Identify the statement (s) which is/are INCORRECT?
A
→a×[→a×(→a×→b)]=(→a×→b)(→a2)
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B
If →a,→b,→c are non coplanar vectors and →v.→a=→v.→b=→v.→c=0, then →v must be a null vector
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C
If →a and →b lie in a plane normal to the plane containing the vectors →c and →d, then (→a×→b)×(→c×→d)=0
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D
If →a,→b,→c and →a′,→b′,→c′ are reciprocal system of vectors, then →a.→b′+→b.→c′+→c.→a′=3
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Solution
The correct options are A→a×[→a×(→a×→b)]=(→a×→b)(→a2) C If →a and →b lie in a plane normal to the plane containing the vectors →c and →d, then (→a×→b)×(→c×→d)=0 D If →a,→b,→c and →a′,→b′,→c′ are reciprocal system of vectors, then →a.→b′+→b.→c′+→c.→a′=3 A) →a×[a×(→a×→b)]=→a×[(a.b)→a−(→a.→a)→b]=0−(→a)2(→a×→b) False
B) →a,→b,→c are non-coplanar
→v→a=0→v→b=0→v→c=0⎫⎪
⎪
⎪⎬⎪
⎪
⎪⎭⇒→v(→a+→b+→c)=0 But →a+→b+→c≠0⇒→v=0 i.e. null vector which is true
C) →a×→b & →c×→d are perpendicular so (→a×→b)×(→c×→d)≠0 False
D) a′=→b×→c[abc],b′=→c×→a[abc],c′=→a×→b[abc] is valid only if →a,→b,→c are non coplanar,
The scalar product of any vector of one system with a vector of other system which does not correspond to it is zero