First slide

Dot product or scalar product of vectors

Question

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

Moderate

A

2
B

$\sqrt{7}$
C

$\sqrt{14}$
D

14

Solution

$\vec{v} \cdot \overset{\land}{u} = \vec{w} \cdot \overset{\land}{u}$
$\vec{v} ⊥ \vec{w} \Rightarrow \vec{v} \cdot \vec{w} = 0$
$\underline{\underline{Now,}} | \vec{u} - \vec{v} + \vec{w} |^{2} = {\vec{u}}^{2} + {\vec{v}}^{2} + {\vec{w}}^{2} - 2 \vec{u} \cdot \vec{v} - 2 \vec{w} \cdot \vec{v} + 2 \vec{u} \cdot \vec{w} = 1 + 4 + 9$
$\underline{\underline{So,}} | \vec{u} - \vec{v} + \vec{w} | = \sqrt{14}$

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