Solveeit Logo
Login

Question

Question: If three non-zero vectors are \(\mathbf { a } = a _ { 1 } \mathbf { i } + a _ { 2 } \mathbf { j } + ...

If three non-zero vectors are a=a1i+a2j+a3k\mathbf { a } = a _ { 1 } \mathbf { i } + a _ { 2 } \mathbf { j } + a _ { 3 } \mathbf { k }, b=b1i+b2j+b3k\mathbf { b } = b _ { 1 } \mathbf { i } + b _ { 2 } \mathbf { j } + b _ { 3 } \mathbf { k } and c=c1i+c2j+c3k\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 π6\frac { \pi } { 6 } , then ∣a1a2a3b1b2b3c1c2c3∣2\left| \begin{array} { l l l } 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

0

B

3(Σa12)(Σb12)(Σc12)4\frac { 3 \left( \Sigma a _ { 1 } ^ { 2 } \right) \left( \Sigma b _ { 1 } ^ { 2 } \right) \left( \Sigma c _ { 1 } ^ { 2 } \right) } { 4 }

C

1

D

(Σa12)(Σb12)4\frac { \left( \Sigma a _ { 1 } ^ { 2 } \right) \left( \Sigma b _ { 1 } ^ { 2 } \right) } { 4 }

Answer

(Σa12)(Σb12)4\frac { \left( \Sigma a _ { 1 } ^ { 2 } \right) \left( \Sigma b _ { 1 } ^ { 2 } \right) } { 4 }

Explanation

Solution

As C\mathbf { C } is the unit vector perpendicular to a\mathbf { a } and , we have

Now, ∣a1a2a3b1b2b3c1c2c3∣2=∣a1a2a3b1b2b3c1c2c3∣∣a1a2a3b1b2b3c1c2c3∣\left| \begin{array} { l l l } 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} { l l l } a _ { 1 } & a _ { 2 } & a _ { 3 } \\ b _ { 1 } & b _ { 2 } & b _ { 3 } \\ c _ { 1 } & c _ { 2 } & c _ { 3 } \end{array} \right| \left| \begin{array} { l l l } a _ { 1 } & a _ { 2 } & a _ { 3 } \\ b _ { 1 } & b _ { 2 } & b _ { 3 } \\ c _ { 1 } & c _ { 2 } & c _ { 3 } \end{array} \right|

= ∣a12+a22+a32a1b1+a2b2+a3b3a1c1+a2c2+a3c3a1b1+a2b2+a3b3b12+b22+b32b1c1+b2c2+b3c3a1c1+a2c2+a3c3b1c1+b2c2+b3c3c12+c22+c32∣\left| \begin{array} { c c c } a _ { 1 } ^ { 2 } + a _ { 2 } ^ { 2 } + a _ { 3 } ^ { 2 } & a _ { 1 } b _ { 1 } + a _ { 2 } b _ { 2 } + a _ { 3 } b _ { 3 } & a _ { 1 } c _ { 1 } + a _ { 2 } c _ { 2 } + a _ { 3 } c _ { 3 } \\ a _ { 1 } b _ { 1 } + a _ { 2 } b _ { 2 } + a _ { 3 } b _ { 3 } & b _ { 1 } ^ { 2 } + b _ { 2 } ^ { 2 } + b _ { 3 } ^ { 2 } & b _ { 1 } c _ { 1 } + b _ { 2 } c _ { 2 } + b _ { 3 } c _ { 3 } \\ a _ { 1 } c _ { 1 } + a _ { 2 } c _ { 2 } + a _ { 3 } c _ { 3 } & b _ { 1 } c _ { 1 } + b _ { 2 } c _ { 2 } + b _ { 3 } c _ { 3 } & c _ { 1 } ^ { 2 } + c _ { 2 } ^ { 2 } + c _ { 3 } ^ { 2 } \end{array} \right|

= ∣a∣2∣b∣2−(∣a∣∣b∣cos⁡π6)2=∣a∣2∣b∣2(1−34)| \mathbf { a } | ^ { 2 } | \mathbf { b } | ^ { 2 } - \left( | \mathbf { a } | | \mathbf { b } | \cos \frac { \pi } { 6 } \right) ^ { 2 } = | \mathbf { a } | ^ { 2 } | \mathbf { b } | ^ { 2 } \left( 1 - \frac { 3 } { 4 } \right)

=