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JEE Advance - Mathematics (2008 - Paper 2 Offline - No. 6)

Let two non-collinear unit vectors $$\widehat a$$ and $$\widehat b$$ form an acute angle. A point $$P$$ moves so that at any time $$t$$ the position vector $$\overrightarrow {OP} $$ (where $$O$$ is the origin) is given by $$\widehat a\cos t + \widehat b\sin t.$$ When $$P$$ is farthest from origin $$O,$$ let $$M$$ be the length of $$\overrightarrow {OP} $$ and $$\widehat u$$ be the unit vector along $$\overrightarrow {OP} $$. Then :
$$\widehat u = {{\widehat a + \widehat b} \over {\left| {\widehat a + \widehat b} \right|}}\,\,and\,\,M = {\left( {1 + \widehat a.\,\widehat b} \right)^{1/2}}$$
$$\widehat u = {{\widehat a - \widehat b} \over {\left| {\widehat a - \widehat b} \right|}}\,\,and\,\,M = {\left( {1 + \widehat a.\,\widehat b} \right)^{1/2}}$$
$$\widehat u = {{\widehat a + \widehat b} \over {\left| {\widehat a + \widehat b} \right|}}\,\,and\,\,M = {\left( {1 + 2\widehat a.\,\widehat b} \right)^{1/2}}$$
$$\widehat u = {{\widehat a - \widehat b} \over {\left| {\widehat a - \widehat b} \right|}}\,\,and\,\,M = {\left( {1 + 2\widehat a.\,\widehat b} \right)^{1/2}}$$

Explanation

$$|\overrightarrow {OP} | = |\widehat a\cos t + \widehat b\sin t|$$

$$|\overrightarrow {OP} {|^2} = (\widehat a\cos t + \widehat b\sin t)\,.\,(\widehat a\cos t + \widehat b\sin t)$$

$$ = |\widehat a{|^2}{\cos ^2}t + 2\widehat a\,.\,\widehat b\cos t\sin t + |\widehat b{|^2}{\sin ^2}t$$

$$ = {\cos ^2}t + \widehat a\,.\,\widehat b\sin 2t + {\sin ^2}t$$

$$ = 1 + \widehat a\,.\,\widehat b\sin 2t$$

The greatest value of $$|\overrightarrow {OP} |$$ is reached when $$\sin 2t = 1$$

i.e., $$t = {\pi \over 4}$$

$$M = |\overrightarrow {OP} |$$ greatest $$ = \sqrt {1 + \widehat a\,.\,\widehat b} $$

The direction of $$\overrightarrow {OP} $$ is given by

$$\widehat u = {{\widehat a + \widehat b} \over {\sqrt 2 \,.\,{{|\widehat a + \widehat b|} \over {\sqrt 2 }}}} = {{\widehat a + \widehat b} \over {|\widehat a + \widehat b|}}$$

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