Question 20 - CBSE Class 12 Sample Paper for 2018 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards
Last updated at Sept. 24, 2021 by

If a, b, c are three vectors such that a + b + c = 0 ,Β then prove that a Γ b = b Γ c = c Γ a, and hence show that [a b c] = 0.
This is a question of CBSE Sample Paper - Class 12 - 2017/18.

You can download the question paper hereΒ
https://www.teachoo.com/cbse/sample-papers/

Transcript

Question 20
If π β, π β, π β are three vectors such that π β + π β + π β = 0 β , then prove that π β Γ π β = π β Γ π β = π β Γ π β, and hence show that [π β" " π β" " π β ] = 0.
Theory
Here [π β" " π β" " π β ] = π β.(π β Γ π β )
Given
π β + π β + π β = 0 β
π βΓ(π β+π β+π β )= π βΓ0 β
π βΓπ β+π βΓπ β+π βΓπ β= 0 β
Since π βΓπ β=0
" " 0+π βΓπ β+π βΓπ β=" " 0 β
π βΓπ β+π βΓπ β=" " 0 β
π βΓπ β=βπ βΓπ β
Since βπ βΓπ β = π βΓπ β
π βΓπ β=π βΓπ β
Similarly,
π β + π β + π β = 0 β
π βΓ(π β+π β+π β )= π βΓ0 β
π βΓπ β+π βΓπ β+π βΓπ β= 0 β
Since π βΓπ β=0
π βΓπ β+0+π βΓπ β= 0 β
π βΓπ β+π βΓπ β=" " 0 β
π βΓπ β=βπ βΓπ β
π βΓπ β=βπ βΓπ β
Since βπ βΓπ β = π βΓπ β
π βΓπ β=π βΓπ β
Thus,
π βΓπ β=π βΓπ β
& π βΓπ β=π βΓπ β
β΄ π βΓπ β=π βΓπ β=π βΓπ β
Now, we need to show that show that [π β" " π β" " π β ] = 0
[π β π β π β ]=π β . (π βΓπ β )
From (1): π βΓπ β = π βΓπ β
=π β . (π βΓπ β )
Now, π βΓπ β will be a vector perpendicular to π β
And dot product of π β with a vector perpendicular to π β will be 0
as angle is 90Β° and cos 90Β° = 0
β΄ [π β π β π β ]=π β . (π βΓπ β ) = 0
Hence proved

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