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3-1.Vectors
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If $\overrightarrow{ A }=(2 \hat{ i }+3 \hat{ j }-\hat{ k }) \;m$ and $\overrightarrow{ B }=(\hat{ i }+2 \hat{ j }+2 \hat{ k })\; m$. The magnitude of component of vector $\overrightarrow{ A }$ along vector $\vec{B}$ will be $......m$.
A
$2$
B
$1$
C
$3$
D
$4$
(JEE MAIN-2022)
Solution
$\overrightarrow{ A }=(2 \hat{ i }+3 \hat{ j }-\hat{ k }) m \text { and } \overrightarrow{ B }=(\hat{ i }+2 \hat{ j }+2 \hat{ k }) m$
Component of $\overrightarrow{ A }$ along $\overrightarrow{ B }=\overrightarrow{ A } \cdot \hat{ B }$
$=\frac{\overrightarrow{ A } \overrightarrow{ B }}{|\overrightarrow{ B }|}$ $=\frac{2+6-2}{\sqrt{1^{2}+2^{2}+2^{2}}}$
$=\frac{6}{3}=2$
Standard 11
Physics
