JEE MAIN - Mathematics (2021 - 17th March Evening Shift - No. 17)
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 __________.
Answer
486
Explanation
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
$$ \therefore $$ $$\overrightarrow x = 8\left( {{{13} \over 8}\widehat i - {{14} \over 8}\widehat j + {{11} \over 8}\widehat k} \right)$$
$$ = 13\widehat i - 14\widehat j + 11\widehat k$$
$$|\overrightarrow x {|^2} = 169 + 196 + 121 = 486$$
$$\overrightarrow x$$ 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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