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Let $\vec{u},\vec{v},\vec{w}$ be such that $\left | \vec{u} \right |=1,\left |\vec{v} \right |=2,\left | \vec{w} \right |=3.$ If the projection $\vec{v} \; along\; \; \vec{u}$ is equal to that of $\vec{w} \; along\; \; \vec{u}\; and\; \vec{v}, \vec{w}$ are perpendicular to each other then $\left |\vec{u}- \vec{v}+ \vec{w}\right |$ equals

Option 1)
$\sqrt{14}\;$

Option 2)
$\; \sqrt{7}$

Option 3)
2

Option 4)
14

Answers (1)

best_answer

As we have learned

Projection of vector b on vector a -

$\vec{b}\cos \Theta = \frac{\vec{a}.\vec{b}}{\left | \vec{a} \right |}$

- wherein

dot product

Scalar Product of two vectors -

$\vec{a}.\vec{b}> 0 \:an\: acute\: angle$

$\vec{a}.\vec{b}< 0 \:an\: obtuse\: angle$

$\vec{a}.\vec{b}= 0 \:a\:right\: angle$

- wherein

$\Theta$ is the angle between the vectors $\vec{a}\:and\:\vec{b}$

$\vec{v} \cdot \frac{\vec{u}}{|\vec{u}|} = \vec{w} \cdot \frac{\vec{u}}{|\vec{u}|}$

$\Rightarrow \vec{v}\cdot \vec{u} = \vec{w} \cdot \vec{u}$

Also , $\Rightarrow \vec{v}\cdot \vec{w} = 0$

$\Rightarrow|\vec{u} \vec{v}+ \vec{w}| = |\vec{u}|^2+ |\vec{v}|^2 + |\vec{w}|^2\\ = 1+4+9 = 14$

Option 1)

$\sqrt{14}\;$

Option 2)

$\; \sqrt{7}$

Option 3)

2

Option 4)

14

Posted by

Himanshu

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