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KEAM 2010 Math Paper

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Question : 7 of 120

Marks: +1, -0

The vectors of magnitude

a, 2 a, 3 a

meet at a point and their directions are along the diagonals of three adjacent faces of a cube. Then, the magnitude of their resultant is

$5 a$
$6 a$
$10 a$
$9 a$
$7 a$

Solution:

Suppose that the sides of cube are unity and unit vector along

O A, O B

and

O C

are

\hat{i}, \hat{j}, \hat{k}

respectively. OR, OS, OT are diagonals of cube having corresponding vector

a, 2 a

and

3 a

(Magnitude) respectively.

∴ Unit vector along

O R = \; \frac{\widehat{j + \hat{k}}}{\sqrt{2}}

∴ Vector along

\overrightarrow{OR} = a\left(\; \frac{\widehat{j + \hat{k}}}{\sqrt{2}}\right)

Similarly, vector along

\overrightarrow{OS} = 2a\left(\; \frac{\widehat{k + \hat{i}}}{\sqrt{2}}\right)

and vector along

\overrightarrow{OT} = 3a\left(\; \frac{\widehat{i + \hat{j}}}{\sqrt{2}}\right)

∴ Resultant

R = \overrightarrow{OR} + \overrightarrow{OS} + \overrightarrow{OT}

= a\left( \frac{\widehat{i + \hat{k}}}{\sqrt{2}} \right) + 2a\left( \frac{\widehat{k + \hat{i}}}{\sqrt{2}} \right) + 3a\left( \frac{\widehat{i + \hat{j}}}{\sqrt{2}} \right)

= \frac{5a}{\sqrt{2}} \hat{i} + \frac{4a}{\sqrt{2}} \hat{j} + \frac{3a}{\sqrt{2}} \hat{k}

\therefore \; |R| = \sqrt{ \; \frac{25 a^{2}}{2} + \; \frac{16 a^{2}}{2} + \; \frac{9 a^{2}}{2} } = 5 a

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