Question

If x = 2p – 3q, y = p + 2q and z = q – 3p then x² + y² + z² + 2yz + 2zx + 2xy is ………………….. 

A) 0 

B) 1 

C) xyz 

D) pqr

Answer 1

Correct option is (A) 0

\(x = 2p-3q,\)

\(y=p+2q\)

and \(z = q - 3p\) 

\(\therefore\) x+y+z = (2p - 3q) + (p + 2q) + (q - 3p)

= (2p + p - 3p) + (-3q + 2q + q)

= (3p - 3p) + (-3q + 3q)

= 0+0 = 0

Now, \(x^2+y^2+z^2+2yz+2zx+2xy\) \(=(x+y+z)^2\)

\(=0^2=0\)

Answer 2

Correct option is A) 0