Vector Algebra Question 153
Question: If three non-zero vectors are $ \mathbf{a}=a_1\mathbf{i}+a_2\mathbf{j}+a_3\mathbf{k}, $ $ \mathbf{b}=b_1\mathbf{i}+b_2\mathbf{j}+b_3\mathbf{k} $ and $ \mathbf{c}=c_1\mathbf{i}+c_2\mathbf{j}+c_3\mathbf{k}. $ If c is the unit vector perpendicular to the vectors a and b and the angle between a and b is $ \frac{\pi }{6}, $ then $ {{| ,\begin{matrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \\ \end{matrix}, |}^{2}} $ is equal to
[IIT 1986]
Options:
A) 0
B) $ \frac{3,(\Sigma a_1^{2}),(\Sigma b_1^{2}),(\Sigma c_1^{2})}{4} $
C) 1
D) $ \frac{(\Sigma a_1^{2}),(\Sigma b_1^{2})}{4} $
Show Answer
Answer:
Correct Answer: D
Solution:
- $ |\mathbf{c}|,=1, $ we have $ |\mathbf{c}{{|}^{2}}=1 $ or $ c_1^{2}+c_2^{2}+c_3^{2}=1 $ …..(i)
Again, since $ \mathbf{c},\bot ,\mathbf{a} $ and $ \mathbf{c},\bot ,\mathbf{b}, $ we have $ \mathbf{c},.,\mathbf{a}=0 $
$ \Rightarrow a_1c_1+a_2c_2+a_3c_3=0 $ …..(ii)
and $ \mathbf{c},.,\mathbf{b}=0\Rightarrow b_1c_1+b_2c_2+b_3c_3=0 $ …(iii)
Also since angle between $ \mathbf{a} $ and $ \mathbf{b} $ is $ \frac{\pi }{6}, $ we have $ \mathbf{a},.,\mathbf{b}=a_1b_1+a_2b_2+a_3b_3 $
Þ $ |\mathbf{a}|,|\mathbf{b}|\cos \frac{\pi }{6}=a_1b_1+a_2b_2+a_3b_3 $
Þ $ \frac{3}{4}(a_1^{2}+a_2^{2}+a_3^{2})(b_1^{2}+b_2^{2}+b_3^{2}) $ $ ={{(a_1b_1+a_2b_2+a_3b_3)}^{2}} $ …..(iv)
Now, $ {{ \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \\end{vmatrix} }^{2}}= \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \\end{vmatrix} , \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \\ \end{vmatrix} , $
$ = \begin{vmatrix} a_1^{2}+a_2^{2}+a_3^{2} & a_1b_1+a_2b_2+a_3b_3 & 0 \\ b_1a_1+b_2a_2+b_3a_3 & b_1^{2}+b_2^{2}+b_3^{2} & 0 \\ 0 & 0 & 1 \\ \end{vmatrix} $
{Using (i), (ii) and (iii)}
$ =\frac{1}{4}(a_1^{2}+a_2^{2}+a_3^{2})(b_1^{2}+b_2^{2}+b_3^{2}) $ , {Using (iv)}
$ =\frac{(\Sigma a_1^{2})(\Sigma b_1^{2})}{4} $ ,
where $ \Sigma a_1^{2}=a_1^{2}+a_2^{2}+a_3^{2} $ and $ \Sigma b_1^{2}=b_1^{2}+b_2^{2}+b_3^{2}. $
BETA
