Vector-Algebra Question 406

Question: Let $ \overrightarrow{A}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k},\overrightarrow{B}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k} $ and $ \overrightarrow{C}=c_1\hat{i}+c_2\hat{j}+c_3\hat{k} $ be three non-zero vectors such that $ \overrightarrow{C} $ is a unit vector perpendicular to both the vectors $ \overrightarrow{A} $ and $ \overrightarrow{B} $ .If the angle between $ \overrightarrow{A} $ and $ \overrightarrow{B} $ is $ \frac{\pi }{6} $ , then.

Options:

A) 0

B) 1

C) $ \frac{1}{4}(a_1^{2}+a_2^{2}+a_3^{2})(b_1^{2}+b_3^{2}) $

D) $ \frac{3}{4}(a_1^{2}+a_2^{2}+a_3^{2})(b_1^{2}+b_2^{2}+b_3^{2})(c_1^{2}+c_3^{2}) $

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Answer:

Correct Answer: C

Solution:

  • [c] $ {{ \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \\ \end{vmatrix} }^{2}}={{[\overset{\to }{\mathop{A}},\overset{\to }{\mathop{B}},\overset{\to }{\mathop{C}},]}^{2}}={{((\overset{\to }{\mathop{A}},\times \overset{\to }{\mathop{B}},).\overset{\to }{\mathop{C}},)}^{2}} $ $ ={{{ |\overset{\to }{\mathop{A}},||\overset{\to }{\mathop{B}},|sin\frac{\pi }{6}(\overset{\to }{\mathop{C}},).\overset{\to }{\mathop{C}}, }}^{2}} $ $ =|\overset{\to }{\mathop{A}},{{|}^{2}}|\overset{\to }{\mathop{B}},{{|}^{2}}{{( \frac{1}{2} )}^{2}}|\overset{\to }{\mathop{C}},{{|}^{4}}=\frac{1}{4}|\overset{\to }{\mathop{A}},{{|}^{2}}|\overset{\to }{\mathop{B}},{{|}^{2}} $ $ =\frac{1}{4}(a_1^{2}+a_2^{2}+a_3^{2})(b_1^{2}+b_2^{2}+b_3^{2}) $