# Given A = i(x+2y) - j(y+3z) +k (3x-y), how do you determine a unit vector parallel to A at point P(1,-1,2)?

Jul 15, 2017

The unit vector is $= < - \frac{1}{\sqrt{42}} , - \frac{5}{\sqrt{42}} , \frac{4}{\sqrt{42}} >$

#### Explanation:

The vector is

$\vec{A} \left(x , y , z\right) = \vec{i} \left(x + 2 y\right) + \vec{j} \left(- y - 3 z\right) + \vec{k} \left(3 x - y\right)$

At the point , $P = \left(1 , - 1 , 2\right)$

$\vec{A} \left(1 , - 1 , 2\right) = \vec{i} \left(- 1\right) + \vec{j} \left(- 5\right) + \vec{k} \left(4\right)$

The unit vector in the direction of $\vec{A} \left(1 , - 1 , 2\right)$ is

$\frac{\vec{A} \left(1 , - 1 , 2\right)}{| | \vec{A} \left(1 , - 1 , 2\right) | |}$

The modulus of $\vec{A \left(1 , - 1 , 2\right)}$ is

$| | \vec{A} \left(1 , - 1 , 2\right) | | = | | \vec{i} \left(- 1\right) + \vec{j} \left(- 5\right) + \vec{k} \left(4\right) | |$

$= \sqrt{1 + 25 + 16}$

$= \sqrt{42}$

The unit vector is $= < - \frac{1}{\sqrt{42}} , - \frac{5}{\sqrt{42}} , \frac{4}{\sqrt{42}} >$