}{{16}}\)


Therefore, the required probability distribution is as follows.

 

 

Mean and Variance of a Random variable:

Let X be a random variable assuming values x1, x2,...., xn with probabilities p1, p2, ..., pn, respectively such that pi ≥ 0,  

Mean of X, denoted by μ [or expected value of X denoted by E (X)] is defined as:



and variance denoted by σ2 , is defined as


Standard deviation of the random variable X is defined as


Let us take the above table

Now mean is

 
= X1P1 + X2P2 + X3P3
= (0X¼ ) + (1 X ½)+ (2X ¼)
= 0 + ½ + ½
=1


Now Variance = σx2 = E(X2) − (E(X))2


so to find Variance let us find E(X2) and (E(X))2


= 0 + ½ + 1
= 3/2
= 1.5


and (E(X))2 = 12=1


∴ Var(X)= E(X2) − (E(X))2 = 1.5 − 1 = 0.5


now standard deviation of X = √(variance (X ))
= √0.5
≈ 0.707


Example: In a meeting, 70% of the members favour and 30% oppose a certain proposal. A member is selected at random and we take X = 0 if he opposed,
and X = 1 if he is in favour. Find E(X) and Var(X).
Solution:
It is given that P(X = 0) = 30% = 30/100 = 0.3
P(X=1)=70%= 70/100 = 0.7

Therefore, the probability distribution is as follows.

x 0 1
P(x) 0.3 0.7




Then, Mean= E(X) =

 
= 0×0.3+1×0.7
=0.7

= 02×0.3+12×0.7 =0.7


Now,
∴ Var(X)= E(X2)-(E(X))2
= 0.7−(0.7)2
= 0.7 − 0.49
= 0.21


Example:The mean of the numbers obtained on throwing a dice having written 1 on three faces, 2 on two faces and 5 on one face is (A) 1 (B) 2 (C) 5 (D) 8/3

Solution:
Let X be the random variable representing a number on the dice.

The total number of observations is six.
∴P(X=1)= 3/6 = 1/2

P(X=2)= 2/6 = 1/3

P(X=5)= 1/6


∴ The probability distribution is as follows.

X 1 2 5
P(X) 1/2 1/3 1/6


Now, Mean=E(X)

 
=1×1/2 +2×1/3 +5×1/6
= 1/2 +2/3 +5/6
= 12/6
= 2


∴ The correct option is B.

 

Bernoulli Trials and Binomial Distribution:

Bernoulli Trials:
Trials of a random experiment are called Bernoulli trials, if they satisfy the following
conditions:
(i) There should be a finite number of trials
(ii) The trials should be independent
(iii) Each trial has exactly two outcomes: success or failure
(iv) The probability of success(or failure) remains the same in each trial.

Example: A tossed coin shows a ‘head’ or ‘tail’. If the occurrence of head is considered as success, then the occurrence of tail is a failure .so, if we consider tossing a coin two times as Bernoulli trials, because 

(i)There are finite number of trials that is 2
(ii)tossing a coin second time is completely independent of the result of first time tossing
(iii)each trial has only two outcomes namely either head or tail and
(iv)The probability of getting head or tail remains the same in each trial.

 

 

Binomial Distribution:

A random variable X taking values 0, 1, 2, ..., n is said to have a binomial distribution
with parameters n and p, if its probability distribution is given by 

P (X = r) = ncr pr qn–r,
where q = 1 – p and r = 0, 1, 2, ..., n.


Example: If a coin is tossed 10 times, find the probability of exactly 5 heads.


Solution: let X denote the number of heads.
So X has the binomial distribution with n = 10 an p = ½
∴ P(X = x) = nCx qn-x px , x = 1,2,3....,n
∴ P(X = x) = 10cx(1/2)10-x(1/2)x = 10cx(1/2)10.
so P(X = 6) = 10c5(1/2)10= (10 !)/ (5 !×5!) ×( 1/210) = 63/ 256


Example:A bag consists of 10 balls each marked with one of the digits 0 to 9. If four balls are drawn successively with replacement from the bag, what is the probability that none is marked with the digit 0? 

Solution:
Let X denote the number of balls marked with the digit 0 among the 4 balls drawn.
Since the balls are drawn with replacement, the trials are Bernoulli trials.
X has a binomial distribution with n = 4 and P= 1/10


∴ Q = 1 − P =1 − 1/10 = 9/10


∴ P(X = x) = nCx qn-x px , x = 1,2,3....,n
= 4Cx ( 9/10 )n−x( 1/10 )x

P (none marked with 0) = P (X = 0)
= 4C0 ( 9/10 )4 ( 1/10 )0
= 1×( 9/10)4 

= ( 9/10 )4

 



Example: Find the probability of getting 5 exactly twice in 7 throws of a dice.

Solution:
The repeated tossing of a dice are Bernoulli trials. Let X represent the number of times of getting 5 in 7 throws of the dice.
Probability of getting 5 in a single throw of the dice is P=16
∴ Q = 1−P = 1 − 1/6 = 5/6


Clearly, X has the probability distribution with n = 7 and P=16
∴P(X = x) = nCx qn-x px
= 7Cx (5/6)7−x (1/6)x


P (getting 5 exactly twice) = P(X = 2)
=7C2 (5/6)5 (1/6)2
= 21×(5/6)5× 1/36
= 7 /12 ×(5/6)5

 

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