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  • Let \underset{u}{\rightarrow} be a vector coplanar with the vectors \underset{a}{\rightarrow}= 2\hat{i}+3\hat{j}-\hat{k} and \underset{b}{\

Let \underset{u}{\rightarrow} be a vector coplanar with the vectors \underset{a}{\rightarrow}= 2\hat{i}+3\hat{j}-\hat{k}  and \underset{b}{\rightarrow}= \hat{j}+\hat{k} . If \underset{u}{\rightarrow} is perpendicular to \underset{a}{\rightarrow} and \underset{u}{\rightarrow}•\underset{b}{\rightarrow}= 24, then \left | \vec{u} \right |^{2} is equal to :

Option 1)  84

Option 2)  336

Option 3)  315

Option 4)  256

Answers (1)

As we have learned.

Vector Product of two vectors(cross product) -

If \vec{a} and \vec{b} are two vectors and \Theta is the angle between them , then \vec{a}\times \vec{b}=\left |\vec{a} \left | \right |\vec{b} \right |Sin\Theta \hat{n}

- wherein

\hat{n} is unit vector perpendicular to both \vec{a} \: and \: \vec{b}

 

Scalar Product of two vectors (dot product) -

\vec{a}\vec{b}=\left | a \right |\left | b \right |Cos\theta

- wherein

\Theta is the angle between the vectors\vec{a}\: and\:\vec{b}

 

 

Here \left | \hat{a} \right |^{2}= 14\: \: \hat{a}\cdot \dot{b}= 2\: \: \left | \dot{b} \right |^{2}= 2

 

Here \hat{u}= \lambda \left ( \hat{a} \times \hat{b}\right )\times \hat{a}

\\*\hat{u}= \lambda \left ( \left ( \hat{a} .\hat{a}\right ) \hat{b}-\left ( \hat{b}.\hat{a} \right )\hat{a}\right )\\* =\lambda \left ( 14\left ( \hat{j}+\dot{k} \right )-2\left ( 2\hat{i}+3\hat{j}-\hat{k} \right ) \right ) \\*-4\lambda \left ( \hat{i}-2\hat{j} -4\hat{k}\right )\\* ={\lambda }'\left ( \hat{i}-2\hat{j} -4\hat{k}\right )\\*

\\*Also\: \: \: \hat{u}\cdot \hat{b}= 24\\* \Rightarrow {\lambda }'\left ( 0-2-4 \right )= 24\\* \Rightarrow \: {\lambda }'= -4

Thus

\\*\hat{u}= -4\hat{i}+8\hat{j}+16\hat{k}\\* \left | \hat{u} \right |^{2}=16+64+256\\* = 336

 

Posted by

lovekush

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