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Question

If the points whose position vectors are 2i+j+k,6ij+2k and 14i5j+pk are collinear, then the value of p is?

A
2
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B
4
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C
6
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D
8
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Solution

The correct option is B 4
Positivevectorare2i+j+k,6^i+^j+2^kand4i5j+pkarecollinearthenP=2ifthreepositionvectorsarecollinearthen,itsdeterminantsis(0)∣ ∣211612145P∣ ∣=02(P+10)1(6P28)+(30+14)=02P+206P+2816=08P+32=0P=4

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