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Let $\vec{a}=\hat{j}-\hat{k}\; and\; \vec{c}=\hat{i}-\hat{j}-\hat{k}.$ Then the vector $\vec{b}$ satisfying $\vec{a}\times \vec{b}+ \vec{c}=0\; and\; \vec{a}\cdot \vec{b}=3\; is$

Option 1)
$-\hat{i}+\hat{j}-2\hat{k}$

Option 2)
$2\hat{i}-\hat{j}+2\hat{k}$

Option 3)
$\hat{i}-\hat{j}-2\hat{k}$

Option 4)
$\hat{i}+\hat{j}-2\hat{k}$

Answers (1)

best_answer

As we learnt in

Unit vector perpendicular to the plane of vector a and b -

$\frac{\vec{a}\times \vec{b}}{\left | \vec{a}\times \vec{b} \right |}$

- wherein

Here $\vec{a}$ and $\vec{b}$ are two vectors.

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 $\vec{a}\: and\:\vec{b}$

$\vec{c}=\vec{b}\times \vec{a}=> \vec{b}\:.\vec{c}=0$

$(b_{1}\hat{i}+b_{2}\hat{j}+b_{3}\hat{k}).(\hat{i}-\hat{j}-\hat{k})=0$

$b_{1}-b_{2}-b_{3}=0$

$\vec{a}\:.\:\vec{b}=3$

$b_{2}-b_{3}=3$

$b_{1}=b_{2}+b_{3}=3+2b_{3}$

$b_{3}=\:-2$

$\vec{b}=(3+2b_{3})\hat{i}+(3+b_{3})\hat{j}+b_{3}\hat{k}$

So $\vec{b}=-\hat{i}+\hat{j}-2\hat{k}$

Option 1)

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

This option is correct.

Option 2)

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

This option is incorrect.

Option 3)

$\hat{i}-\hat{j}-2\hat{k}$

This option is incorrect.

Option 4)

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

This option is incorrect.

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prateek

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