Chapter 6, Problem 45CR

To determine

To find: The sum of the vector $u+v$ .

Answer to Problem 45CR

The sum of the vector $u+v$ is $\langle 4,4\rangle$ .

Explanation of Solution

Given information: $u=\langle 4,5\rangle \&v=\langle 0,-1\rangle$

Formula used: $u+v=\langle u_1+v_1,u_2+v_2\rangle$

Calculation: Since $u+v=\langle u_1+v_1,u_2+v_2\rangle$

Hence, put the value of u and v,

  $\begin{array}{l}u+v=\langle \left(4\right)+\left(0\right),\left(5\right)+\left(-1\right)\rangle \\ u+v=\langle 4,4\rangle \end{array}$

Therefore, $u+v=\langle 4,4\rangle$

To determine

To find: The difference of the vector $u-v$ .

Answer to Problem 45CR

The difference of the vector $u-v$ is $\langle 4,6\rangle$ .

Explanation of Solution

Given information: $u=\langle 4,5\rangle \&v=\langle 0,-1\rangle$

Formula used: $u-v=\langle u_1-v_1,u_2-v_2\rangle$

Calculation: Since $u-v=\langle u_1-v_1,u_2-v_2\rangle$

Hence, put the value of u and v,

  $\begin{array}{l}u-v=\langle \left(4\right)-\left(0\right),\left(5\right)-\left(-1\right)\rangle \\ u-v=\langle 4,6\rangle \end{array}$

Therefore, $u-v=\langle 4,6\rangle$

To determine

To find: The three times multiple of u.

Answer to Problem 45CR

The answer is $\langle 12,15\rangle$ .

Explanation of Solution

Given information: $u=\langle 4,5\rangle \&v=\langle 0,-1\rangle$

Formula used: $ku=\langle k{u}_1,k{u}_2\rangle$

Calculation: Since $ku=\langle k{u}_1,k{u}_2\rangle$

Hence, put the value of u,

  $\begin{array}{l}3u=3\langle \left(4\right),\left(5\right)\rangle \\ 3u=\langle 3\left(4\right),3\left(5\right)\rangle \\ =\langle 12,15\rangle \end{array}$

Therefore, $3u=\langle 12,15\rangle$

To determine

To find: The value of $2v+5u$

Answer to Problem 45CR

The answer is $\langle 20,23\rangle$ .

Explanation of Solution

Given information: $u=\langle 4,5\rangle \&v=\langle 0,-1\rangle$

Formula used: $ku=\langle k{u}_1,k{u}_2\rangle$ and $u+v=\langle u_1+v_1,u_2+v_2\rangle$

Calculation: Since $ku=\langle k{u}_1,k{u}_2\rangle$ and $u+v=\langle u_1+v_1,u_2+v_2\rangle$

Hence, put the value of u and v,

  $\begin{array}{l}2v+5u=2\langle \left(0,-1\right),5\left(4,5\right)\rangle \\ 2v+5u=\langle 0,-2\rangle +\langle 20,25\rangle \\ 2v+5u=\langle \left(0\right)+\left(20\right),\left(-2\right)+\left(25\right)\rangle \\ 2v+5u=\langle 20,23\rangle \end{array}$

Therefore, $2v+5u=\langle 20,23\rangle$