# If  bb(ul u) = << −1,0,2 >> ,  bb(ul v) = << 3,1,2 >>  and  bb(ul w) = << 1,−2,−2 >>  then find?

## (a)  3bb(ul v)−2bb(ul u)  (b) $k \boldsymbol{\underline{u}} + \boldsymbol{\underline{v}} + k \boldsymbol{\underline{w}}$ (c) $| | \boldsymbol{\vec{P Q}} | |$ where $\boldsymbol{\vec{O P}} = - 3 \boldsymbol{\underline{u}}$ and $\boldsymbol{\vec{O Q}} = \boldsymbol{\underline{v}} + 5 \boldsymbol{\underline{w}}$ (d) Area of the parallelogram with sides $\boldsymbol{\underline{u}}$ and $2 \boldsymbol{\underline{v}}$

Aug 24, 2017

We have:

 bb(ul u) = << −1,0,2 >>
$\boldsymbol{\underline{v}} = \left\langle3 , 1 , 2\right\rangle$
 bb(ul w) = << 1,−2,−2 >>

Then:

Part (a):

 3bb(ul v)−2bb(ul u) = 3 << 3,1,2 >> - 2<< −1,0,2 >>
 " " = << 9,3,6 >> - << −2,0,4 >>
$\text{ } = \left\langle11 , 3 , 2\right\rangle$

Part (b):

 kbb(ul u) + bb(ul v) + kbb(ul w) = k<< −1,0,2 >> + << 3,1,2 >> + k<< 1,−2,−2 >>

 " " = << −k,0,2k >> + << 3,1,2 >> + << k,−2k,−2k >>
 " " = << −k+3+k,0+1-2k,2k+2-2k >>
$\text{ } = \left\langle3 , 1 - 2 k , 2\right\rangle$

Part (c):

Let:

$\boldsymbol{\vec{O P}} = - 3 \boldsymbol{\underline{u}}$
 " " = -3<< −1,0,2 >>
$\text{ } = \left\langle3 , 0 , - 6\right\rangle$

$\boldsymbol{\vec{O Q}} = \boldsymbol{\underline{v}} + 5 \boldsymbol{\underline{w}}$
 " " = << 3,1,2 >> + 5<< 1,−2,−2 >>
 " " = << 3,1,2 >> + << 5,−10,−10 >>
$\text{ } = \left\langle8 , - 9 , - 8\right\rangle$

Then:

$\boldsymbol{\vec{P Q}} = \boldsymbol{\vec{O Q}} - \boldsymbol{\vec{O P}}$
$\text{ } = \left\langle8 , - 9 , - 8\right\rangle - \left\langle3 , 0 , - 6\right\rangle$
$\text{ } = \left\langle5 , - 9 , - 2\right\rangle$

Finally:

$| | \boldsymbol{\vec{P Q}} | | = | | \left\langle5 , - 9 , - 2\right\rangle | |$
$\text{ } = \sqrt{{5}^{2} + {\left(- 9\right)}^{2} + {\left(- 2\right)}^{2}}$
$\text{ } = \sqrt{25 + 81 + 4}$
$\text{ } = \sqrt{110}$

Part (c):

The area is given by the magnitude of the cross product:

$A = | | \boldsymbol{\underline{u}} \times \boldsymbol{\underline{v}} | |$

we calculate the cross product using:

 bb(ul u) xx bb(ul v) = << −1,0,2 >> xx << 3,1,2 >>
 \ \ \ = | ( bb(ul(hat i)),bb(ul(hat j)),bb(ul(hat k))), (−1,0,2), (3,1,2) |

$\setminus \setminus \setminus = | \left(0 , 2\right) , \left(1 , 2\right) | \boldsymbol{\underline{\hat{i}}} - | \left(- 1 , 2\right) , \left(3 , 2\right) | \boldsymbol{\underline{\hat{j}}} + | \left(- 1 , 0\right) , \left(3 , 1\right) | \boldsymbol{\underline{\hat{k}}}$

$\setminus \setminus \setminus = \left(0 - 2\right) \boldsymbol{\underline{\hat{i}}} - \left(- 2 - 6\right) \boldsymbol{\underline{\hat{j}}} + \left(- 1 - 0\right) \boldsymbol{\underline{\hat{k}}}$

$\setminus \setminus \setminus = - 2 \boldsymbol{\underline{\hat{i}}} + 8 \boldsymbol{\underline{\hat{j}}} - \boldsymbol{\underline{\hat{k}}}$

Hence,

$A = | | \boldsymbol{\underline{u}} \times \boldsymbol{\underline{v}} | |$
$\setminus \setminus \setminus = \sqrt{{\left(- 2\right)}^{2} + {8}^{2} + {\left(- 1\right)}^{2}}$
$\setminus \setminus \setminus = \sqrt{4 + 64 + 1} | |$
$\setminus \setminus \setminus = \sqrt{69}$