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Solve! - Vector Algebra - JEE Main-4

Let \vec{a} =\hat{i}+\hat{j}+\hat{k}

\vec{c}= \hat{j}-\hat{k}   and a vector \vec{b}   be such that \vec{a}\times \vec{b}= \vec{c}

and\vec{a}\cdot \vec{b}= 3 then \left | \vec{b} \right |  equals 

  • Option 1)

    11/3

  • Option 2)

    11/\sqrt3

  • Option 3)

    \sqrt(11/3)

  • Option 4)

    \sqrt11/3

 
Answers (2)
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As we have learned

Vector Product of two vectors -

Vector product

- wherein

\vec{a}= (a_{1}\hat{i}+a_{2}\hat{j}+a_{3}\hat{k})

\vec{b}= (b_{1}\hat{i}+b_{2}\hat{j}+b_{3}\hat{k})

 

 

Properties of Scalar Product -

\vec{i}.\vec{i}=\vec{j}.\vec{j}=\vec{k}.\vec{k}=1

\vec{i}.\vec{j}=\vec{j}.\vec{k}=\vec{k}.\vec{i}=0

-

 

  

\hat{a}= \hat{i}+\hat{j}+\hat{k}

\hat{b}= \hat{xi}+\hat{yj}+\hat{zk}

\hat{a}-\hat{b}=3\Rightarrow x+y+z=3

\hat{a}\times \hat{b}=\hat{j}-\hat{k}

thus  z-y=0 \Rightarrow z=y

and x-z=1

y-x=-1

also x+2y=3

y=2/3 and z=  2/3  and x= 5/3

\left | b^{-1} \right |= \sqrt{(5/3)^{2}+(2/3)^{2}+(2/3)^{2}}=\sqrt{\frac{33}{9}}

\sqrt{\frac{11}{3}}

 

 

 

 

 

 


Option 1)

11/3

This is incorrect 

Option 2)

11/\sqrt3

This is incorrect 

Option 3)

\sqrt(11/3)

This is correct 

Option 4)

\sqrt11/3

This is incorrect 

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