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Let $\underset{a}{\rightarrow}$ $=\: 2\: \hat{i}\: +\: \hat{j}\: -2\: \hat{k}$ and $\underset{b}{\rightarrow}$ $= \hat{i}\: +\hat{j}.$

Let $\underset{c}{\rightarrow}$ be a vector such that

and the angle between $\underset{c}{\rightarrow}$ and

thenis equal to:

Option 1)
$2$

Option 2)
$5$

Option 3)
$\frac{1}{8}$

Option 4)
$\frac{25}{8}$

Answers (1)

best_answer

As we learnt in

Vector Triple Product -

$\vec{a}\times \left ( \vec{b} \times \vec{c}\right )= x\vec{b}+y\vec{c}$

- wherein

x and y are scalars.

and

Scalar Product of two vectors -

$\vec{a}.\vec{b}> 0 \:an\: acute\: angle$

$\vec{a}.\vec{b}< 0 \:an\: obtuse\: angle$

$\vec{a}.\vec{b}= 0 \:a\:right\: angle$

- wherein

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

$(\vec{c}-\vec{a})=3, \ |(\vec{a}\times \vec{b})\times \vec{c} |=3$

Also, $\vec{a}\times \vec{b}=\begin{vmatrix} \hat{i} & \hat{j} &\hat{k} \\ 2 & 1 &-2 \\ 1& 1 & 0 \end{vmatrix}$

$=2\hat{i}+2\hat{j}+\hat{k}$

$\left | \left ( \vec{a}\times \vec{b} \right ) \times \vec{c}\right |= \left | \left ( \vec{a}\times \vec{b} \right ) \right |\left | \vec{c} \right |sin 30^{\circ}$

$3=3\times \left | \vec{c} \right |\times \frac{1}{2}$

$\Rightarrow \left | \vec{c} \right |=2$

Also, $\left | \vec{c}-\vec{a} \right |=3$

So, $\left | \vec{c} \right |^{2}+\left | \vec{a} \right |^{2}-2\vec{c}\cdot \vec{a}=9$

$4+9-2\vec{c}\cdot \vec{a}=9$

$\vec{c}\cdot \vec{a}=2$

Option 1)

$2$

This option is correct.

Option 2)

$5$

This option is incorrect.

Option 3)

$\frac{1}{8}$

This option is incorrect.

Option 4)

$\frac{25}{8}$

This option is incorrect.

Posted by

divya.saini

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