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Given `a_(i)^(2) + b_(i)^(2) + c_(i)^(2) = 1, i = 1, 2, 3 and a_(i) a_(j) + b_(i) b_(j) + c_(i) c_(j) = 0 (i !=j, i, j =1, 2, 3)`, then the value of the determinant `|(a_(1),a_(2),a_(3)),(b_(1),b_(2),b_(3)),(c_(1),c_(2),c_(3))|`, isA. `(1)/(2)`B. 0C. 2D. 1 |
Answer» Correct Answer - D We have, `|(a_(1),a_(2),a_(3)),(b_(1),b_(2),b_(3)),(c_(1),c_(2),c_(3))|^(2)` `= |(a_(1),a_(2),a_(3)),(b_(1),b_(2),b_(3)),(c_(1),c_(2),c_(3))||(a_(1),a_(2),a_(3)),(b_(1),b_(2),b_(3)),(c_(1),c_(2),c_(3))| = |(Sigmaa_(i)^(2),Sigmaa_(i)b_(j),Sigma a_(i) c_(j)),(Sigma a_(i) b_(j),Sigma b_(i)^(2),Sigma b_(i) c_(j)),(Sigma a_(i) c_(j),Sigma b_(i) c_(j),Sigma c_(i)^(2))|` `= |(1,0,0),(0,1,0),(0,0,1)| =1` `:. |(a_(1),a_(2),a_(3)),(b_(1),b_(2),b_(3)),(c_(1),c_(2),c_(3))| = +- 1` |
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