Let a = 4 i - j + 3 k , b = 10 i + 2 j - k and a vector c be such that 2( a b ) + 3( b c ) = 0 . If a c = 15 , then c ( i + j -3 k ) is equal to:

Step-by-step solution

Given 2( a b ) + 3( b c ) = 0 2( a b ) - 3( c b ) = 0 (2 a - 3 c ) b = 0 Since the cross product is zero, the vectors are collinear: 2 a - 3 c = b 3 c = 2 a - b Taking the dot product with a on both sides: 3( a c ) = 2| a |^2 - ( a b ) We have a = 4 i - j + 3 k and b = 10 i + 2 j - k . | a |^2 = 4^2 + (-1)^2 + 3^2 = 26 a b = 4(10) + (-1)(2) + 3(-1) = 40 - 2 - 3 = 35 Given a c = 15 , substituting these values: 3(15) = 2(26) - 35 45 = 52 - 35 35 = 7 = 1 5 Substituting back into the equation for c : 3 c = 2 a - 1 5 b