question_answer

ABCD is parallelogram. The position vectors of A and C are respectively,\[3\hat{i}+3\hat{j}+5\hat{k}\] and\[\hat{i}-5\hat{j}-5\hat{k}\].If M is the midpoint of the diagonal\[\overset{\to }{\mathop{OM}}\,\]then the magnitude of the projection of on\[\overset{\to }{\mathop{OC}}\,,\] where O is the origin, is   JEE Main Online Paper (Held On 07 May 2012)

A) \[7\sqrt{51}\]                   

B)                        \[\frac{7}{\sqrt{50}}\]

C)                        \[7\sqrt{50}\]                   

D)                        \[\frac{7}{\sqrt{51}}\]

Correct Answer: D

Solution :

                In a parallelogram, diagonals bisect each other. So, mid-point of DB is also the mid- point of AC. Mid-point of \[M=2\hat{i}-\hat{j}\] Direction ratio of OC = (1, - 5, - 5) Direction ratio of OM= (2, -1, 0) Angle \[\theta \]between OM and OC is given by \[\cos \theta =\frac{\left( 1\times 2 \right)+\left( -5 \right)\left( -1 \right)+\left( -5 \right)\left( 0 \right)}{\sqrt{{{2}^{2}}+{{\left( -1 \right)}^{2}}}\sqrt{{{\left( 1 \right)}^{2}}+{{\left( -5 \right)}^{2}}+{{\left( -5 \right)}^{2}}}}\] \[=\frac{2+5}{\sqrt{5}\sqrt{51}}=\frac{7}{\sqrt{5}\sqrt{51}}\] Projection of \[\overset{\to }{\mathop{OM}}\,\]on \[\overset{\to }{\mathop{OC}}\,\] is given by \[\left| OM \right|.\cos \theta =\sqrt{5}\times \frac{7}{\sqrt{5}\times \sqrt{51}}=\frac{7}{\sqrt{51}}\]