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Question: If a<sub>i</sub><sup>2</sup> + b<sub>i</sub><sup>2</sup> + c<sub>i</sub><sup>2</sup> = 1(i = 1,2,3) ...

If ai2 + bi2 + ci2 = 1(i = 1,2,3) and ai aj + bi bj + ci cj = 0(i ¹ j, i, j = 1,2,3) , then value of a1a2a3b1b2b3c1c2c32\left| \begin{matrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{matrix} \right|^{2} is

A

0

B

½

C

1

D

2

Answer

1

Explanation

Solution

a1a2a3b1b2b3c1c2c32\left| \begin{matrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{matrix} \right|^{2}= a1b1c1a2b2c2a3b3c3\left| \begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix} \right| a1b1c1a2b2c2a3b3c3\left| \begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix} \right|

=a12+b12+c12a1a2+b1b2+c1c2a1a3+b1b3+c1c3a2a1+b2b1+c2c1a22+b22+c22a2a3+b2b3+c2c3a3a1+b3b1+c3c1a3a2+b3b2+c3c2a32+b32+c32\left| \begin{matrix} a_{1}^{2} + b_{1}^{2} + c_{1}^{2} & a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} & a_{1}a_{3} + b_{1}b_{3} + c_{1}c_{3} \\ a_{2}a_{1} + b_{2}b_{1} + c_{2}c_{1} & a_{2}^{2} + b_{2}^{2} + c_{2}^{2} & a_{2}a_{3} + b_{2}b_{3} + c_{2}c_{3} \\ a_{3}a_{1} + b_{3}b_{1} + c_{3}c_{1} & a_{3}a_{2} + b_{3}b_{2} + c_{3}c_{2} & a_{3}^{2} + b_{3}^{2} + c_{3}^{2} \end{matrix} \right|

= 100010001\left| \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right| = 1