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For \($\bold{v} = \begin{pmatrix} \phantom -2 \\ \phantom -3 \\ -1 \end{pmatrix}$ and $\bold{w} = \begin{pmatrix} \phantom -2 \\ -1 \\ \phantom -0 \end{pmatrix}$ \) compute the projection of $\mathbf v$ onto $\mathbf w.$

i literally dont know how to solve this

 Apr 18, 2022

Best Answer 

 #1
avatar+9369 
+3

We have the formula:

\(\operatorname{Proj}_{\mathbf{w}} (\mathbf v) = \dfrac{\langle\mathbf v, \mathbf w\rangle}{\|\mathbf w\|^2}\mathbf{w}\)

 

Now, we compute \(\langle\mathbf v, \mathbf w\rangle\) and \(\|\mathbf w\|^2\).

 

\(\begin{array}{cl} &\langle\mathbf v, \mathbf w\rangle\\ =& 2(2) + 3(-1) + (-1)(0)\\ =& 1\\ &\|\mathbf w\|^2\\ =& 2^2 + (-1)^2 + 0^2\\ =&5 \end{array}\)

 

Therefore, substituting back into the formula, we have

 

\(\operatorname{Proj}_{\mathbf{w}} (\mathbf v) = \dfrac15\mathbf{w} = \begin{pmatrix}\frac25\\-\frac15\\0\end{pmatrix}\)

 Apr 18, 2022
 #1
avatar+9369 
+3
Best Answer

We have the formula:

\(\operatorname{Proj}_{\mathbf{w}} (\mathbf v) = \dfrac{\langle\mathbf v, \mathbf w\rangle}{\|\mathbf w\|^2}\mathbf{w}\)

 

Now, we compute \(\langle\mathbf v, \mathbf w\rangle\) and \(\|\mathbf w\|^2\).

 

\(\begin{array}{cl} &\langle\mathbf v, \mathbf w\rangle\\ =& 2(2) + 3(-1) + (-1)(0)\\ =& 1\\ &\|\mathbf w\|^2\\ =& 2^2 + (-1)^2 + 0^2\\ =&5 \end{array}\)

 

Therefore, substituting back into the formula, we have

 

\(\operatorname{Proj}_{\mathbf{w}} (\mathbf v) = \dfrac15\mathbf{w} = \begin{pmatrix}\frac25\\-\frac15\\0\end{pmatrix}\)

MaxWong Apr 18, 2022

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