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QJEEM22Phe8c6: JEE Main 2022 Physics - Class 11 JEEM22Phe8c6 Solution: If and . The magnitude of component of vector along vector will …

If A⃗=(2i^+3j^−k^) m\vec{A}=(2 \hat{i}+3 \hat{j}-\hat{k})\, \mathrm{m} and B⃗=(i^+2j^+2k^) m\vec{B}=(\hat{i}+2 \hat{j}+2 \hat{k}) \,\mathrm{m} . The magnitude of component of vector A⃗\vec{A} along vector B⃗\vec{B} will be ____________ m\mathrm{m} .

Step-by-Step Verified Solution

amp;A⃗⋅B⃗=(2i^+3j^−k^)⋅(i^+2j^+2k^)amp;⇒A⃗⋅B⃗=2×1+3×2−1×2amp;⇒A⃗⋅B⃗=6\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 ∣B⃗∣=12+22+22=9=3|\vec{B}|=\sqrt{1^2+2^2+2^2}=\sqrt{9}=3

Magnitude of component of

A⃗\vec{A} along B⃗=A⃗⋅B⃗∣B∣=63=2\vec{B}=\frac{\vec{A} \cdot \vec{B}}{|B|}=\frac{6}{3}=2

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