Chapter 9, Problem 2T

(a)

To determine

The vector $u-3v$ for the vectors $u=\langle 1,3\rangle$ and $v=\langle -6,2\rangle$.

Answer to Problem 2T

The vector $u-3v$ is $\langle 19,-3\rangle$.

Explanation of Solution

Given:

The two vectors are $u=\langle 1,3\rangle$ and $v=\langle -6,2\rangle$.

Formula used: The scalar multiplication of a vector $u=\langle a_1,a_2\rangle$ with the scalar $c$ is given below,

$cu=\langle c{a}_1,c{a}_2\rangle c\in R$ (1)

The subtraction of the vectors $u=\langle a_1,a_2\rangle$ and $v=\langle b_1,b_2\rangle$ is given below,

$u-v=\langle a_1-b_1,a_2-b_2\rangle$ (2)

Calculation:

Substitute $\langle 1,3\rangle$ for u and $\langle -6,2\rangle$ for v in equation $u-3v$, to find the vector $u-3v$,

$\begin{array}{c}u-3v=\langle 1,3\rangle -3\langle -6,2\rangle \\ =\langle 1,3\rangle -\langle -18,6\rangle \left(\because \text{From}\text{equation}\left(1\right)\right)\\ =\langle 1-\left(-18\right),3-6\rangle \left(\because \text{From}\text{equation}\left(\text{2}\right)\right)\\ =\langle 19,-3\rangle \end{array}$

Thus, the vector $u-3v$ is $\langle 19,-3\rangle$.

(b)

To determine

The value of $\left|u+v\right|$.

Answer to Problem 2T

The value of $\left|u+v\right|$ is $5\sqrt{2}$.

Explanation of Solution

Given:

The two vectors are $u=\langle 1,3\rangle$ and $v=\langle -6,2\rangle$.

Formula used: The addition of the vectors $u=\langle a_1,a_2\rangle$ and $v=\langle b_1,b_2\rangle$ is given below,

$u+v=\langle a_1+b_1,a_2+b_2\rangle$ (3)

The magnitude of a vector $v=\langle a_1,a_2\rangle$ is given below,

$\left|v\right|=\sqrt{a_1^2+a_2^2}$ (4)

Calculation:

Substitute 1 for $a_1$, 3 for $a_2$, $-6$ for $b_1$ and 2 for $b_2$ in equation (4), to find the vector $u+v$,

$\begin{array}{c}u+v=\langle 1+\left(-6\right),3+2\rangle \\ =\langle 1-6,5\rangle \\ =\langle -5,5\rangle \end{array}$

The vector $u+v$ is $\langle -5,5\rangle$.

Substitute $-5$ for $a_1$ and 5 for $a_2$ in equation (4), to find the value of $\left|u+v\right|$,

$\begin{array}{c}\left|u+v\right|=\sqrt{{\left(-5\right)}^2+{\left(5\right)}^2}\\ =\sqrt{25+25}\\ =\sqrt{50}\\ =5\sqrt{2}\end{array}$

Thus, the value of $\left|u+v\right|$ is $5\sqrt{2}$.

(c)

To determine

The value of $u\cdot v$.

Answer to Problem 2T

The value of $u\cdot v$ is 0.

Explanation of Solution

Given:

The two vectors are $u=\langle 1,3\rangle$ and $v=\langle -6,2\rangle$.

Formula used: The Dot Product of the vectors $u=\langle a_1,a_2\rangle$ and the vector $v=\langle b_1,b_2\rangle$ is given below,

$u\cdot v=a_1{b}_1+a_2{b}_2$ (5)

Calculation:

Substitute 1 for $a_1$, 3 for $a_2$, $-6$ for $b_1$ and 2 for $b_2$ in equation (5), to find the value of $u\cdot v$,

$\begin{array}{c}u\cdot v=\left(1\right)\left(-6\right)+\left(3\right)\left(2\right)\\ =-6+6\\ =0\end{array}$

Thus, the value of $u\cdot v$ is 0.

(d)

To determine

Whether the vectors $u=\langle 1,3\rangle$ and $v=\langle -6,2\rangle$ are perpendicular to each other.

Answer to Problem 2T

The vectors $u=\langle 1,3\rangle$ and $v=\langle -6,2\rangle$ are perpendicular to each other.

Explanation of Solution

The two vectors are perpendicular if and only if the dot product of the vectors is 0, i.e., the vectors u and v are perpendicular if $u\cdot v=0$.

From part (c), the value of $u\cdot v$ is 0.

Thus, the vectors $u=\langle 1,3\rangle$ and $v=\langle -6,2\rangle$ are perpendicular to each other.