Question

If $|\overrightarrow{A}|=21$, direction of $\overrightarrow{A}$ is East, direction of $\overrightarrow{B}$ is ${37}^{\circ }$ West of North and direction of $\overrightarrow{C}$ is North-East, then the magnitude of $\overrightarrow{C}$ is (Given, $\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}$ )

• A. $6\sqrt{2}$
• B. $10\sqrt{2}$
• C. $12\sqrt{2}$
• D. $12$

Answer 1

The correct option is C $12\sqrt{2}$

The given condition of the vectors can be shown as

Let $|\overrightarrow{B}|$ be $x$, $|\overrightarrow{C}|$ be $y$
Given, $\overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}$
Now, resolving the vectors along the axes we get
$|\overrightarrow{C_x}|=21-x\sin {37}^{\circ }$
$\Rightarrow |\overrightarrow{C_x}|=21-\frac{3x}{5}$
Similarly, $|\overrightarrow{C_y}|=x\cos {37}^{\circ }$
$\Rightarrow |\overrightarrow{C_x}|=\frac{4x}{5}$


Thus, $\tan {45}^{\circ }=\frac{|\overrightarrow{C_y}|}{|\overrightarrow{C_x}|}=\frac{\frac{4x}{5}}{21-\frac{3x}{5}}=1$
$\Rightarrow x=15$
So, $|\overrightarrow{C_x}|=21-9=12$ and $|\overrightarrow{C_y}|=12$
Thus, $y=\sqrt{{|\overrightarrow{C_x}|}^2+{|\overrightarrow{C_y}|}^2}=\sqrt{{12}^2+{12}^2}=12\sqrt{2}$