Vector Algebra · Question 5

Consider the vectors x= i+2 j+3 k, y=2 i+3 j+ k, z=3 i+ j+2 k. For two distinct positive real numbers and , define X= x+ y- z, Y= y+ z- x, Z= z+ x- y. If the vectors X, Y, and Z lie in a plane, then the value of + -3 is .

Answer

-2

Solution

[ X Y Z]= vmatrix & & -1\ -1 & & \ & -1 & vmatrix[ x y z].

[ x y z]= vmatrix1&2&3\\2&3&1\\3&1&2 vmatrix=-18 0.

The circulant determinant with a= , b= , c=-1 equals a3+b3+c3-3abc= 12(a+b+c) [(a-b)2+(b-c)2+(c-a)2 ].

Since , the bracket is non-zero, so coplanarity needs + -1=0.

+ -3=-2

Answer: -2