Chapter 11, Problem 7CT

(a)

To determine

To find: a unit vector in the direction of $u$ .

Answer to Problem 7CT

  $\widehat{u}=\langle -\frac{1}{\sqrt{19}},\frac{3}{\sqrt{19}},-\frac{3}{\sqrt{19}}\rangle$

Explanation of Solution

Given points : $u$ and $v$ are the vectors from A(8,-2,5) to B(6,4,-1) and from A to C(-4,3,0) respectively.

Calculation:

We know that if $\left(x_1,y_1,z_1\right)$ and $\left(x_2,y_2,z_2\right)$ are initial and terminal points of a vector $a$ respectively, then component form of vector

  $a$ is

  $a=\langle x_2-x_1,y_2-y_1,z_2-z_1\rangle$

Therefore,

  $\begin{array}{l}u=\langle 6-8,4+2,-1-5\rangle \\ u=\langle -2,6,-6\rangle \end{array}$

Now, unit vector in the direction of $u$ is

  $\begin{array}{l}\widehat{u}=\frac{u}{\left|u\right|}=\frac{\langle -2,6,-6\rangle }{\sqrt{{\left(-2\right)}^2+{\left(6\right)}^2+{\left(-6\right)}^2}}=\frac{\langle -2,6,-6\rangle }{2\sqrt{19}}\\ \widehat{u}=\langle -\frac{1}{\sqrt{19}},\frac{3}{\sqrt{19}},-\frac{3}{\sqrt{19}}\rangle \end{array}$

(b)

To determine

To find: a unit vector in the direction of $v$ .

Answer to Problem 7CT

  $\widehat{v}=\frac{1}{\sqrt{197}}\langle -12,5,-5\rangle$

Explanation of Solution

Given points : $u$ and $v$ are the vectors from A(8,-2,5) to B(6,4,-1) and from A to C(-4,3,0) respectively.

Calculation:

We know that if $\left(x_1,y_1,z_1\right)$ and $\left(x_2,y_2,z_2\right)$ are initial and terminal points of a vector $a$ respectively, then component form of vector

  $a$ is

  $a=\langle x_2-x_1,y_2-y_1,z_2-z_1\rangle$

Therefore,

  $\begin{array}{l}v=\langle -4-8,3+2,0-5\rangle \\ \Rightarrow v=\langle -12,5,-5\rangle \end{array}$

Now, unit vector in the direction of $v$ is

  $\begin{array}{l}\widehat{v}=\frac{v}{\left|v\right|}=\frac{\langle -12,5,-5\rangle }{\sqrt{{\left(-12\right)}^2+{\left(5\right)}^2+{\left(-5\right)}^2}}=\frac{\langle -12,5,-5\rangle }{\sqrt{194}}\\ \widehat{v}=\frac{1}{\sqrt{197}}\langle -12,5,-5\rangle \end{array}$