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
+9674 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
\(\operatorname{Proj}_{\mathbf{w}} (\mathbf v) = \dfrac15\mathbf{w} = \begin{pmatrix}\frac25\\-\frac15\\0\end{pmatrix}\)
+9674 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
\(\operatorname{Proj}_{\mathbf{w}} (\mathbf v) = \dfrac15\mathbf{w} = \begin{pmatrix}\frac25\\-\frac15\\0\end{pmatrix}\)
