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Vector Algebra Part 3

Section: Mathematics

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Question : 36 of 72

Marks: +1, -0

Let

\vec{a}=3 \hat{i}-\hat{j}+2 \hat{k}, \vec{b}=\vec{a}\times(\hat{i}-2 \hat{k})

\vec{c}=\vec{b}\times \hat{k}

Then the projection of

\vec{c}=2 \hat{j}

\vec{a}

[24 Jan 2025 Shift 2]

$\sqrt{14}$
$2 \sqrt{7}$
$2 \sqrt{14}$
$3 \sqrt{7}$

Solution:

Given

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

\vec{b}=\vec{a}\times(\hat{i}-2 \hat{k}) \Rightarrow \vec{b}=2 \hat{i}+8 \hat{j}+\hat{k}

\vec{c}=\vec{b}\times \hat{k} \Rightarrow \vec{c}=8 \hat{i}-2 \hat{j}

We need projection of

\vec{c}-2 \hat{j}

on

\vec{a}

:

\vec{c}-2 \hat{j}=(8 \hat{i}-2 \hat{j})-2 \hat{j}=8 \hat{i}-4 \hat{j}

Scalar projection formula

\text{ Projection } = \frac{\vec{(c-2 \hat{j})} \cdot \vec{a}}{\vec{|a|}}

Dot product:

(8,-4,0) \cdot (3,-1,2)=24+4=28

Magnitude:

|\vec{a}|=\sqrt{14}

Final Answer

\frac{28}{\sqrt{14}}=2 \sqrt{14}

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