Let → a = 2 ^ i + 7 ^ j - ^ k , ^ b = 3 ^ i + 5 ^ k and → c = ^ i - ^ j + 2 ^ k Let → d be a vector which is perpendicular to both → a and → b , and → c · → d = 12 . Then (- ^ i + ^ j - ^ k ) · ( → c × → d ) is equal to

Correct answer: B. 44 Given, → a = 2 ^ i + 7 ^ j - ^ k , ^ b = 3 ^ i + 5 ^ k and → c = ^ i - ^ j + 2 ^ k And → d be a vector which is perpendicular to both → a and → b , So, → d = λ( → a × → b ) Now finding, → a × → b = | cases ^ i & ^ j & ^ k 2 & 7 & - 1 3 & 0 & 5 cases | ⇒ → a × → b = 35 ^ i - 13 ^ j - 21 ^ k So, → d = λ(35 ^ i - 13 ^ j - 21 ^ k ) Also given, → c · → d = 12 ⇒ → c · → d = ( ^ i - ^ j + 2 ^ k ) · λ(35 ^ i - 13 ^ j - 21 ^ k ) ⇒ λ ( 35 + 13 - 42 ) = 12 ⇒ λ = 2 So, → c × → d = 2 × | cases ^ i & ^ j & ^ k 1 & - 1 & 2 35 & - 13 & - 21 cases | ⇒ → c × → d = 2(47 ^ i + 91 ^ j + 22 ^ k ) Now finding, (- ^ i + ^ j - ^ k ) · ( → c × → d ) = 2 (- ^ i + ^ j - ^ k )(47 ^ i + 91 ^ j + 22 ^ k ) ⇒ (- ^ i + ^ j - ^ k ) · ( → c × → d ) = 2 (- 47 + 91 - 22) = 44

JEE Main 202310 AprilEvening Shift

Exam: JEE Main
Year: 2023
Date: 10 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Vector Algebra

JEE Main 2023 · 10 April · Evening Shift · Mathematics · Vector Algebra

Let → a = 2 ^ i + 7 ^ j - ^ k , ^ b = 3 ^ i + 5 ^ k and → c = ^ i -...

Let

\overset{\to}{a} = 2\overset{^}{i} + 7\overset{^}{j} - \overset{^}{k}, \overset{^}{b} = 3\overset{^}{i} + 5\overset{^}{k}

and

\overset{\to}{c} = \overset{^}{i} - \overset{^}{j} + 2\overset{^}{k}

Let

\overset{\to}{d}

be a vector which is perpendicular to both

\overset{\to}{a} and \overset{\to}{b}

, and

\overset{\to}{c} \cdot \overset{\to}{d} = 12

. Then \left(- \overset{^}{i} + \overset{^}{j} - \overset{^}{k}\right) \cdot \left(\overset{\to}{c} \times \overset{\to}{d}\right) is equal to

Options

A $24$
B $44$
C $42$
D $48$

Correct answer

B. $44$

Step-by-step solution

Given,
$\overset{\to}{a} = 2\overset{^}{i} + 7\overset{^}{j} - \overset{^}{k}, \overset{^}{b} = 3\overset{^}{i} + 5\overset{^}{k}$ and $\overset{\to}{c} = \overset{^}{i} - \overset{^}{j} + 2\overset{^}{k}$
And $\overset{\to}{d}$ be a vector which is perpendicular to both $\overset{\to}{a} and \overset{\to}{b}$ ,
So, $\overset{\to}{d} = \lambda\left(\overset{\to}{a} \times \overset{\to}{b}\right)Now finding,$ \overset{\to}{a} \times \overset{\to}{b} = \left| $\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 2 & 7 & - 1 \\ 3 & 0 & 5\end{cases}$ \right| $$ ⇒ \overset{\to}{a} \times \overset{\to}{b} = 35\overset{^}{i} - 13\overset{^}{j} - 21\overset{^}{k} $So,$ \overset{\to}{d} = \lambda\left(35\overset{^}{i} - 13\overset{^}{j} - 21\overset{^}{k}\right)
Also given,
$\overset{\to}{c} \cdot \overset{\to}{d} = 12$
$⇒ \overset{\to}{c} \cdot \overset{\to}{d} = \left(\overset{^}{i} - \overset{^}{j} + 2\overset{^}{k}\right) \cdot \lambda\left(35\overset{^}{i} - 13\overset{^}{j} - 21\overset{^}{k}\right)$ ⇒ \lambda ( 35 + 13 - 42 ) = 12 $$ ⇒ \lambda = 2 $So,$ \overset{\to}{c} \times \overset{\to}{d} = 2 \times \left| $\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 1 & - 1 & 2 \\ 35 & - 13 & - 21\end{cases}$ \right| $$ ⇒ \overset{\to}{c} \times \overset{\to}{d} = 2\left(47\overset{^}{i} + 91\overset{^}{j} + 22\overset{^}{k}\right)
Now finding,
\left(- \overset{^}{i} + \overset{^}{j} - \overset{^}{k}\right) \cdot \left(\overset{\to}{c} \times \overset{\to}{d}\right) = 2 \left(- \overset{^}{i} + \overset{^}{j} - \overset{^}{k}\right)\left(47\overset{^}{i} + 91\overset{^}{j} + 22\overset{^}{k}\right) $$ ⇒ \left(- \overset{^}{i} + \overset{^}{j} - \overset{^}{k}\right) \cdot \left(\overset{\to}{c} \times \overset{\to}{d}\right) = 2 \left(- 47 + 91 - 22\right) = 44

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