Chapter 11, Problem 1RE

Writing Vectors in Different Forms In Exercises 1 and 2, let $u=\stackrel{\rightharpoonup }{PQ}$ and $v=\stackrel{\rightharpoonup }{PR}$ and (a) write u and v in component form, (b) write u and v as the linear combination of the standard unit vectors i and j, (c) find the magnitudes of u and v, and (d) find $-3u+v$.

$P=(1,2),Q=(4,1),R=(5,4)$

To determine

To calculate: For given $P=\left(1,2\right),Q=\left(4,1\right),R=\left(5,4\right)$, $u=\overrightarrow{PQ},v=\overrightarrow{PR}$ Find acomponent form of u and v.

Answer to Problem 1RE

Solution:

For given $P=\left(1,2\right),Q=\left(4,1\right),R=\left(5,4\right)$, $u=\overrightarrow{PQ},v=\overrightarrow{PR}$

Find a component form of u and vis $\langle 3,-1\rangle$ and $\langle 4,2\rangle$ respectively.

Explanation of Solution

Given:

$P=\left(1,2\right),Q=\left(4,1\right),R=\left(5,4\right)$, $u=\overrightarrow{PQ},v=\overrightarrow{PR}$

Formula used:

If $P\left(p_1,p_2\right)$ and $Q\left(q_1,q_2\right)$ are the initial and terminal points of a directed line segment, then the component form of the vector u represented by $\overrightarrow{PQ}$ is

$\langle u_1,u_2\rangle =\langle q_1-p_1,q_2-p_2\rangle$

Calculation:

If

$\begin{array}{c}{u}_2=q_2-p_2\\ =1-2\\ =-1\end{array}$

And

u $=\overrightarrow{PQ}$ and $P=\left(1,2\right),Q=\left(4,1\right)$, then:

$\begin{array}{c}{u}_1=q_1-p_1\\ =4-1\\ =3\end{array}$

The component form of u is $\langle 3,-1\rangle$.

If v $=\overrightarrow{PR}$ and $P=\left(1,2\right),R=\left(5,4\right)$, then:

$\begin{array}{c}{v}_2=r_2-p_2\\ =4-2\\ =2\end{array}$

And

$\begin{array}{c}{v}_1=r_1-p_1\\ =5-1\\ =4\end{array}$

The component form of v is $\langle 4,2\rangle$.

To determine

To calculate: For given $P=\left(1,2\right),Q=\left(4,1\right),R=\left(5,4\right)$, $u=\overrightarrow{PQ},v=\overrightarrow{PR}$, find u and vas the linear combination of the standard unit vectors i and j.

Answer to Problem 1RE

Solution: For given $P=\left(1,2\right),Q=\left(4,1\right),R=\left(5,4\right)$, $u=\overrightarrow{PQ},v=\overrightarrow{PR}$, u and vas the linear combination of the standard unit vectors i and j are the vectors $u=3\text{i}-\text{j}$ and $v=4\text{i}+2\text{j}$$$

Explanation of Solution

Given:

$P=\left(1,2\right),Q=\left(4,1\right),R=\left(5,4\right)$, $u=\overrightarrow{PQ},v=\overrightarrow{PR}$

Formula used:

If $\langle v_1,v_2\rangle$ is the component form of v, then the vector $v=v_1\text{i}+v_2\text{j}$ is called a linear combination of i and j.

Calculation:

The vectors $u=3\text{i}-\text{j}$ and $v=4\text{i}+2\text{j}$ are the linear combinations of i and j.

As according to the calculation of part (a), a component form of u and vis $\langle 3,-1\rangle$ and $\langle 4,2\rangle$.

To determine

To calculate: Forgiven $P=\left(1,2\right),Q=\left(4,1\right),R=\left(5,4\right)$, $u=\overrightarrow{PQ},v=\overrightarrow{PR}$. Find magnitudes of u and v.

Answer to Problem 1RE

Solution:

Forgiven $P=\left(1,2\right),Q=\left(4,1\right),R=\left(5,4\right)$, $u=\overrightarrow{PQ},v=\overrightarrow{PR}$

Magnitudes of u and vu is $\Vert u\Vert =\sqrt{10}$ and magnitude of v is $\Vert v\Vert =2\sqrt{5}$.

Explanation of Solution

Given:

$P=\left(1,2\right),Q=\left(4,1\right),R=\left(5,4\right)$, $u=\overrightarrow{PQ},v=\overrightarrow{PR}$

Formula used:

According to the Distance Formula, the length (or magnitude) of vectoru is:

$\begin{array}{c}\Vert u\Vert =\sqrt{{\left(q_1-p_1\right)}^2+{\left(q_2-p_2\right)}^2}\\ =\sqrt{u_1{}^2+u_2{}^2}\end{array}$

Calculation:

As per part (a),

v=$\langle 4,2\rangle$. The length of vector v is:

$\begin{array}{c}\Vert v\Vert =\sqrt{v_1{}^2+v_2{}^2}\\ =\sqrt{4^2+2^2}\\ =\sqrt{16+4}\\ =\sqrt{20}\end{array}$

$\text{u}=\langle 3,-1\rangle$. The length of vector u is:

$\begin{array}{c}\Vert u\Vert =\sqrt{u_1{}^2+u_2{}^2}\\ =\sqrt{3^2+{\left(-1\right)}^2}\\ =\sqrt{9+1}\\ =\sqrt{10}\end{array}$

and

Therefore, the magnitude of u is $\Vert u\Vert =\sqrt{10}$ and magnitude of v is $\Vert v\Vert =2\sqrt{5}$.

To determine

To calculate: For given $P=\left(1,2\right),Q=\left(4,1\right),R=\left(5,4\right)$, $u=\overrightarrow{PQ},v=\overrightarrow{PR}$. Find the value of $\text{-3u}+\text{v}$.

Answer to Problem 1RE

Solution:

For given $P=\left(1,2\right),Q=\left(4,1\right),R=\left(5,4\right)$, $u=\overrightarrow{PQ},v=\overrightarrow{PR}$. The value of $\text{-3u}+\text{v}$ is $\langle -5,5\rangle$.

Explanation of Solution

Given:

$P=\left(1,2\right),Q=\left(4,1\right),R=\left(5,4\right)$, $u=\overrightarrow{PQ},v=\overrightarrow{PR}$

Formula used:

The vector sum of u and v is the vector:

$\text{u}+\text{v}=\langle u_1+v_1,u_2+v_2\rangle$

The scalar multiple of c and u is the vector:

$cu=\langle c{u}_1,c{u}_2\rangle$

Calculation:

Value of $\text{-3u}+\text{v}$ using scalar multiplication and vector sum formulae:

As according to the calculation of part (a), a component form of u and v is $\langle 3,-1\rangle$ and $\langle 4,2\rangle$.

$\begin{array}{c}\text{-3u}+\text{v}=-3\langle 3,-1\rangle +\langle 4,2\rangle \\ =\langle -9,3\rangle +\langle 4,2\rangle \\ =\langle -5,5\rangle \end{array}$