+0

# dumb precalculus

0
162
1

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
+9461
+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+0 Answers

#1
+9461
+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