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If the position vectors of the points P and Q are 2 i + 3j + 4k and 3 i – 2 j – 3 k , find the direction cosines of the vector vector PQ and hence prove that,Cos2α + Cos2β + Cos2γ = 1 |
Answer» Let O be the fixed point vector OP = position vector of P = 2 i + 3j + 4k = ( 2 , 3 , 4 ) vector OQ = position vector of Q = 3 i – 2 j – 3 k = ( 3 , - 2 , -3) vector PQ = vector (OQ – OP) = (1 , -5 , -7) vector │PQ│= √( 1 + 25 + 49 ) = √75 Direction cosines of vector PQ are Cosα = 1/√ 75 , Cosβ = -5/√ 75 & cosγ = - 7 /√ 75. Consider, Cos2α + Cos2β + Cos2γ = 1/75 + 25 /75 + 49 /75 = 1 |
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