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  • Need clarity, kindly explain! - Vector Algebra - JEE Main-2

if \left | \vec{c} \right |^{2}=60\; and\; \vec{c}\times (\hat{i}+2\hat{j}+5\hat{k})=\vec{0}, then a value of  \left \vec{c} \right \cdot (-7\hat{i}+2\hat{j}+3\hat{k})\; is\, \; :   

  • Option 1)

    4\sqrt{2}

  • Option 2)

    12

  • Option 3)

    24

  • Option 4)

    12\sqrt{2}

 

Answers (1)

best_answer

As we learnt in 

Collinear Vectors -

Two vectors are said to be collinear if and only if there exists a scalar m such as that \vec{a}=m\vec{b}

- wherein

m is a Scalar.

 

 \vec{c}\times (\vec{i}+\vec{2j}+\vec{5k})=\vec{0}

So\:\vec{c}\:and (\vec{i}+\vec{2j}+\vec{5k})\:are\:collinear

\vec{c}=P(\vec{i}+\vec{2j}+\vec{5k})

\left |\vec{c} \right |^{2}=P^{2}\times (1^{2}+2^{2}+5^{2})=60\Rightarrow P^{2}=2

value= \vec{c}.(-7\vec{i}+2\vec{j}+3\vec{k})

=\sqrt{2}(\vec{i}+2\vec{j}+5\vec{k}).(-7\vec{i}+2\vec{j}+3\vec{k})

=\sqrt{2}(-7+4+15)=12\sqrt{2}

Correct option is 4.


Option 1)

4\sqrt{2}

This option is incorrect 

Option 2)

12

This option is correct 

Option 3)

24

This option is incorrect 

Option 4)

12\sqrt{2}

This option is  incorrect 

Posted by

Aadil

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