# How do you find the scalar and vector projections of b onto a given a = i + j + k, b = i - j + k?

Apr 29, 2016

Saclar projection$\frac{1}{\sqrt{3}}$ and Vector projection $\frac{1}{3} \left(\hat{i} + \hat{j} + \hat{k}\right)$

#### Explanation:

We have been given two vectors $\vec{a} \mathmr{and} \vec{b}$ , we are to find out the scalar and vector projection of $\vec{b}$ onto $\vec{a}$
we have $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$
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The scalar projection of $\vec{b}$ onto $\vec{a}$ means the magnitude of resolved component of $\vec{b}$ inthe direction of $\vec{a}$ and is given by
The scalar projection of $\vec{b}$ onto $\vec{a}$ $= \frac{\vec{b} \cdot \vec{a}}{|} \vec{a} |$
$= \frac{\left(\hat{i} + \hat{j} + \hat{k}\right) \cdot \left(\hat{i} - \hat{j} + \hat{k}\right)}{\sqrt{{1}^{2} + {1}^{1} + {1}^{2}}}$
$= \frac{{1}^{2} - {1}^{2} + {1}^{2}}{\sqrt{3}} = \frac{1}{\sqrt{3}}$

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The Vector projection of $\vec{b}$ onto $\vec{a}$ means the resolved component of $\vec{b}$ in the direction of $\vec{a}$ and is given by
The vector projection of $\vec{b}$ onto $\vec{a}$ $= \frac{\vec{b} \cdot \vec{a}}{|} \vec{a} {|}^{2.} \left(\hat{i} + \hat{j} + \hat{k}\right)$
$= \frac{\left(\hat{i} + \hat{j} + \hat{k}\right) \cdot \left(\hat{i} - \hat{j} + \hat{k}\right)}{{\left(\sqrt{{1}^{2} + {1}^{1} + {1}^{2}}\right)}^{2}} . \left(\hat{i} + \hat{j} + \hat{k}\right)$
$= \frac{{1}^{2} - {1}^{2} + {1}^{2}}{3.} \left(\hat{i} + \hat{j} + \hat{k}\right) = \frac{1}{3} \left(\hat{i} + \hat{j} + \hat{k}\right)$