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NIMCET 2025 Paper

Section: Mathematics

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Question : 3 of 120

Marks: +1, -0

The length of the projection of

\vec{a}=2\hat{i}+3\hat{j}+\hat{k}

\vec{b}=-2\hat{i}+\hat{j}+2\hat{k},

$\; \frac{2}{3}$
$\; \frac{-1}{3}$
$\; \frac{1}{3}$
$\; \frac{-2}{3}$

Solution:

Concept:

Length of projection of

\vec{a}

on

\vec{b}

is

\left|\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|}\right|

(magnitude of the scalar projection).

Formula:

Projection length

= \left|\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|}\right|

Solution:

\vec{a}\cdot\vec{b}= (2)(-2)+(3)(1)+(1)(2) = -4+3+2=1

|\vec{b}|= \sqrt{(-2)^2+1^2+2^2}= \sqrt{4+1+4}=3

Projection length

= \left|\frac{1}{3}\right| = \frac{1}{3}

Answer:

C.

\frac{1}{3}

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