Chapter 13, Problem 14CR

To determine

To find vector sum of (2,6) and 3(1,-2)

Answer to Problem 14CR

(5,0)

Explanation of Solution

Given points

Let, $\text{P:(2,6) Q:3(1,-2)}$

i.e $\text{P:(2,6) Q:(3,-6)}$

Let $\text{S}$ be the coordinate of vector sum of $\text{P}$ and $\text{Q}$

If ${\text{A:(x}}_{\text{1}}{\text{,y}}_{\text{1}}\text{)}$ and ${\text{B:(x}}_2{\text{,y}}_2\text{)}$

Then the vector sum of $\text{A}$ and ${\text{B : (x}}_{\text{1}}+{\text{x}}_2,{\text{ y}}_{\text{1}}+{\text{y}}_2)$

Here,

Vector sum of $\text{P}$ and $\text{Q}$ , $\text{S: (2+3 , 6-6)}$

  

i.e. $\text{S : (5,0)}$

In the figure . $\overline{\text{OP}}$ represents the position vectors $\text{(2,6)}$

In the figure . $\overline{\text{OR}}$ represents the position vectors $\text{(1,-2)}$

In the figure . $\overline{\text{OQ}}$ represents the position vectors $\left(\text{3(1,-2)}\right)$

By parallelogram law of vector addition,

Diagonal of parallelogram $\text{OPSQ}$ gives the vector sum of $\overline{\text{OP}}$ and $\overline{\text{OQ}}$

i.e., vector sum of $\overline{\text{OP}}$ and $\overline{\text{OQ}}$ is vector $\overline{\text{OS}}$

From figure,

  $\text{S : (5,0)}$

So,

Vector sum of P and Q is $\text{(5,0)}$