Respuesta :
[tex]\text { Saclar projection } \frac{1}{\sqrt{3}} \text { and Vector projection } \frac{1}{3}(\hat{i}+\hat{j}+\hat{k})[/tex]
We have been given two vectors [tex]$\vec{a}$[/tex] and [tex]$\vec{b}$[/tex], we are to find out the scalar and vector projection of [tex]$\vec{b}$[/tex] onto [tex]$\vec{a}$[/tex]
we have [tex]$\vec{a}=\hat{i}+\hat{j}+\hat{k}$[/tex] and [tex]$\vec{b}=\hat{i}-\hat{j}+\hat{k}$[/tex]
The scalar projection of[tex]$\vec{b}$[/tex]onto [tex]$\vec{a}$[/tex]means the magnitude of the resolved component of [tex]$\vec{b}$[/tex] the direction of [tex]$\vec{a}$[/tex] and is given by
The scalar projection of [tex]$\vec{b}$[/tex]onto
[tex]$\vec{a}=\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|}$[/tex]
[tex]$\begin{aligned}&=\frac{(\hat{i}+\hat{j}+\hat{k}) \cdot(\hat{i}-\hat{j}+\hat{k})}{\sqrt{1^2+1^1+1^2}} \\&=\frac{1^2-1^2+1^2}{\sqrt{3}}=\frac{1}{\sqrt{3}}\end{aligned}$[/tex]
The Vector projection of [tex]$\vec{b}$[/tex] onto [tex]$\vec{a}$[/tex] means the resolved component of [tex]$\vec{b}$[/tex] in the direction of [tex]$\vec{a}$[/tex] and is given by
The vector projection of [tex]$\vec{b}$[/tex] onto
[tex]$\vec{a}=\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} \cdot(\hat{i}+\hat{j}+\hat{k})$[/tex]
[tex]$\begin{aligned}&=\frac{(\hat{i}+\hat{j}+\hat{k}) \cdot(\hat{i}-\hat{j}+\hat{k})}{\left(\sqrt{1^2+1^1+1^2}\right)^2} \cdot(\hat{i}+\hat{j}+\hat{k}) \\&=\frac{1^2-1^2+1^2}{3} \cdot(\hat{i}+\hat{j}+\hat{k})=\frac{1}{3}(\hat{i}+\hat{j}+\hat{k})\end{aligned}$[/tex]
To learn more about scalar and vector projection visit:https://brainly.com/question/21925479
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