JEE MAIN - Physics (2022 - 26th July Evening Shift - No. 19)
If $$\vec{A}=(2 \hat{i}+3 \hat{j}-\hat{k})\, \mathrm{m}$$ and $$\vec{B}=(\hat{i}+2 \hat{j}+2 \hat{k}) \,\mathrm{m}$$. The magnitude of component of vector $$\vec{A}$$ along vector $$\vec{B}$$ will be ____________ $$\mathrm{m}$$.
Answer
2
Explanation
$$
\begin{aligned}
& \vec{A} \cdot \vec{B}=(2 \hat{i}+3 \hat{j}-\hat{k}) \cdot(\hat{i}+2 \hat{j}+2 \hat{k}) \\\\
& \Rightarrow \vec{A} \cdot \vec{B}=2 \times 1+3 \times 2-1 \times 2 \\\\
& \Rightarrow \vec{A} \cdot \vec{B}=6
\end{aligned}
$$
And $$ |\vec{B}|=\sqrt{1^2+2^2+2^2}=\sqrt{9}=3 $$
Magnitude of component of
$\vec{A}$ along $\vec{B}=\frac{\vec{A} \cdot \vec{B}}{|B|}=\frac{6}{3}=2$
And $$ |\vec{B}|=\sqrt{1^2+2^2+2^2}=\sqrt{9}=3 $$
Magnitude of component of
$\vec{A}$ along $\vec{B}=\frac{\vec{A} \cdot \vec{B}}{|B|}=\frac{6}{3}=2$
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