Question

How do you find a) u+v, b) u-v, c) 2u-3v given $u=\left\langle 2,3 \right\rangle ,v=\left\langle 4,0 \right\rangle $ ?

Answer 1

Hint: To find $u+v$ , we will have to add the corresponding values. If $u=\left\langle {{x}_{1}},{{y}_{1}} \right\rangle ,v=\left\langle {{x}_{2}},{{y}_{2}} \right\rangle $ then $u+v=\left\langle {{x}_{1}}+{{x}_{2}},{{y}_{1}}+{{y}_{2}} \right\rangle $ . Similarly, we can find the value of $u-v$ . If $u=\left\langle {{x}_{1}},{{y}_{1}} \right\rangle ,v=\left\langle {{x}_{2}},{{y}_{2}} \right\rangle $ then $u-v=\left\langle {{x}_{1}}-{{x}_{2}},{{y}_{1}}-{{y}_{2}} \right\rangle $ . To find the value of $2u-3v$ , we will have to multiply 2 with $u=\left\langle 2,3 \right\rangle $ and 3 with $v=\left\langle 4,0 \right\rangle $ and then take the difference.
If $u=\left\langle {{x}_{1}},{{y}_{1}} \right\rangle $ , then $au=a\left\langle {{x}_{1}},{{y}_{1}} \right\rangle =\left\langle a{{x}_{1}},a{{y}_{1}} \right\rangle $ .

Complete step-by-step solution:
We are given that $u=\left\langle 2,3 \right\rangle ,v=\left\langle 4,0 \right\rangle $ . Let us move to the first part of the question.
a) To find $u+v$ , we will have to add the corresponding values.
If $u=\left\langle {{x}_{1}},{{y}_{1}} \right\rangle ,v=\left\langle {{x}_{2}},{{y}_{2}} \right\rangle $ then $u+v=\left\langle {{x}_{1}}+{{x}_{2}},{{y}_{1}}+{{y}_{2}} \right\rangle $
Therefore, we can write
 $u+v=\left\langle 2+4,3+0 \right\rangle $
Let us now add the values.
$u+v=\left\langle 6,3 \right\rangle $
b) Let us now evaluate $u-v$ . We will have to take the difference of the corresponding values.
If $u=\left\langle {{x}_{1}},{{y}_{1}} \right\rangle ,v=\left\langle {{x}_{2}},{{y}_{2}} \right\rangle $ then $u-v=\left\langle {{x}_{1}}-{{x}_{2}},{{y}_{1}}-{{y}_{2}} \right\rangle $ .
Therefore, we can write
 $u-v=\left\langle 2-4,3-0 \right\rangle $
Let us now take the difference of the values.
$u-v=\left\langle -2,3 \right\rangle $
c) To find the value of $2u-3v$ , we will have to multiply 2 with $u=\left\langle 2,3 \right\rangle $ and 3 with $v=\left\langle 4,0 \right\rangle $ and then take the difference.
If $u=\left\langle {{x}_{1}},{{y}_{1}} \right\rangle $ , then $au=a\left\langle {{x}_{1}},{{y}_{1}} \right\rangle =\left\langle a{{x}_{1}},a{{y}_{1}} \right\rangle $ .
Hence, the value of $2u-3v$ can be written as
$\begin{align}
  & 2u-3v=2\left\langle 2,3 \right\rangle -3\left\langle 4,0 \right\rangle \\
 & =\left\langle 4,6 \right\rangle -\left\langle 12,0 \right\rangle \\
 & =\left\langle 4-12,6-0 \right\rangle \\
 & =\left\langle -8,6 \right\rangle \\
\end{align}$
Hence the answer is a) $u+v=\left\langle 6,3 \right\rangle $ , b) $u-v=\left\langle -2,3 \right\rangle $ , c) $2u-3v=\left\langle -8,6 \right\rangle $

Note: Students have a chance of making mistakes when finding $u-v$ . They may write $u-v=\left\langle {{x}_{2}}-{{x}_{1}},{{y}_{2}}-{{y}_{1}} \right\rangle $ . You can also find the value of say, $\dfrac{u}{2}$ , similar to 2u except that here you will have to divide u by 2, that is, if $u=\left\langle {{x}_{1}},{{y}_{1}} \right\rangle $ , then $\dfrac{u}{2}=\dfrac{1}{2}\left\langle {{x}_{1}},{{y}_{1}} \right\rangle =\left\langle \dfrac{{{x}_{1}}}{2},\dfrac{{{y}_{1}}}{2} \right\rangle $ .