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)
P Prateek Shrivastava

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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