O is the origin and A is a fixed point on the circle of radius a with centre O . The vector O A → is denoted by a → . A variable point lies on the tangent at A and O P → = r → . Show that a → . r → = a 2 . Hence, if P ( x , y ) and A ( x 1 , y 1 ) deduce the equation of tangent at A to this circle.

Assume A ( x 1 , y 1 ) in the x y – plane. ∴ O A → = a → = x 1 i ^ + y 1 j ^ OP → = r → = x i ^ + y j ^ Now A P = position vector of P - position vector of A AP → = ( r → – a → ) ∵ Point P lies on the tangent to the circle. ∴ O A is perpendicular to A P . ⇒ O A → · A P → = 0 ⇒ a → · ( r → - a → ) = 0 ⇒ a → · r → - a → . a → = 0 ⇒ a → · r → - a → 2 = 0 ⇒ a → · r → = a → 2 ⇒ x i ^ + y j ^ . x 1 i ^ + y 1 j ^ = a 2 ⇒ x x 1 + y y 1 = a 2 This is the equation of the tangent to the circle at the point A .

If r → and s → are nonzero constant vectors and the scalar b is chosen such that r → + b s → is minimum, then show that the value of b s → 2 + r → + b s → 2 is equal to r → 2 .

r → + b s → is minimum when r → + b s → 2 is minimum. Now, r → + b s → 2 = r → 2 + b s → 2 + 2 r → · b s → = r → 2 + b s → 2 + 2 r → b s → cos θ , where θ is the angle between r → & b s → . Since minimum value of cos θ = - 1 , so min r → + b s → 2 = min r → 2 + b s → 2 - 2 r → b s → = min r → - b s → 2 = 0 when r → = b s → i.e. minimum value of r → + b s → = 0 and r → = b s → So, b s → 2 + r → + b s → 2 = r → 2 .

Find a unit vector a → which makes an angle of π 4 with the z -axis and it is such that ( a → + i ^ + j ^ ) is a unit vector.

Let a → = x i ^ + y j ^ + z k ^ Given | a → | = 1 , therefore x 2 + y 2 + z 2 = 1 . . . ( i ) Angle between a → and z -axis is π / 4 ; therefore cos π 4 = a → · k ^ | a → | | k ^ | ⇒ z = 1 2 Now a → + i ^ + j ^ = ( x + 1 ) i ^ + ( y + 1 ) j ^ + z k ^ Given that a → + i ^ + j ^ is a unit vector. Therefore, | a → + i ^ + j ^ | = ( x + 1 ) 2 + ( y + 1 ) 2 + z 2 = 1 or x 2 + y 2 + z 2 + 2 x + 2 y + 1 = 0 or 1 + 2 x + 2 y + 1 = 0 or y = - ( x + 1 ) Putting in ( i ) , we get x 2 + ( x + 1 ) 2 + 1 2 2 = 1 ⇒ 4 x 2 + 4 x + 1 = 0 or ( 2 x + 1 ) 2 = 0 ⇒ x = - 1 2 ⇒ y = - 1 2 Hence, a → = - 1 2 i ^ - 1 2 j ^ + 1 2 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 → ^ = 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

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IMPORTANT

Earn 100

Two medians of a triangle are equal, then using vector show that the triangle is isosceles.

Important Questions on Vector Algebra

MEDIUM

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 21 - Vector Algebra>EXERCISE-4>Q 13

O

is the origin and

A

is a fixed point on the circle of radius

a

with centre

O

. The vector

\vec{O A}

is denoted by

\vec{a}

. A variable point lies on the tangent at

A

and

\vec{O P} = \vec{r}

. Show that

\vec{a} . \vec{r} = a^{2}

. Hence, if

P (x, y)

and

A (x_{1}, y_{1})

deduce the equation of tangent at

A

to this circle.

MEDIUM

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 21 - Vector Algebra>EXERCISE-4>Q 14

\vec{r}

and

\vec{s}

are nonzero constant vectors and the scalar

b

is chosen such that

|\vec{r} + b \vec{s}|

is minimum, then show that the value of

{|b \vec{s}|}^{2} + {|\vec{r} + b \vec{s}|}^{2}

is equal to

{|\vec{r}|}^{2}

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 21 - Vector Algebra>EXERCISE-4>Q 15

Find a unit vector

\vec{a}

which makes an angle of

\frac{π}{4}

with the

z

-axis and it is such that

(\vec{a} + \hat{i} + \hat{j})

is a unit vector.

MEDIUM

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 21 - Vector Algebra>EXERCISE-4>Q 17

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)

HARD

JEE Main/Advance

IMPORTANT

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

HARD

JEE Main/Advance

IMPORTANT

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)

MEDIUM

JEE Main/Advance

IMPORTANT

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

JEE Main/Advance

IMPORTANT

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})]}

Two medians of a triangle are equal, then using vector show that the triangle is isosceles.

Important Questions on Vector Algebra

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