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Class 9 Chapter 7 (Triangles) Class Notes

 

Triangles

Triangle

A closed figure with three sides is called a Triangle. It has three vertex, sides and Angles.

atriangle


Types of Triangle

1. There are three types of triangles on the basis of the length of the sides.





Name of Triangle Property Image
Scalene Length of all sides are different
Isosceles Length of two sides are equal
Equilateral Length of all three sides are equal Equilateral triangle

2. There are three types of triangles on the basis of angles.





Name of Triangle Property Image
Acute All the three angles are less than 90°  

Obtuse One angle is greater than 90°
Right One angle is equal to 90°

Congruence

If the shape and size of two figures are same then these are called Congruent.

1. Two circles are congruent if their radii are same.


2. Two squares are congruent if their sides are equal.


Congruence of Triangles

A triangle will be congruent if its corresponding sides and angles are equal.

The symbol of congruent is ”.


AB = DE, BC = EF, AC = DF

m∠A = m∠D, m∠B = m∠E, m∠C = m∠F

Here ∆ABC ≅ ∆DEF

Criteria for Congruence of Triangles

S.No. Rule Meaning Figure
1.

SAS (Side-Angle-Side) Congruence rule

If the two sides and the including angle of one triangle is equal to another triangle then they are called congruent triangles.

2.

ASA (Angle-Side-Angle) Congruence rule

If the two angles and the including side of one triangle is equal to another triangle then they are called congruent triangles.

3.

AAS (Angle-Angle-Side) Congruence rule

If any two pairs of angles and a pair of the corresponding side is equal in two triangles then these are called congruent triangles.

4.

SSS (Side-Side-Side) Congruence rule

If all the three sides of a triangle are equal with the three corresponding sides of another triangle then these are called congruent triangles.

5.

RHS (Right angle-Hypotenuse-Side) Congruence rule

If there are two right-angled triangles then they will be congruent if their hypotenuse and any one side are equal. 

Remark

1. SSA and ASS do not show the congruency of triangles.

2. AAA is also not the right condition to prove that the triangles are congruent.

Criteria for Congruence of Triangles:
(i) SAS Congruence Rule:
Statement: Two triangles are congruent if two sides and the included angle of one triangle are equal to the sides and the included angle of the other triangle.

For example: Prove Δ AOD ≅ Δ BOC.

From figure, we can see that

OA = OB and OC = OD
Also, we can see that, ∠ AOD and ∠ BOC form a pair of vertically opposite angles,
∠ AOD = ∠ BOC
Now, since two sides and an included angle of triangle are equal, by SAS congruence rule, we can write that Δ AOD ≅ Δ BOC.

(ii) ASA Congruence Rule:
Statement: Two triangles are congruent if two angles and the included side of one triangle are equal to two angles and the included side of other triangle.
Proof: Suppose we have two triangles ABC and DEF, such that ∠ B = ∠ E, ∠ C = ∠ F, and BC = EF.
We need to prove that Δ ABC ≅ Δ DEF.
Case 1: Suppose AB = DE.

From the assumption, AB = DE and given that ∠ B = ∠ E, BC = EF, we can say that Δ ABC ≅ Δ DEF as per the SAS rule.
Case 2: Suppose AB > DE or AB < DE.

Let us take a point P on AB such that PB = DE as shown in the figure.

Now, from the assumption, PB = DE and given that ∠ B = ∠ E, BC = EF, we can say that Δ PBC ≅ Δ DEF as per the SAS rule.
Now, since triangles are congruent, their corresponding parts will be equal. Hence, ∠ PCB = ∠DFE
We are given that ∠ ACB = ∠ DFE, which implies that ∠ ACB = ∠PCB
This thing is possible only if P are A are same points or BA = ED.
Thus, Δ ABC ≅ Δ DEF as per the SAS rule.
On similar arguments, for AB < DE, it can be proved that Δ ABC ≅ Δ DEF.

For Example: AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB.

From the figure, we can see that,

∠ AOD = ∠ BOC (Vertically opposite angles)
∠ CBO = ∠ DAO (Both are of 90o)
BC = AD (Given)
Now, as per AAS Congruence Rule, we can say that Δ AOD ≅ Δ BOC.
Hence, BO = AO which means CD bisects AB.

(iii) SSS Congruence Rule:
Statement: If three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent.

For ExampleTwo sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of Δ PQR. Show that Δ ABM ≅ Δ PQN.

From the figure, we can see that, AM is the median to BC.

So, BM = ½ BC.
Similarly, PN is median to QR. So, QN = ½ QR.
Now, BC = QR.
So, ½ BC = ½ QR i.e. BM = QN
Given that, AB = PQ, AM = QN and AM = PN.
Therefore, Δ ABM ≅ Δ PQN by SSS Congruence Rule.

 

Example

Find the ∠P, ∠R, ∠N and ∠M if ∆LMN ≅ ∆PQR.

Solution

If ∆ LMN ≅ ∆PQR, then

∠L=∠P

∠M =∠Q

∠N =∠R

So,

∠L=∠P = 105°

∠M =∠Q = 45°

∠M + ∠N + ∠L = 180° (Sum of three angles of a triangle is 180°)

45° + 105° + ∠N = 180°

∠N = 180°- 45° + 105°

∠N = 30°

∠N = ∠R = 30°

Some Properties of a Triangle

If a triangle has two equal sides then it is called an Isosceles Triangle.

1. Two angles opposite to the two equal sides of an isosceles triangle are also equal.

2. Two sides opposite to the equal angles of the isosceles triangle are also equal. This is the converse of the above theorem.

Some Properties of a Triangle:
Theorem 1: Angles opposite to equal sides of an isosceles triangle are equal.
Proof: Suppose we are given isosceles triangle ABC having AB = AC.
We need to prove that ∠ B = ∠C

Firstly, we will draw bisector of ∠ A which intersects BC at point D.

For the Δ BAD and Δ CAD, given that AB = AC, from the figure ∠ BAD = ∠ CAD and AD = AD.
Thus, by SAS rule Δ BAD ≅ Δ CAD.
Therefore, ∠ ABD = ∠ ACD, since they are corresponding angles of congruent triangles.
Hence, ∠ B = ∠C.

For Example: In ∆ ABC, AB = AC, D and E are points on BC such that BE = CD. Show that AD = AE.

From the figure, we can see that in Δ ABD and Δ ACE,
AB = AC and
∠ B = ∠ C (Angles opposite to equal sides)
Given that BE = CD.
Subtracting DE from both the sides, we have,
BE – DE = CD – DE i.e. BD = CE.
Now, using SAS rule, we can say that Δ ABD ≅ Δ ACE
Therefore, by CPCT, AD = AE.

Theorem 2: The sides opposite to equal angles of a triangle are equal.
For ExampleIn Δ ABC, the bisector AD of ∠ A is perpendicular to side BC. Show that AB = AC and Δ ABC is isosceles.

From the figure, we can see that in Δ ABD and Δ ACD,
It is given that, ∠ BAD = ∠ CAD
AD = AD (Common side)
∠ ADB = ∠ ADC = 90°
So, Δ ABD ≅ Δ ACD by ASA congruence rule.
Therefore, by CPCT, AB = AC (CPCT) or in other words Δ ABC is an isosceles triangle.

 

Inequalities in a Triangle

Inequalities in a Triangle

Theorem 1: In a given triangle if two sides are unequal then the angle opposite to the longer side will be larger.

a > b, if and only if ∠A > ∠B

Longer sides correspond to larger angles.

Theorem 2: In the given triangle, the side opposite to the larger angle will always be longer. This is the converse of above theorem.

Theorem 3: The sum of any two sides of a triangle will always be greater than the third side.

Example

Show whether the inequality theorem is applicable to this triangle or not?

Solution

The three sides are given as 7, 8 and 9.

According to inequality theorem, the sum of any two sides of a triangle will always be greater than the third side.

Let’s check it

7 + 8 > 9

8 + 9 > 7

9 + 7 > 8

This shows that this theorem is applicable to all the triangles irrespective of the type of triangle.

For ExampleFor the given figure, PR > PQ and PS bisects ∠ QPR. Prove that ∠ PSR > ∠ PSQ.

Given, PR > PQ.
Therefore, ∠ PQR > ∠ PRQ (As per angle opposite to larger side is larger) – (1)
Also, PS bisects QPR, so, ∠ QPS = ∠ RPS – (2)
Now, ∠ PSR = ∠ PQR + ∠ QPS, since exterior angle of a triangle is equal to the sum of opposite interior angles. – (3)
Similarly, ∠ PSQ = ∠ PRQ + ∠ RPS, since exterior angle of a triangle is equal to the sum of opposite interior angles. – (4)
Adding (1) and (2), we get,
∠ PQR + ∠ QPS > ∠ PRQ + ∠ RPS
Now, from 3 & 4, we get,
∠ PSR > ∠ PSQ.

For ExampleD is a point on side BC of Δ ABC such that AD = AC. Show that AB > AD.

Given that AD = AC,

Hence, ∠ ADC = ∠ ACD as they are angles opposite to equal sides.
Now, ∠ ADC is an exterior angle for ΔABD. Therefore, ∠ ADC > ∠ ABD or, ∠ ACD > ∠ ABD or, ∠ ACB > ∠ ABC.
So, AB > AC since side opposite to larger angle in Δ ABC.
In other words, AB > AD (AD = AC).

 

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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 \(\overrightarrow a \) ⊥ \(\overrightarrow b \).

( θ = 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 = π, 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 + ĵ.

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

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 ĵ +7 k̂) on the vector (7î - ĵ +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 ĵ +3 k̂, b ⃗=-î +2 ĵ +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 ĵ +3 k̂,  b ⃗=-î +2 ĵ +k̂ and c ⃗ = 3î + ĵ

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

= (2- λ) î + (2+2 λ) ĵ + (3+ λ) k̂

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

(a ⃗+ λ b ⃗). c ⃗=0.

[(2- λ) î +(2+2 λ) ĵ+(3+ λ) k̂].(3 î+ ĵ)=0

(2- λ) (3) + (2+2 λ) (1) + (3+ λ) (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)} î + (2-0) ĵ + (3-0) k̂ = 2î +2 ĵ +3 k̂

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

Therefore (BA ⃗. BC ⃗) = (2î +2 ĵ +3 k̂). (î + ĵ +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) î + (6-2) ĵ + (3-7) k̂ = î+ 4 ĵ -4 k̂

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

AC ⃗ = (3-1) î + (10-2) ĵ + (-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 = (π/2) then ( a-> x  b->) =|  a->  ||  b-> | nˆ.

îx î =  ĵ x ĵ =   k̂ x k̂ =0  and  î x ĵ = k̂ ,  ĵ x k̂ = î  and  k̂ x î  = 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 ĵ +a3 k̂) and (b1 î +b2 ĵ +b3 k̂) respectively.

To Prove:-

 

Using ( a-> x b->) = (a1 î +a2 ĵ +a3 k̂) x (b1 î +b2 ĵ +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 ĵ - a2 b1 k̂ + a2 b3 î + a3 b1 ĵ - a3 b2 î

Problem: -

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

Now | c ⃗| = √ (4+16+4) = √24 = 2√6.

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

= (-1/√6) î + (2/ √6) ĵ – (1/√6) k̂

 

Problem:-

If a unit vector a⃗ makes  an  angle (π/3)  with î,  (π/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 (π/3) with î and (π/4) ĵ with, an

acute angle θ with k̂.

Then we have:

cos (π/3) = (a1/|a|)

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

cos (π/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 (π/3) = (1/2)

Hence, θ = (π/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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