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Let \overrightarrow{\mathrm{a}}=2 \hat{i}-\hat{j}+2 \hat{k}$ and $\overrightarrow{\mathrm{b}}=\hat{i}+2 \hat{j}-\hat{k}. Let a vector \vec{v}be in the plane containing \vec{a}$ and $\vec{b}. If \vec{v} is perpendicular to the vector 3 \hat{i}+2 \hat{j}-\hat{k} and its projection on \vec{a} is 19 units, then |2 \vec{v}|^{2} is equal to___________
 

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Let\, \vec{\vartheta }= x\vec{a}+y\vec{b}
        \vec{\vartheta }\cdot \left ( 3\hat{i}+2\hat{j}-\hat{k} \right )= 0
And\: \vec{\vartheta }\cdot \hat{a}= 19
Let\,\: 3\hat{i}+2\hat{j}-\hat{k}= \vec{c}

Let\,\: \vec{\vartheta } = \lambda \vec{c}\times \left ( \vec{a}\times \vec{b} \right )
(a vector in the plane of \vec{a} & \vec{b} and perpendicular  to \vec{c})
\vec{c}\times \left ( \vec{a}\times \vec{b} \right )= 14\hat{i}-12\hat{j}+18\hat{k}
\vec{\vartheta }= \lambda\left ( 14\hat{i}-12\hat{j}+18\hat{k} \right )
Also\,\; \vec{\vartheta }\cdot \hat{a}= 19
\Rightarrow \lambda \left ( 14\hat{i}-12\hat{j}+18\hat{k} \right )\cdot \frac{\left ( 2\hat{i} -\hat{j}+2\hat{k}\right )}{\sqrt{2^{2}+1^{1}+2^{2}}}= 19
\Rightarrow \lambda = \frac{19\times 3}{28+12+36}= \frac{19\times 3}{76}= \frac{3}{4}
so\: \left | 2\vartheta ^{2} \right |= \left | 2\times \frac{3}{4}\left ( 14\hat{i}-12\hat{j}+18\hat{k} \right ) \right |^{2}
                  = 9\left ( 7^{2}+6^{2}+9^{2} \right )
                  = 1494

Posted by

Kuldeep Maurya

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