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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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