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 \[\left| OM \right|.\cos \theta =\sqrt{5}\times \frac{7}{\sqrt{5}\times \sqrt{51}}=\frac{7}{\sqrt{51}}\]You need to login to perform this action.
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