Position vectors of A , B , C are given by a → , b → , c → , where a → × b → + b → × c → + c → × a → = 0 . If A C → = 2 i ^ - 3 j ^ + 6 k ^ , then find B C → if B C = 14 .

Given position vectors of A , B , C are a → , b → , c → , respectively. And a → × b → + b → × c → + c → × a → = 0 Let A be the origin. Then a → = 0 → , b → = A B → and c → = A C → . We get 0 → × b → + b → × c → + c → × 0 → = 0 . ⇒ b → × c → = 0 ⇒ b → and c → are parallel or collinear. ∴ b → = λ c → , where λ is any scalar. Now B C → = A C → - A B → = c → - b → = c → - λ c → = 1 - λ c → . Given, A C → = 2 i ^ - 3 j ^ + 3 k ^ . ∴ B C → = 1 - λ 2 i ^ - 3 j ^ + 6 k ^ ⇒ B C → = 1 - λ 2 i ^ - 3 j ^ + 6 k ^ ⇒ 14 = 1 - λ 2 2 + - 3 2 + 6 2 ⇒ 14 = 1 - λ = 7 ⇒ 1 - λ = 2 ⇒ 1 - λ = 2 , 1 - λ = - 2 ⇒ λ = - 1 , λ = 3 Case 1 : When λ = - 1 ∴ B C → = 1 + 1 2 i ^ - 3 j ^ + 6 k ^ = 2 2 i ^ - 3 j ^ + 6 k ^ Case 2 : When λ = 3 ⇒ B C → = 1 - 3 2 i ^ - 3 j ^ + 6 k ^ = - 2 2 i ^ - 3 j ^ + 6 k ^ Therefore, B C → = ± 2 2 i ^ - 3 j ^ + 6 k ^ = ± 4 i ^ - 6 j ^ + 12 k ^ .

a A vector c → is perpendicular to the vectors 2 i ^ + 3 j ^ - k ^ , i ^ - 2 j ^ + 3 k ^ and satisfies the condition c → · 2 i ^ - j ^ + k ^ + 6 = 0 . Find the vector c → .

Given, a → = i ^ + 3 j ^ - k ^ and b → = i ^ - 2 j ^ + 3 k ^ Also c → is perpendicular to both a → & b → . Let c → = λ a → × b → . Now a → × b → = i ^ j ^ k ^ 2 3 - 1 1 - 2 3 = 9 - 2 i ^ - 6 + 1 j ^ + - 4 - 3 k ^ = 7 i ^ - 7 j ^ - 7 k ^ . ∴ c → = λ 7 i ^ - 7 j ^ - 7 k ^ Also, given c → · 2 i ^ - j ^ + k ^ + 6 = 0 . ⇒ λ 7 i ^ - 7 j ^ - 7 k ^ · 2 i ^ - j ^ + k ^ + 6 = 0 ⇒ λ 14 + 7 - 7 + 6 = 0 ⇒ 14 λ + 6 = 0 ⇒ λ = - 6 14 ⇒ λ = - 3 7 ∴ c → = - 3 7 7 i ^ - 7 j ^ - 7 k ^ = - 3 i ^ - j ^ - k ^ = 3 - i ^ + j ^ + k ^

a Show that the perpendicular distance of the point c → from the line joining a → and b → is b → × c → + c → × a → + a → × b → b → - a → .

Let A B C be a triangle where a → , b → & c → are the position vectors of A , B & C , respectively. Here B C = p , the perpendicular distance of point c → from the line joining a → & b → . We know that, Area of ∆ A B C = 1 2 × Base × Height = 1 2 A B → × p ∴ We get 1 2 × A B → × p = 1 2 A B → × A C → ⇒ p = A B → × A C → A B → = ( b → - a → ) × ( c → - a → ) | b → - a → | = | ( b → × c → ) - ( a → × c → ) - ( b → × a ¯ ) + a → × a → | | b → - a → | = b → × c → + c → × a → + a → × b → b → - a → ∵ a → × a → = 0 → & a → × b → = - b → × a →

Show that { ( a → + b → + c → ) × ( c → - b → ) } · a → = 2 a → b → c → .

a → + b → + c → × c → - b → . a → = a → × c → + b → × c → + c → × c → - a → × b → - b → × b → - c → × b → . a → = a → × c → + b → × c → + 0 - a → × b → - 0 - c → × b → . a → = a → × c → + b → × c → - a → × b → + b → × c → . a → = a → × c → + 2 b → × c → - a → × b → . a → = a → c → a → + 2 b → c → a → - a → b → a → = 0 + 2 a → b → c → - 0 = 2 a → b → c → Hence proved

a Given unit vectors m ^ , n ^ and p ^ such that m ^ n ^ ^ = p ^ m ^ × n ^ = α , then find value of n ^ p ^ m ^ in terms of α .

Given m ^ , n ^ & p ^ are unit vectors. ∴ m ^ = n ^ = p ^ = 1 Also m ^ n ^ ^ = p ^ m ^ × n ^ = α i.e. Angle between m ^ & n ^ is α and angle between p ^ and m ^ × n ^ is also α . ∴ n ^ p ^ m ^ = p ^ m ^ n ^ · ∵ a → b → c → = b → c → a → = p ^ · m ^ × n ^ , ∵ a → b → c → = a → · b → × c → = p ^ m ^ × n ^ cos α = 1 · m ^ n ^ sin α · cos α = 1 · 1 · 1 sin α cos α = sin α cos α

Examine for coplanarity of the following sets of points a 4 i ^ + 8 j ^ + 12 k ^ , 2 i ^ + 4 j ^ + 6 k ^ , 3 i ^ + 5 j ^ + 4 k ^ , 5 i ^ + 8 j ^ + 5 k ^ .

Given, A → = 4 i ^ + 8 j ^ + 12 k ^ , B → = 2 i ^ + 4 j ^ + 6 k ^ , C → = 3 i ^ + 5 j ^ + 4 k ^ , D → = 5 i ^ + 8 j ^ + 5 k ^ So, A B → = B → - A → = 2 i ^ - 4 j ^ - 6 k ^ A C → = C → - A → = - i ^ - 3 j ^ - 8 k ^ A D → = D → - A → = i ^ - 7 k ^ Now, A B → A C → A D → = - 2 - 4 - 6 - 1 - 3 - 8 1 0 - 7 ⇒ - 2 21 - 0 + 4 7 + 8 - 6 - 3 ⇒ - 42 + 60 - 18 = 60 - 60 = 0 ∵ A B → A C → A D → = 0 ∴ A B → , A C → & A D → lie on same plane. Hence, A → , B → , C → & D → all are coplanar.

E M B I B E

Alpha Question Bank for Engineering: Mathematics>Chapter 20 - Vector Algebra>Exercise-1>Q 18

MEDIUM

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IMPORTANT

Earn 100

Prove using vectors: If two medians of a triangle are equal in magnitude, then it is isosceles.

Important Questions on Vector Algebra

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 20 - Vector Algebra>Exercise-1>Q 19

Position vectors of

A, B, C

are given by

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

, where

\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} = 0

. If

\vec{A C} = 2 \hat{i} - 3 \hat{j} + 6 \hat{k}

, then find

\vec{B C}

if

|B C| = 14 .

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 20 - Vector Algebra>Exercise-1>Q 20

(a)

A vector

\vec{c}

is perpendicular to the vectors

2 \hat{i} + 3 \hat{j} - \hat{k},

\hat{i} - 2 \hat{j} + 3 \hat{k}

and satisfies the condition

\vec{c} \cdot (2 \hat{i} - \hat{j} + \hat{k}) + 6 = 0

. Find the vector

\vec{c}

.

MEDIUM

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IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 20 - Vector Algebra>Exercise-1>Q 21

(a)

Show that the perpendicular distance of the point

\vec{c}

from the line joining

\vec{a}

and

\vec{b}

is

\frac{|\vec{b} \times \vec{c} + \vec{c} \times \vec{a} + \vec{a} \times \vec{b}|}{|\vec{b} - \vec{a}|}

.

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 20 - Vector Algebra>Exercise-1>Q 23

P, Q

are the mid-points of the non-parallel sides

B C

and

A D

of a trapezium

A B C D

. Show that

∆ A P D = Δ C Q B .

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 20 - Vector Algebra>Exercise-1>Q 25

Prove that the shortest distance between the diagonals of a rectangular parallelepiped whose sides are

a, b, c

and the edges not meeting it are

\frac{b c}{\sqrt{b^{2} + c^{2}}}, \frac{c a}{\sqrt{c^{2} + a^{2}}}, \frac{a b}{\sqrt{a^{2} + b^{2}}}

.

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 20 - Vector Algebra>Exercise-1>Q 26

Show that

{(\vec{a} + \vec{b} + \vec{c}) \times (\vec{c} - \vec{b})} \cdot \vec{a} = 2 [\begin{array}{l} \vec{a} & \vec{b} & \vec{c} \end{array}]

.

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 20 - Vector Algebra>Exercise-1>Q 27

(a)

Given unit vectors

\hat{m}, \hat{n}

and

\hat{p}

such that

(\hat{\hat{m} \hat{n}}) = \hat{p} (\hat{m} \times \hat{n}) = α

, then find value of

[\hat{n} \hat{p} \hat{m}]

in terms of

α .

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 20 - Vector Algebra>Exercise-1>Q 29

Examine for coplanarity of the following sets of points

$(a)$ $4 \hat{i} + 8 \hat{j} + 12 \hat{k},$ $2 \hat{i} + 4 \hat{j} + 6 \hat{k},$ $3 \hat{i} + 5 \hat{j} + 4 \hat{k},$ $5 \hat{i} + 8 \hat{j} + 5 \hat{k} .$

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