Let → a be a vector which is perpendicular to the vector 3 ^ i + (1)/(2) ^ j + 2 ^ k . If → a × (2 ^ i + ^ k ) = 2 ^ i - 13 ^ j - 4 ^ k , then the projection of the vector → a on the vector 2 ^ i + 2 ^ j + ^ k is

Correct answer: C. (5)/(3) Given, → a × (2 ^ i + ^ k ) = 2 ^ i - 13 ^ j - 4 ^ k Now multiplying both side by (3 ^ i + (1)/(2) ^ j + 2 ^ k ) we get, ⇒ ( → a × (2 ^ i + ^ k )) × (3 ^ i + (1)/(2) ^ j + 2 ^ k ) = (2 ^ i - 13 ^ j - 4 ^ k ) × (3 ^ i + (1)/(2) ^ j + 2 ^ k ) ⇒ ((3 ^ i + (1)/(2) ^ j + 2 ^ k ) . → a )(2 ^ i + ^ k ) - (6 + 2) → a = | cases ^ i & ^ j & ^ k 2 & - 13 & - 4 3 & (1)/(2) & 2 cases | ⇒ 0 - (6 + 2) → a = | cases ^ i & ^ j & ^ k 2 & - 13 & - 4 3 & (1)/(2) & 2 cases | → a = 3 ^ i + 2 ^ j - 5 ^ k Now, projection of → a on vector 2 ^ i + 2 ^ j + ^ k is → a · (2 ^ i + 2 ^ j + ^ k ) 3 = (5)/(3)

JEE Main 202228 JuneEvening Shift

Exam: JEE Main
Year: 2022
Date: 28 June
Shift: Evening Shift
Subject: Mathematics
Chapter: Vector Algebra

JEE Main 2022 · 28 June · Evening Shift · Mathematics · Vector Algebra

Let → a be a vector which is perpendicular to the vector 3 ^ i +...

Let

\overset{\to}{a}

be a vector which is perpendicular to the vector

3\overset{^}{i} + \dfrac{1}{2}\overset{^}{j} + 2\overset{^}{k}

. If

\overset{\to}{a} \times \left(2\overset{^}{i} + \overset{^}{k}\right) = 2\overset{^}{i} - 13\overset{^}{j} - 4\overset{^}{k}

, then the projection of the vector

\overset{\to}{a}

on the vector

2\overset{^}{i} + 2\overset{^}{j} + \overset{^}{k}

Options

A $\dfrac{1}{3}$
B $1$
C $\dfrac{5}{3}$
D $\dfrac{7}{3}$

Correct answer

C. $\dfrac{5}{3}$

Step-by-step solution

Given, $\overset{\to}{a} \times$ \left(2\overset{^}{i} + \overset{^}{k}\right) = 2\overset{^}{i} - 13\overset{^}{j} - 4\overset{^}{k} $</p><p>Now multiplying both side by \left(3\overset{^}{i} + \dfrac{1}{2}\overset{^}{j} + 2\overset{^}{k}\right)$ we get,

$⇒ \left(\overset{\to}{a} \times \left(2\overset{^}{i} + \overset{^}{k}\right)\right) \times \left(3\overset{^}{i} + \dfrac{1}{2}\overset{^}{j} + 2\overset{^}{k}\right)$

$= \left(2\overset{^}{i} - 13\overset{^}{j} - 4\overset{^}{k}\right) \times \left(3\overset{^}{i} + \dfrac{1}{2}\overset{^}{j} + 2\overset{^}{k}\right)$

$⇒ \left(\left(3\overset{^}{i} + \dfrac{1}{2}\overset{^}{j} + 2\overset{^}{k}\right) . \overset{\to}{a}\right)\left(2\overset{^}{i} + \overset{^}{k}\right) - \left(6 + 2\right)\overset{\to}{a} = \left|$ \begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 2 & - 13 & - 4 \\ 3 & \dfrac{1}{2} & 2\end{cases} $\right|$

$⇒ 0 - \left(6 + 2\right)\overset{\to}{a} = \left|$ \begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 2 & - 13 & - 4 \\ 3 & \dfrac{1}{2} & 2\end{cases} $\right|$

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

Now, projection of $\overset{\to}{a}$ on vector $2\overset{^}{i} + 2\overset{^}{j} + \overset{^}{k}$ is $\overset{\to}{a} \cdot \dfrac{\left(2\overset{^}{i} + 2\overset{^}{j} + \overset{^}{k}\right)}{3} = \dfrac{5}{3}$

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