Class 12 Chapter 10 (Vector Algebra) Class Notes Part II ~ The Math Centre

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 = (F_x Displacement + F_y Displacement), Where F_X= Force in x direction and F_Y = 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:-

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

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:- (a⃗× b⃗) =| a⃗ || b⃗ | 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:-

a⃗× (b⃗ + c ⃗) = a⃗ × b⃗ + a⃗ × c ⃗.

λ(a⃗× b⃗) = (λ a⃗) x (b⃗) = a⃗ x (λ b⃗)

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 = (A_x, A_y, A_z) , B = (B_x, B_y, B_z) , and C = (C_x, C_y, C_z) 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 A, B, 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)