1.

Let `a_i^2+b_i^2+c_i^2=1` (for i=1,2,3) and `a_ia_j+b_ib_j+c_ic_j=0 (i != j;i,j=1,2,3).` Then the absolute value of determinant `|(a_1,a_2,a_3),(b_1,b_2,b_3),(c_1,c_2,c_3)|` (A) `1/2` (B) 0 (C) 1 (D) 2

Answer» Let `vecp = a_1hati+b_1hatj+c_1hatk`,
`vecq = a_2hati+b_2hatj+c_2hatk`,
`vecr = a_3hati+b_3hatj+c_3hatk`
Here, we are given, for `(i=1,2,3)`,
`a_i^2+b_i^2+c_i^2 = 1`
It means, `|vecp|=|vecp|=|vecp| = 1`
Also, we are given,`a_ia_j+b_ib_j+c_ic_j = 0`
`:. vecp*vecq = a_1a_2+b_1b_2+c_1c_2 = 0`
So, `vecp _|_vecq`.
Similarly, `vecq _|_vecr` and `vecr _|_vecp`.
Now, we have to find the value of
`A=|[a_1,a_2,a_3],[b_1,b_2,b_3],[c_1,c_2,c_3]|`
Transpose of this determinant will be same. So, we can write,
`A = |[a_1,b_1,c_1],[a_2,b_2,c_2],[a_3,b_2,c_3]|`
`A = [vecp*vecq*vecr] `
As `|vecp|=|vecp|=|vecp| = 1`
`:. A =1`


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