If `vec(a)=ht(i)-hat(k), vec(b)=xhat(i)+hat(j)+(1-x)hat(k)`
`vec(c)=yhat(i)+xhat(j)+(1+x-y)hat(k)`.
then `vec(a).(vec(b) xx vec (c))` depends on

Options

A. x only
B. y only
C. Both x and y
D. Neither x nor y

Answer

D

Explanation

Let `vec(a)=hat(i)-hat(k), vec(b)=xhat(i)+hat(j)+(1-x)hat(k)` and `vec(c)=yhat(i)+x hat(j)+(1+x-y)hat(k)`
Now, `(vec(b)xxvec(c))=|(hat(i), hat(j), hat(k)),(x, 1, (1-x)), (y, x, (1+x-y))|`
`=hat(i) (1+x-y-x+x^(2))-hat(j)(x+x^(2)-xy-y)+hat(k)(x^(2)-y)`
`=hat(i)(1-y+x^(2))-hat(j)(x+x^(2)-xy-y)+hat(k)(x^(2)-y)`
`=hat(i)(1-y+x^(2))-hat(j)(x+x^(2)-xy-y)+hat(k)(x^(2)-y)`
Now, `vec(a). (vec(b)xxvec(c))=1(1-y+x^(2))+0(x+x^(2)-xy-y)-1(x^(2)-y)`
`=1-y+x^(2)-x^(2)+y`
= 1 which shows that `vec(a).(vec(b)xxvec(c))` does not depend on x an y.