QJEEM21Ma6948: JEE Main 2021 Physics - Class 11 JEEM21Ma6948 Solution: Let be a vector in the plane containing vectors and . If the…

Question

Verified Solution • 2025–26 Session moderate

Let

\overrightarrow x

be a vector in the plane containing vectors

\overrightarrow a = 2\widehat i - \widehat j + \widehat k

and

\overrightarrow b = \widehat i + 2\widehat j - \widehat k

. If the vector

\overrightarrow x

is perpendicular to

\left( {3\widehat i + 2\widehat j - \widehat k} \right)

and its projection on

\overrightarrow a

is

{{17\sqrt 6 } \over 2}

, then the value of

|\overrightarrow x {|^2}

is equal to __________.

Accepted Answer

Let,

\overrightarrow x = k(\overrightarrow a + \lambda \overrightarrow b )

\overrightarrow x

is perpendicular to

3\widehat i + 2\widehat j - \widehat k

I. k{(2 +

\lambda

)3 + (2

\lambda

-

1)2 + (1

-

\lambda

)(

-

1) = 0

\Rightarrow

8

\lambda

+ 3 = 0

\lambda = {{ - 3} \over 8}

II. Also projection of

\overrightarrow x

on

\overrightarrow a

is

{{17\sqrt 6 } \over 2}

therefore

{{\overrightarrow x .\overrightarrow a } \over {|\overrightarrow a |}} = {{17\sqrt 6 } \over 2}

\Rightarrow k\left\{ {{{(\overrightarrow a + \lambda \overrightarrow b ).\overrightarrow a } \over {\sqrt 6 }}} \right\} = {{17\sqrt 6 } \over 2}

\Rightarrow k\left\{ {6 + \left( {{3 \over 8}} \right)} \right\} = {{17 \times 6} \over 2}

\Rightarrow k = {{51} \over {51}} \times 8

k = 8

\therefore

\overrightarrow x = 8\left( {{{13} \over 8}\widehat i - {{14} \over 8}\widehat j + {{11} \over 8}\widehat k} \right)

= 13\widehat i - 14\widehat j + 11\widehat k

|\overrightarrow x {|^2} = 169 + 196 + 121 = 486

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