The component of vec{i} + vec{j} along vec{j} | vedclass

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Question

(B) Let $\vec{a} = \vec{i} + \vec{j}$ and $\vec{b} = \vec{j} + \vec{k}$.The component of vector $\vec{a}$ along vector $\vec{b}$ is given by the formu

Vedclass · Accepted Answer

$\frac{\vec{j} + \vec{k}}{2}$

Vedclass · Answer

(B) Let $\vec{a} = \vec{i} + \vec{j}$ and $\vec{b} = \vec{j} + \vec{k}$.The component of vector $\vec{a}$ along vector $\vec{b}$ is given by the formula: $\left( \vec{a} \cdot \hat{b} \right) \hat{b}$,where $\hat{b} = \frac{\vec{b}}{|\vec{b}|}$.First,calculate the magnitude of $\vec{b}$: $|\vec{b}| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2}$.So,$\hat{b} = \frac{\vec{j} + \vec{k}}{\sqrt{2}}$.Now,calculate the dot product $\vec{a} \cdot \hat{b} = (\vec{i} + \vec{j}) \cdot \left( \frac{\vec{j} + \vec{k}}{\sqrt{2}} \right) = \frac{1}{\sqrt{2}} (\vec{i} \cdot \vec{j} + \vec{i} \cdot \vec{k} + \vec{j} \cdot \vec{j} + \vec{j} \cdot \vec{k}) = \frac{1}{\sqrt{2}} (0 + 0 + 1 + 0) = \frac{1}{\sqrt{2}}$.Finally,the component is $\left( \frac{1}{\sqrt{2}} \right) \left( \frac{\vec{j} + \vec{k}}{\sqrt{2}} \right) = \frac{\vec{j} + \vec{k}}{2}$.

The component of $\vec{i} + \vec{j}$ along $\vec{j} + \vec{k}$ is:

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