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$\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ are three vectors such that $|\mathbf{a}|=1,|\mathbf{b}|=2,|\mathbf{c}|=3 \quad$ and $\mathbf{b}, \mathbf{c} \quad$ are perpendicular. If projection of $\mathbf{b}$ on $\mathbf{a}$ is the same as the projection of $\mathbf{c}$ on $\mathbf{a}$, then $|\mathbf{a}-\mathbf{b}+\mathbf{c}|$ is equal to
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Verified Answer
The correct answer is:
$\sqrt{14}$
Given that,
$|\mathbf{a}|=1,|\mathbf{b}|=2,|\mathbf{c}|=3$
$\because b$ and $c$ are perpendicular.
$\therefore \mathrm{b} \cdot \mathrm{c}=0$
And projection of $b$ on $a=\frac{a \cdot b}{|\mathbf{a}|}$
Projection of $\mathbf{c}$ on $\mathbf{a}=\frac{\mathbf{a} \cdot \mathbf{c}}{|\mathbf{a}|}$
$\because$ Both projections are same.
$\begin{aligned}
& \therefore \quad \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|}=\frac{\mathbf{a} \cdot \mathbf{c}}{|\mathbf{a}|} \\
& \Rightarrow \quad \mathbf{a} \cdot \mathbf{b}=\mathbf{a} \cdot \mathbf{c} \\
& \text {Then, }|\mathbf{a}-\mathbf{b}+\mathbf{c}|^2=|\mathbf{a}|^2+|\mathbf{b}|^2+|\mathbf{c}|^2-2 \mathbf{a} \cdot \mathbf{b} \\
& =(1)^2+(2)^2+(3)^2-2 \mathbf{a} \cdot \mathbf{b}-0+2 \mathbf{a} \cdot \mathbf{c} \\
& =1+4+9-2 \mathbf{a} \cdot \mathbf{b}+2 \mathbf{a} \cdot \mathbf{b}=14
\end{aligned}$
[from Eqs. (i) and (ii)]
$\therefore \quad|\mathbf{a}-\mathbf{b}+\mathbf{c}|=\sqrt{14}$
$|\mathbf{a}|=1,|\mathbf{b}|=2,|\mathbf{c}|=3$
$\because b$ and $c$ are perpendicular.
$\therefore \mathrm{b} \cdot \mathrm{c}=0$
And projection of $b$ on $a=\frac{a \cdot b}{|\mathbf{a}|}$
Projection of $\mathbf{c}$ on $\mathbf{a}=\frac{\mathbf{a} \cdot \mathbf{c}}{|\mathbf{a}|}$
$\because$ Both projections are same.
$\begin{aligned}
& \therefore \quad \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|}=\frac{\mathbf{a} \cdot \mathbf{c}}{|\mathbf{a}|} \\
& \Rightarrow \quad \mathbf{a} \cdot \mathbf{b}=\mathbf{a} \cdot \mathbf{c} \\
& \text {Then, }|\mathbf{a}-\mathbf{b}+\mathbf{c}|^2=|\mathbf{a}|^2+|\mathbf{b}|^2+|\mathbf{c}|^2-2 \mathbf{a} \cdot \mathbf{b} \\
& =(1)^2+(2)^2+(3)^2-2 \mathbf{a} \cdot \mathbf{b}-0+2 \mathbf{a} \cdot \mathbf{c} \\
& =1+4+9-2 \mathbf{a} \cdot \mathbf{b}+2 \mathbf{a} \cdot \mathbf{b}=14
\end{aligned}$
[from Eqs. (i) and (ii)]
$\therefore \quad|\mathbf{a}-\mathbf{b}+\mathbf{c}|=\sqrt{14}$
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