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 C0.5 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[6−8]−^j[4−4]+^k[4−3] ∴→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