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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Vector Algebra
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If a person standing at the point $(1,2) $ walks $5$ units towards east, then $3$ units towards north and $\sqrt 2$ units along the direction of $\hat i+\hat j$ and then walks $\sqrt 2 $ unit along the direction $\perp$ to this vector towards the starting point. What is the displacement of the person from origin?

(A) $(6+\sqrt 2)\hat i+(7+\sqrt 2)\hat j$

(B)$7\hat i+6\hat j$

(C) $6\hat i+7\hat j$

(D) $(7+\sqrt 2)\hat i+(6+\sqrt 2)\hat j$

Can you answer this question?

1 Answer

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Let the starting point be $A(1,2)\:\Rightarrow\:\overrightarrow {OA}=\hat i+2\hat j$
where $O$ is origin.
Then he moves 5 units towards east to $B$ (say) $\therefore \:\overrightarrow {OB}=6\hat i+2\hat j$
Then he moves 3 units north to $C$. $\therefore\:\overrightarrow {OC}=6\hat i+5\hat j$
Then he moves $\sqrt 2$ units towards the direction $\hat i+\hat j$ to $D$.
$\overrightarrow {CD}=\sqrt 2 (\large\frac{\hat i}{\sqrt 2}+\frac{\hat j}{\sqrt2})$$=\hat i+\hat j$
$\therefore\:$ In $\Delta OCD$, $\overrightarrow {OC}+\overrightarrow {CD}=\overrightarrow {OD}$
$\Rightarrow\:\overrightarrow {OD}=(6\hat i+5\hat j)+(\hat i+\hat j)=7\hat i+6\hat j$
Then he walks $\sqrt 2$ init $\perp$ to $\hat i+\hat j$ towards the starting point to $E$.
$\therefore\: $ the direction of $\overrightarrow {DE}$ is along $-\hat i+\hat j$
$\Rightarrow\:\overrightarrow {DE}=\sqrt 2(\large\frac{-\hat i}{\sqrt 2}+\frac{\hat j}{\sqrt 2})=-\hat i+\hat j$
$\therefore\:$ In $\Delta ODE$, $\overrightarrow {OD}+\overrightarrow {DE}=\overrightarrow {OE}$
$\therefore\:\overrightarrow {OE}=(7\hat i+6\hat j)+(-\hat i+\hat j)$
$=6\hat i+7\hat j$
answered Dec 10, 2013 by rvidyagovindarajan_1
edited Mar 13, 2014 by sreemathi.v

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