# How do you find u=1/2v-w+2z given v=<4,-3,5>, w=<2,6,-1> and z=<3,0,4>?

May 19, 2017

$u = < 6 , - \frac{15}{2} , \frac{23}{2} >$

#### Explanation:

As $v = < 4 , - 3 , 5 > , w = < 2 , 6 , - 1 >$ and $z = < 3 , 0 , 4 >$

$u = \frac{1}{2} v - w + 2 z$

$= < \left(\frac{1}{2} \times 4 - 2 + 2 \times 3\right) , \left(\frac{1}{2} \left(- 3\right) - 6 + 2 \times 0\right) , \left(\frac{1}{2} \times 5 - \left(- 1\right) + 2 \times 4\right) >$

$= < 6 , - \frac{15}{2} , \frac{23}{2} >$

May 19, 2017

$< 6 , - \frac{15}{2} , \frac{23}{2} >$

#### Explanation:

$\text{Notation} < x , y , z > \equiv \left(\begin{matrix}x \\ y \\ z\end{matrix}\right)$

To multiply a vector by a scalar quantity, multiply each component of the vector by the scalar.

$\textcolor{red}{a} \left(\begin{matrix}x \\ y \\ z\end{matrix}\right) = \left(\begin{matrix}\textcolor{red}{a} x \\ \textcolor{red}{a} y \\ \textcolor{red}{a} z\end{matrix}\right)$

$\left(\begin{matrix}{x}_{1} \\ {y}_{1} \\ {z}_{1}\end{matrix}\right) \pm \left(\begin{matrix}{x}_{2} \\ {y}_{2} \\ {z}_{2}\end{matrix}\right) = \left(\begin{matrix}{x}_{1} \pm {x}_{2} \\ {y}_{1} \pm {y}_{2} \\ {z}_{1} \pm {z}_{2}\end{matrix}\right)$

$\Rightarrow \frac{1}{2} \underline{v} - \underline{w} + 2 \underline{z}$

$= \frac{1}{2} \left(\begin{matrix}4 \\ - 3 \\ 5\end{matrix}\right) - \left(\begin{matrix}2 \\ 6 \\ - 1\end{matrix}\right) + 2 \left(\begin{matrix}3 \\ 0 \\ 4\end{matrix}\right)$

$= \left(\begin{matrix}2 \\ - \frac{3}{2} \\ \frac{5}{2}\end{matrix}\right) - \left(\begin{matrix}2 \\ 6 \\ - 1\end{matrix}\right) + \left(\begin{matrix}6 \\ 0 \\ 8\end{matrix}\right)$

$= \left(\begin{matrix}6 \\ - \frac{15}{2} \\ \frac{23}{2}\end{matrix}\right)$