Question

The $x$ and $y$ components of vector $\overrightarrow{A}$ are $4m$ and $6m$ respectively. The $x$ and $y$ components of vector $\overrightarrow{A}+\overrightarrow{B}$ are $10m$ and $9m$ respectively. For the vector $\overrightarrow{B}$ calculate the following
(i)$x$ and $y$ components
(ii) length and
(iii) the angle it makes with x-axis

• A. (i)$6m,3m$
  (ii)$\sqrt{45}m$
  (iii)$ta{n}^{-1}\left(\begin{array}{c}\frac{1}{2}\end{array}\right)$
• B. (i)$4m,2m$ (ii)$\sqrt{90}m$ (iii)$ta{n}^{-1}\left(\begin{array}{c}\frac{1}{2}\end{array}\right)$
• C. (i)$3m,6m$ (ii)$\sqrt{45}m$ (iii)$ta{n}^{-1}\left(\begin{array}{c}\frac{3}{2}\end{array}\right)$
• D. (i)$4m,3m$ (ii)$\sqrt{60}m$ (iii)$ta{n}^{-1}\left(\begin{array}{c}\frac{5}{2}\end{array}\right)$

Answer 1

The correct option is A (i)$6m,3m$

(ii)$\sqrt{45}m$
(iii)$ta{n}^{-1}\left(\begin{array}{c}\frac{1}{2}\end{array}\right)$

$\because A_x=4,A_y=6$ So $A_x+B_x=10$ and $A_y+B_y=9$

(i)$B_x=10-4=6m$ and $B_y=9-6=3m$

(ii) Length $=\sqrt{B_x^2+B_y^2}=\sqrt{36+9}=\sqrt{45}m$

(iii)$\theta =ta{n}^{-1}\left(\begin{array}{c}\frac{B_y}{B_x}\end{array}\right)=ta{n}^{-1}\left(\begin{array}{c}\frac{3}{6}\end{array}\right)=ta{n}^{-1}\left(\begin{array}{c}\frac{1}{2}\end{array}\right)$