If the points whose position vectors are 2i+j+k,6i−j+2k and 14i−5j+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 B4 Positivevectorare2i+j+k,6^i+^j+2^kand4i−5j+pkarecollinearthenP=2ifthreepositionvectorsarecollinearthen,itsdeterminantsis(0)⇒∣∣
∣∣2116−1214−5P∣∣
∣∣=0⇒2(−P+10)−1(6P−28)+(−30+14)=0⇒−2P+20−6P+28−16=0⇒−8P+32=0∴P=4