Given four non zero vectors a → , b → , c → and d → . The vectors a → , b → and c → are coplanar but not collinear pair by pair and vector d → is not coplanar with vectors a → , b → and c → and a → b → ^ = b → c → ^ = π 3 , d → a → ^ = α , d → b → ^ = β then prove that d → c → ^ = cos - 1 cos β - cos α

Since a → , b → , c → are coplanar, then we have c ^ = x a ^ + y b ^ , where a ^ , b ^ , c ^ are the corresponding unit vectors. Now, c ^ · a ^ = x a ^ · a ^ + y b ^ · a ^ = x + y a ^ · b ^ . . . 1 and c ^ · b ^ = x a ^ · b ^ + y b ^ · b ^ = x a ^ · b ^ + y . . . 2 Since angle between a → & c → is 2 π 3 , so c ^ · a ^ = c ^ a ^ cos 2 π 3 = - 1 2 Similarly, c ^ · b ^ = 1 2 = b ^ · a ^ . Putting these values in 1 and 2 , we get y = 1 , x = - 1 Thus, c ^ = - a ^ + b ^ . Now, d ^ · c ^ = - d ^ · a ^ + d ^ · b ^ ⇒ d ^ c ^ cos θ = - d ^ a ^ cos α + d ^ b ^ cos β ⇒ cos θ = cos β - cos α ⇒ θ = cos - 1 cos β - cos α So, d → c → ^ = cos - 1 cos β - cos α

Given four non zero vectors a → , b → , c → and d → . The vectors a → , b → and c → are coplanar but not collinear pair by pair and vector d → is not coplanar with vectors a → , b → and c → and a → b → ^ = b → c → ^ = π 3 , d → a → ^ = α , d → b → ^ = β then prove that d → c → ^ = cos - 1 cos β - cos α

Since a → , b → , c → are coplanar, then we have c ^ = x a ^ + y b ^ , where a ^ , b ^ , c ^ are the corresponding unit vectors. Now, c ^ · a ^ = x a ^ · a ^ + y b ^ · a ^ = x + y a ^ · b ^ . . . 1 and c ^ · b ^ = x a ^ · b ^ + y b ^ · b ^ = x a ^ · b ^ + y . . . 2 Since angle between a → & c → is 2 π 3 , so c ^ · a ^ = c ^ a ^ cos 2 π 3 = - 1 2 Similarly, c ^ · b ^ = 1 2 = b ^ · a ^ . Putting these values in 1 and 2 , we get y = 1 , x = - 1 Thus, c ^ = - a ^ + b ^ . Now, d ^ · c ^ = - d ^ · a ^ + d ^ · b ^ ⇒ d ^ c ^ cos θ = - d ^ a ^ cos α + d ^ b ^ cos β ⇒ cos θ = cos β - cos α ⇒ θ = cos - 1 cos β - cos α So, d → c → ^ = cos - 1 cos β - cos α

If O is origin of reference, point A a → ; B b → ; C c → ; D a → + b → ; E b → + c → ; F c → + a → ; G a → + b → + c → where a → = a 1 i ^ + a 2 j ^ + a 3 k ^ ; b → = b 1 i ^ + b 2 j ^ + b 3 k ^ and c → = c 1 i ^ + c 2 j ^ + c 3 k ^ , then prove that these points are vertices of a cube having length of its edge equal to unity provided the matrix a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 is orthogonal. Also find the length X Y such that X is the point of intersection of C M and G P ; Y is the point of intersection of O Q and D N , where P , Q , M , N are respectively the mid-point of sides C F , B D , G F and O B .

Let L = a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 Since, L is orthogonal, so L L T = I ⇒ a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 = 1 0 0 0 1 0 0 0 1 ⇒ a 1 2 + a 2 2 + a 3 2 a 1 b 1 + a 2 b 2 + a 3 b 3 a 1 c 1 + a 2 c 2 + a 3 c 3 a 1 b 1 + a 2 b 2 + a 3 b 3 b 1 2 + b 2 2 + b 3 2 b 1 c 1 + b 2 c 2 + b 3 c 3 a 1 c 1 + a 2 c 2 + a 3 c 3 b 1 c 1 + b 2 c 2 + b 3 c 3 c 1 2 + c 2 2 + c 3 2 = 1 0 0 0 1 0 0 0 1 ⇒ a → 2 a → · b → a → · c → a → · b → b → 2 b → · c → a → · c → b → · c → c → 2 = 1 0 0 0 1 0 0 0 1 Hence, a → 2 = b → 2 = c → 2 = 1 and a → · c → = a → · b → = b → · c → = 0 . Therefore, a → , b → & c → are mutually perpendicular unit vectors. Hence, points A , B , C , D , E , F & G forms a cube with edge length unity. O P → = O C → + O F → 2 = 2 c → + a → 2 O M → = O G → + O F → 2 = 2 a → + 2 c → + b → 2 O Q → = O B → + O D → 2 = a → + 2 b → 2 O N → = O B → 2 = b → 2 Hence, C M → = O M → - c = 2 a → + 2 c → + b → 2 - c → = 2 a → + b → 2 G P → = O P → - a → + b → + c = 2 c → + a → 2 - a → + b → + c = - a → - 2 b → 2 D N → = O N → - a → + b → = b → 2 - a → + b → = - 2 a → - b → 2 So, equation of C M is r → = c → + λ 2 a → + 2 c → + b → 2 - c → ⇒ r → = c → + λ 2 a → + b → 2 Equation of G P is r → = a → + b → + c → + μ - a → - 2 b → 2 On solving both lines, we get X → = 2 a → + b → + 3 c → 3 Similarly, equation of O Q is r → = 0 a → + 0 b → + 0 c → + t a → + 2 b → 2 Equation of D N is r → = a → + b → + s - 2 a → - b → 2 On solving both lines, we get Y → = a → + 2 b → 3 X Y → = - a → + b → - 3 c → 3 X Y → = 1 3 a → 2 + b → 2 + 9 c → 2 = 11 3 units

Let a → = 3 i ^ - j ^ and b → = 1 2 i ^ + 3 2 j ^ and x → = a → + q 2 - 3 b → , y → = - p a → + q b → . If x → ⊥ y → , then express p as a function of q , say p = f q , p ≠ 0 and q ≠ 0 and find the intervals of monotonicity of f q .

Since x → ⊥ y → , then x → · y → = 0 ⇒ a → + q 2 - 3 b → · - p a → + q b → = 0 ⇒ - p a → 2 + q q 2 - 3 b → 2 + q - p q 2 - 3 a → · b → = 0 . . . ( 1 ) Now, a → = 3 i ^ - j ^ ⇒ a → = 3 2 + - 1 2 = 2 , b → = 1 2 i ^ + 3 2 j ^ ⇒ b → = 1 2 2 + 3 2 2 = 1 and a → · b → = 3 · 1 2 - 1 · 3 2 = 0 Putting these values in ( 1 ) , we get p = q q 2 - 3 4 = q 3 - 3 q 4 = f q , q ≠ 0 For monotonicity: f ' q = 3 q 2 - 3 4 f ' q < 0 ⇒ 3 q 2 - 3 < 0 ⇒ q - 1 q + 1 < 0 ⇒ - 1 < q < 1 So, f q is decreasing for q ∈ - 1 , 1 , q ≠ 0 and increasing for q ∈ - ∞ , - 1 ∪ 1 , ∞ .

If a → = i ^ + j ^ - k ^ , b → = - i ^ + 2 j ^ + 2 k ^ & c → = - i ^ + 2 j ^ - k ^ , find a unit vectors normal to the vectors a → + b → and b → - c → .

Given that a → = i ^ + j ^ - k ^ , b → = - i ^ + 2 j ^ + 2 k ^ and c → = - i ^ + 2 j ^ - k ^ . So, a → + b → = 3 j ^ + k ^ and b → - c → = 3 k ^ Then d → = a → + b → × b → - c → = 3 j ^ + k ^ × 3 k ^ = 3 j ^ × 3 k ^ + k ^ × 3 k ^ = 9 i ^ Since vector product of two vectors is always normal (i.e. perpendicular) to both vectors, so the required unit vector will be d ^ = ± d → d → = ± i ^

Prove that a → × b → = - b → · a → × a → × b →

Let angle between a → & b → be θ . RHS = - b → · a → × a → × b → = - b → · a → · b → a → - a → · a → b → = - b → · a → · b → a → - a → 2 b → = a → 2 b → · b → - a → · b → 2 = a → 2 b → 2 - a → · b → 2 = a → 2 b → 2 - a → 2 b → 2 cos 2 θ = a → 2 b → 2 sin 2 θ = a → × b → 2 = a → × b → = L H S

If a → , b → , c → are non-coplanar vectors and d → is a unit vector, then find the value of a → · d → b → × c → + b → · d → c → × a → + c → · d → a → × b → independent of d → .

a → · d → b → × c → + b → · d → c → × a → + c → · d → a → × b → = a → · d → b → × c → - c → · d → b → × a → + b → · d → c → × a → ∵ m → × n → = - n → × m → = b → × a → · d → c → - c → · d → a → + b → · d → c → × a → (By distributive property of cross product of vectors) = b → × a → × c → × d → + b → · d → c → × a → = b → · d → a → × c → - b → · a → × c → d → - b → · d → a → × c → = - b → a → c → d → = a → b → c → d → (Using the property of scalar triple product) = a → b → c → ∵ d → = 1

Find the vector r → which is perpendicular to a → = i ^ - 2 j ^ + 5 k ^ and b → = 2 i ^ + 3 j ^ - k ^ and r → · 2 i ^ + j ^ + k ^ + 8 = 0 .

Since r → is perpendicular to both a → & b → and a → × b → is also perpendicular to both the vectors, then r → will be parallel to a → × b → . Thus r → = λ a → × b → , where λ ∈ R . Now, a → × b → = i ^ j ^ k ^ 1 - 2 5 2 3 - 1 = - 13 i ^ + 11 j ^ + 7 k ^ So, r → = λ - 13 i ^ + 11 j ^ + 7 k ^ Now, given that r → · 2 i + j ^ + k ^ + 8 = 0 ⇒ λ - 13 i ^ + 11 j ^ + 7 k ^ · 2 i + j ^ + k ^ + 8 = 0 ⇒ λ - 13 · 2 + 11 · 1 + 7 · 1 + 8 = 0 ⇒ - 8 λ + 8 = 0 ⇒ λ = 1

Two vertices of a triangle are at - i ^ + 3 j ^ and 2 i ^ + 5 j ^ and its orthocentre is at i ^ + 2 j ^ . Find the position vector of third vertex.

Here P is orthocentre with position vector i ^ + 2 j ^ . Then C P → ⊥ A B → ⇒ C P → · A B → = 0 . . ( 1 ) and A P → ⊥ B C → ⇒ A P → · B C → = 0 . . ( 2 ) Using concept of position vectors, C P → = 1 - a i ^ + 2 - b j ^ , A B → = 3 i ^ + 2 j ^ , A P → = 2 i ^ - j ^ and B C → = a - 2 i ^ + b - 5 j ^ So, ( 1 ) becomes 3 1 - a + 2 2 - b = 0 ⇒ 3 a + 2 b = 7 . . . ( 3 ) and 2 becomes 2 a - 2 - b - 5 = 0 ⇒ 2 a - b = - 1 . . ( 4 ) From 3 & 4 , a = 5 7 , b = 17 7 So, position vector of the third vertex C is 5 7 i ^ + 17 7 j ^ .

Find the point R in which the line A B cuts the plane C D E where a → = i ^ + 2 j ^ + k ^ , b → = 2 i ^ + j ^ + 2 k ^ , c → = - 4 j ^ + 4 k ^ , d → = 2 i ^ - 2 j ^ + 2 k ^ and e → = 4 i ^ + j ^ + 2 k ^ .

Here a → , b → , c → , d → , e → are the position vectors of points A , B , C , D , E respectively. Let the position vectors of required point R be x i ^ + y j ^ + z k ^ . Now, equation of line A B ↔ in vector form will be r → = a → + λ A B → ⇒ r → = i ^ + 2 j ^ + k ^ + λ 2 - 1 i ^ + 1 - 2 j + 2 - 1 k ^ ⇒ r → = i ^ + 2 j ^ + k ^ + λ i ^ - j + k ^ ⇒ r → = 1 + λ i ^ + 2 - λ j ^ + 1 + λ k ^ Since R lies on this line, so position vector of R 1 + λ i ^ + 2 - λ j ^ + 1 + λ k ^ . For plane C D E , normal to the plane will be along C D → × C E → C D → = 2 - 0 i ^ + - 2 + 4 j ^ + 2 - 4 k ^ = 2 i ^ + 2 j ^ - 2 k ^ and C E → = 4 - 0 i ^ + 1 + 4 j ^ + 2 - 4 k ^ = 4 i ^ + 5 j ^ - 2 k ^ C D → × C E → = i ^ j ^ k ^ 2 2 - 2 4 5 - 2 = 6 i ^ - 4 j ^ + 2 k ^ , which is the normal, n → , to the plane. So, equation of plane C D E will be r → · n → = m Since c → satisfies this equation ⇒ c → · n → = m ⇒ 6 · 0 + - 4 · - 4 + 4 · 2 = m ⇒ m = 24 Now, R also lies on this plane, so 1 + λ i ^ + 2 - λ j ^ + 1 + λ k ^ · 6 i ^ - 4 j ^ + 2 k ^ = 24 ⇒ 6 1 + λ - 4 2 - λ + 2 1 + λ = 24 ⇒ λ = 2 So, the position vector of required point is R → = 3 i ^ + 3 k ^

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Given four non zero vectors $\vec{a}, \vec{b}, \vec{c}$ and $\vec{d}$ . The vectors $\vec{a}, \vec{b}$ and $\vec{c}$ are coplanar but not collinear pair by pair and vector $\vec{d}$ is not coplanar with vectors $\vec{a}, \vec{b}$ and $\vec{c}$ and $\hat{(\vec{a} \vec{b})} = \hat{(\vec{b} \vec{c})} = \frac{π}{3}, \hat{(\vec{d} \vec{a})} = α, \hat{(\vec{d} \vec{b})} = β$ then prove that $\hat{(\vec{d} \vec{c})} = \cos^{- 1} (\cos β - \cos α)$

Important Questions on Vector Algebra

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Beta Question Bank for Engineering: Mathematics>Chapter 21 - Vector Algebra>EXERCISE-4>Q 18

O

is origin of reference, point

A (\vec{a}); B (\vec{b}); C (\vec{c}); D (\vec{a} + \vec{b}); E (\vec{b} + \vec{c}); F (\vec{c} + \vec{a}); G (\vec{a} + \vec{b} + \vec{c})

where

\vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}; \vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}

and

\vec{c} = c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k},

then prove that these points are vertices of a cube having length of its edge equal to unity provided the matrix

[\begin{matrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{matrix}]

is orthogonal. Also find the length

X Y

such that

X

is the point of intersection of

C M

and

G P; Y

is the point of intersection of

O Q

and

D N

, where

P, Q, M, N

are respectively the mid-point of sides

C F, B D, G F

and

O B

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Beta Question Bank for Engineering: Mathematics>Chapter 21 - Vector Algebra>EXERCISE-4>Q 20

Let

\vec{a} = \sqrt{3} \hat{i} - \hat{j}

and

\vec{b} = \frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j}

and

\vec{x} = \vec{a} + (q^{2} - 3) \vec{b}, \vec{y} = - p \vec{a} + q \vec{b}

. If

\vec{x} ⊥ \vec{y}

, then express

p

as a function of

q

, say

p = f (q), (p \neq 0 and q \neq 0)

and find the intervals of monotonicity of

f (q)

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Beta Question Bank for Engineering: Mathematics>Chapter 21 - Vector Algebra>EXERCISE-4>Q 21

\vec{a} = \hat{i} + \hat{j} - \hat{k}, \vec{b} = - \hat{i} + 2 \hat{j} + 2 \hat{k}

\vec{c} = - \hat{i} + 2 \hat{j} - \hat{k}

, find a unit vectors normal to the vectors

\vec{a} + \vec{b}

and

\vec{b} - \vec{c}

EASY

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Beta Question Bank for Engineering: Mathematics>Chapter 21 - Vector Algebra>EXERCISE-4>Q 22

Prove that

|\vec{a} \times \vec{b}| = \sqrt{- \vec{b} \cdot [\vec{a} \times (\vec{a} \times \vec{b})]}

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Beta Question Bank for Engineering: Mathematics>Chapter 21 - Vector Algebra>EXERCISE-4>Q 23

\vec{a}, \vec{b}, \vec{c}

are non-coplanar vectors and

\vec{d}

is a unit vector, then find the value of

|(\vec{a} \cdot \vec{d}) (\vec{b} \times \vec{c}) + (\vec{b} \cdot \vec{d}) (\vec{c} \times \vec{a}) + (\vec{c} \cdot \vec{d}) (\vec{a} \times \vec{b})|

independent of

\vec{d}

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Beta Question Bank for Engineering: Mathematics>Chapter 21 - Vector Algebra>EXERCISE-4>Q 24

Find the vector

\vec{r}

which is perpendicular to

\vec{a} = \hat{i} - 2 \hat{j} + 5 \hat{k}

and

\vec{b} = 2 \hat{i} + 3 \hat{j} - \hat{k}

and

\vec{r} \cdot (2 \hat{i} + \hat{j} + \hat{k}) + 8 = 0

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Beta Question Bank for Engineering: Mathematics>Chapter 21 - Vector Algebra>EXERCISE-4>Q 25

Two vertices of a triangle are at

- \hat{i} + 3 \hat{j}

and

2 \hat{i} + 5 \hat{j}

and its orthocentre is at

\hat{i} + 2 \hat{j}

. Find the position vector of third vertex.

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Beta Question Bank for Engineering: Mathematics>Chapter 21 - Vector Algebra>EXERCISE-4>Q 26

Find the point

R

in which the line

A B

cuts the plane

C D E

where

\vec{a} = \hat{i} + 2 \hat{j} + \hat{k}, \vec{b} = 2 \hat{i} + \hat{j} + 2 \hat{k}, \vec{c} = - 4 \hat{j} + 4 \hat{k}, \vec{d} = 2 \hat{i} - 2 \hat{j} + 2 \hat{k}

and

\vec{e} = 4 \hat{i} + \hat{j} + 2 \hat{k}

Given four non zero vectors a→,b→,c→ and d→. The vectors a→,b→ and c→ are coplanar but not collinear pair by pair and vector d→ is not coplanar with vectors a→,b→ and c→ and a→b→^=b→c→^=π3, d→a→^=α, d→b→^=β then prove that d→c→^=cos-1cosβ-cosα

Important Questions on Vector Algebra

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