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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)

best_answer

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

Posted by

Himanshu

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