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Questions  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Vector Algebra

A man walks 4 mtrs. towards west and then walks 3 mtrs in the direction of $\large\frac{\pi}{6}\:$ radians east of north. What is the displacement of the man from the starting point?

$\begin{array}{1 1} (A) -\large\frac{11}{4}\hat i+\large\frac{3\sqrt 3}{4}\hat j \\ (B) -\large\frac{11}{2}\hat i+\large\frac{3\sqrt 3}{2}\hat j \\ (C) -\large\frac{11}{2}\hat i+\large\frac{27}{4}\hat j \\ (D) -\large\frac{11}{4}\hat i+\large\frac{3}{2}\hat j\end{array} $

1 Answer

Let the man start from origin. He walks west side 4 metrs.
$\therefore\:\overrightarrow {OA}=-4\hat i$
Then he walks 3 metrs. $\large\frac{\pi}{6}$ east of north and moves to $B$
$\therefore \: |\overrightarrow {AB}|=3$, $< BAO=\large\frac{\pi}{3}$ and $ |\overrightarrow {OA}|=4$
Let $\overrightarrow {OB}=x\hat i+y\hat j$
From triangular law of addition $\overrightarrow {OA}+\overrightarrow {AB}=\overrightarrow {OB}$
$\Rightarrow\:-4\hat i+\overrightarrow {AB}=x\hat i+y\hat k$
$\Rightarrow\:\overrightarrow {AB}=x\hat i+y\hat k+4\hat i=(x+4)\hat i+y\hat k$
Also $cos\large\frac{\pi}{3}$$=\large\frac{\overrightarrow {OA}.\overrightarrow {AB}}{|\overrightarrow {OA}||\overrightarrow {AB}|}=\frac{(-4\hat i).((x+4)\hat i+y\hat j)}{12}$
Now $|\overrightarrow {AB}|=3\Rightarrow\:(x+4)^2+y^2=9$
Substituting the value of $x$ we get $y^2=9-\large\frac{9}{4}$
$\Rightarrow\:y=\large\frac{3\sqrt 3}{2}$
$\therefore $ The required displacement $\overrightarrow {OB}=-\large\frac{11}{2}$$\hat i+\large\frac{3\sqrt 3}{2}$$\hat j$
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