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

Dec 17, 2017

$u = \left[18 , 36 , 2\right]$

Explanation:

Find $u = 6 w + 2 z$ where $w = \left[2 , 6 , - 1\right]$ and $z = \left[3 , 0 , 4\right]$.

This is essentially a substitution problem and you need to remember that when adding vectors you add corresponding components. So $\left[{a}_{1} , {a}_{2} , {a}_{3}\right] + \left[{b}_{1} , {b}_{2} , {b}_{3}\right] = \left[{a}_{1} + {b}_{1} , {a}_{2} + {b}_{2} , {a}_{3} + {b}_{3}\right]$

$6 w = \left[6 \left(2\right) , 6 \left(6\right) , 6 \left(- 1\right)\right] = \left[12 , 36 , - 6\right]$.
$2 z = \left[2 \left(3\right) , 2 \left(0\right) , 2 \left(4\right)\right] = \left[6 , 0 , 8\right]$.

$u = 6 w + 2 z$
$u = \left[12 , 36 , - 6\right] + \left[6 , 0 , 8\right] = \left[18 , 36 , 2\right]$

Dec 17, 2017

$\boldsymbol{\underline{u}} = \left\langle18 , 36 , 2\right\rangle$

Explanation:

We have:

$\boldsymbol{\underline{v}} = \left\langle4 , - 3 , 5\right\rangle$
$\boldsymbol{\underline{w}} = \left\langle2 , 6 , - 1\right\rangle$
$\boldsymbol{\underline{z}} = \left\langle3 , 0 , 4\right\rangle$

Then, to compute $\boldsymbol{\underline{u}}$, we just scale the individual vectors and add the individual components, thus:

$\boldsymbol{\underline{u}} = 6 \boldsymbol{\underline{w}} + 2 \boldsymbol{\underline{z}}$
$\setminus \setminus \setminus = 6 \left\langle2 , 6 , - 1\right\rangle + 2 \left\langle3 , 0 , 4\right\rangle$
$\setminus \setminus \setminus = \left\langle6 \cdot 2 , 6 \cdot 6 , 6 \cdot \left(- 1\right)\right\rangle + \left\langle2 \cdot 3 , 2 \cdot 0 , 2 \cdot 4\right\rangle$
$\setminus \setminus \setminus = \left\langle12 , 36 , - 6\right\rangle + \left\langle6 , 0 , 8\right\rangle$
$\setminus \setminus \setminus = \left\langle12 + 6 , 36 + 0 , - 6 + 8\right\rangle$
$\setminus \setminus \setminus = \left\langle18 , 36 , 2\right\rangle$

Note that $\boldsymbol{\underline{v}}$ is superfluous to the question.