Question

If three non-zero vectors are $a=a_{1}i+a_{2}j+a_{3}k,b=b_{1}i+b_{2}j+b_{3}k$ and $c=c_{1}i+c_{2}j+c_{3}k$ 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 ${\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|}^2$
is equal to

• A. $\frac{3(\sum a_1^2)(\sum b_1^2)(\sum c_1^2)}{4}$
• B. 1
• C. $\frac{(\sum a_1^2)(\sum b_1^2)}{4}$

Answer 1

The correct option is C $\frac{(\sum a_1^2)(\sum b_1^2)}{4}$

$|c|=1$, we have $|c{|}^2=1or{c}_1^2+c_2^2+c_3^2=1$ ....(i)
Again, Since $c\perp a$ and $c\perp b$, we have c.a =0
$\Rightarrow a_1{c}_1+a_2{c}_2+a_3{c}_3=0$ ....(ii)
and c. $b=0\Rightarrow b_1{c}_1+b_2{c}_2+b_3{c}_3=0$ .....(iii)
Also since angle between a and b is $\frac{\pi }{6}$, we have a, b = $a_1{b}_1+a_2{b}_2+a_3{b}_3$
$\Rightarrow |a||b|cos\frac{\pi }{6}=a_1{b}_1+a_2{b}_2+a_3{b}_3$
$\Rightarrow \frac{3}{4}(a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2)=(a_1{b}_1+a_2{b}_2+a_3{b}_3{)}^2$
Now,
${\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|}^2=\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|$

$=\left|\begin{array}{ccc}{a}_1^2+a_2^2+a_3^2& a_1{b}_1+a_2{b}_2+a_3{b}_3& 0\\ b_1{a}_1+b_2{a}_2+b_3{a}_3& b_1^2+b_2^2+b_3^2& 0\\ 0& 0& 1\end{array}\right|$

(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))
$=\frac{(\sum a_1^2)(\sum b_1^2)}{4}$
where $\sum a_1^2=a_1^2+a_2^2+a_3^{2}and\sum b_1^2=b_1^2+b_2^2+b_3^2$