Let a = 2 i + j + 4 k and b = i + j + 3 k . Let c = i - 2 j + k be a vector in the plane of a and b . If c is perpendicular to a , then the value of + is :

Step-by-step solution

Since c is perpendicular to a , their dot product must be zero: c a = 0 (1)(2) + (-2)( ) + (1)(4) = 0 2 - 2 + 4 = 0 2 = 6 = 3 Since c lies in the plane of a and b , the three vectors are coplanar. Therefore, their scalar triple product is zero: [ c a b ] = 0 vmatrix 1 & -2 & 1 2 & & 4 1 & & 3 vmatrix = 0 Substitute = 3 into the determinant: vmatrix 1 & -2 & 1 2 & 3 & 4 1 & & 3 vmatrix = 0 Expanding along the first row: 1(9 - 4 ) - (-2)(6 - 4) + 1(2 - 3) = 0 9 - 4 + 2(2) + 2 - 3 = 0 9 - 4 + 4 + 2 - 3 = 0 10 - 2 = 0