Test Index

Vector Algebra Part 1

Section: Mathematics

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Question : 84 of 100

Marks: +1, -0

Let

\vec{u}, \vec{v}, \vec{w}

|\vec{u}|=1,|\vec{v}| 2,|\vec{w}| 3

. If the projection

\vec{v}

\vec{u}

is equal to that of

\vec{w}

\vec{u}

\vec{v}, \vec{w}

are perpendicular to each other then

|\vec{u}-\vec{v}+\vec{w}|

$14$
$\sqrt{7}$
$\sqrt{14}$
$2$

Solution:

Projection of

\vec{v}

along

\vec{u} = \frac{\vec{v} \cdot \vec{u}}{|\vec{u}|} = \frac{\vec{v} \cdot \vec{u}}{2}

projection of

\vec{w}

along

\vec{u} = \frac{\vec{w} \cdot \vec{u}}{|\vec{u}|} = \frac{\vec{w} \cdot \vec{u}}{2}

Given

\frac{\vec{v} \cdot \vec{u}}{2} = \frac{\vec{w} \cdot \vec{u}}{2}

....(1)

Also,

\vec{v} \cdot \vec{w}=0

.....(2)

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{v} \cdot \vec{w}+2 \vec{u} \cdot \vec{w}

=1+4+9+0[

From (1) and (2)

]=14

\therefore |\vec{u}-\vec{v}+\vec{w}|=\sqrt{14}

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