(C) The vector projection of $\vec{b}$ on $\vec{a}$ is given by the formula: $\left(\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}\right) \vec{a}$.First,ca

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$\frac{3(3 \hat{i}+2 \hat{j}+5 \hat{k})}{19}$

(C) The vector projection of $\vec{b}$ on $\vec{a}$ is given by the formula: $\left(\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}\right) \vec{a}$.First,calculate the dot product $\vec{b} \cdot \vec{a} = (7)(3) + (-5)(2) + (-1)(5) = 21 - 10 - 5 = 6$.Next,calculate the square of the magnitude of $\vec{a}$: $|\vec{a}|^2 = 3^2 + 2^2 + 5^2 = 9 + 4 + 25 = 38$.Now,substitute these values into the formula:Vector projection = $\frac{6}{38} (3 \hat{i} + 2 \hat{j} + 5 \hat{k}) = \frac{3}{19} (3 \hat{i} + 2 \hat{j} + 5 \hat{k})$.

Class 12 Mathematics Vector Algebra Scalar or Dot product of two vectors and its applications

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The vector projection of $\vec{b}$ on $\vec{a}$ where $\vec{a}=3 \hat{i}+2 \hat{j}+5 \hat{k}$ and $\vec{b}=7 \hat{i}-5 \hat{j}-\hat{k}$ is:

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A
$\frac{6(3 \hat{i}+2 \hat{j}+5 \hat{k})}{\sqrt{38}}$
B
$\frac{3(3 \hat{i}+2 \hat{j}+5 \hat{k})}{38}$
C
$\frac{3(3 \hat{i}+2 \hat{j}+5 \hat{k})}{19}$
D
$\frac{3(3 \hat{i}+2 \hat{j}+5 \hat{k})}{\sqrt{38}}$

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The vector projection of $\vec{b}$ on $\vec{a}$ where $\vec{a}=3 \hat{i}+2 \hat{j}+5 \hat{k}$ and $\vec{b}=7 \hat{i}-5 \hat{j}-\hat{k}$ is:

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If $\vec{a} = \hat{i} - 2\hat{j} + 2\hat{k}$ and $\vec{b} = 2\hat{i} - 3\hat{j} + \hat{k}$, then the component of $\vec{b}$ perpendicular to $\vec{a}$ is

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