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Question

Three vectors A, B & C are given as A=2^i+3^j+4^k , B=^i+2^j+2^k and C=m^i+2^j+^k . Find the value of m, for which these vectors are coplanar.

A
1
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B
2
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C
0.5
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D
0.25
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Solution

The correct option is C 0.5
If these vectors are coplanar, then scalar triple product of these vector must equal to zero.
[A B C] =(A×B).C=0.....(i)
A×B=^i^j^k234122

A×B=^i[68]^j[44]+^k[43]
A×B= 2^i+ ^k
Now taking dot product with C:
(A×B).C=(2^i+ ^k).(m^i+2^j+^k)
(A×B).C= 2m+1
From Eq. i:
(A×B).C=0
2m+1=0
m=12=0.5

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