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  • Projection vector of vector a on vector b on is A. vector a . vector b/ vector b ^2 vector b

$
\begin{aligned}
&\text { Q: Projection vector of } \vec{a} \text { on } \vec{b} \text { is }\\
&\begin{aligned}
& \text { A. }\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right) \vec{b} \\
& \text { в. } \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \\
& \text { c. } \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|} \\
& \text { D. } \left.\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}\right) \hat{b} \right\rvert\,
\end{aligned}
\end{aligned}
$

Answers (1)

Solution

Let θ be the angle between \vec {a} $ and $\vec{b}

From figure we can see that,length \mathrm{OL}  is the projection of 

\\\vec{a} on \overrightarrow{\mathrm{b}} and \overrightarrow{\mathrm{O}} \text{is the projection vector of} \overrightarrow{\mathrm{a}} on \overrightarrow{\mathrm{b}} \\ In \Delta OLA, \text{we have}\\ \cos \theta=\frac{O L}{O A}\\ \Rightarrow \mathrm{OL}=\mathrm{OA} \cos \theta

\\ \Rightarrow \mathrm{OL}=|\overrightarrow{\mathrm{a}}| \cos \theta \\ \qquad \mathrm{OL}=|\overrightarrow{\mathrm{a}}|\left\{\frac{(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}})}{|\overrightarrow{\mathrm{a}}||\mathrm{b}|}\right\}\left(\because \cos \theta=\frac{(\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}})}{|\overrightarrow{\mathrm{a}}||\mathrm{b}|}\right) \\ \Rightarrow \mathrm{OL}=\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{b}}|}

\\ \begin{aligned} &\text { Now, }\\ &\overrightarrow{\mathrm{OL}}=(\mathrm{OL}) \hat{\mathrm{b}}\\ &\Rightarrow \overrightarrow{\mathrm{OL}}=\left\{\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{b}}|}\right\} \hat{\mathrm{b}}\\ &\overrightarrow{\mathrm{OL}}=\left\{\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{b}}|}\right\} \frac{\overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{b}}|}=\left\{\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{b}}|^{2}}\right\} \overrightarrow{\mathrm{b}} \end{aligned}

Correct answer option A.

 

Posted by

infoexpert22

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