$\vec{a} = 2 \hat{i} + \hat{j} - 2 \hat{k}$ ভেক্টর বরাবর $\vec{b} = 5 \hat{i} - 3 \hat{j} + 2 \hat{k}$ ভেক্টরের উপাংশের সাংখ্যিক মান কত?

উপাংশ $=\frac{\bar{a} \cdot \bar{b}}{a}=\frac{10-3-4}{\sqrt{2^{2}+1^{2}+2^{2}}}=\frac{3}{3}=1$

To find the scalar component (projection) of the vector $\vec{b}$ along vector $\vec{a}$, we use the formula:

$\text{Projection of } \vec{b} \text{ on } \vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|}$

First, calculate the dot product $\vec{a} \cdot \vec{b}$:

$\vec{a} \cdot \vec{b} = (2 \hat{i} + \hat{j} - 2 \hat{k}) \cdot (5 \hat{i} - 3 \hat{j} + 2 \hat{k})$

$= (2)(5) + (1)(-3) + (-2)(2) = 10 - 3 - 4 = 3$

Next, find the magnitude of $\vec{a}$:

$|\vec{a}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$

Now calculate the projection:

$\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|} = \frac{3}{3} = 1$

The scalar component of $\vec{b}$ along $\vec{a}$ is 1.

Correct Answer: (C) 1