Question

Assertion (A) : A homogeneous equation of second degree in $x , y , z$ represent a pair of planes through origin
Reason (R) : A homogeneous equation of second degree in $x, y, z$ can be factorised into homogeneous linear factors

• A. Both (A) & (R) are individually true & (R) is correct explanation of (A)
• B. Both (A) & (R) are individually true but (R) is not the correct (proper) explanation of (A)
• C. (A) is true but (R) is false
• D. (A) is false but (R) is true

Answer 1

Answer: (A)

A homogeneous second degree equation in $x,y,z$ is given by
$ax^{2}+by^{2}+cz^{2}+2fyz+2gxz+2hxy=0\,\dots(i)$
Let two equations of (i) be $A_{1}\,x+B_{1}\,y+C_{1}\,z=0$ &
$A_{2}\,x+B_{2}y+C_{2}\,z=0$ (due to reason (R ))
So, we have $ax^{2}+by^{2}+cz^{2}+2fyz+2gxz+2hxy$
$\equiv (A_{1}\,x+B_{1}\,y+C_{1}\,z) (A_{2}\,x+B_{2}\,y+C_{2}\,z)$
As equation (i) does not contain any constant term , then these planes must pass through the origin and $A_{1}x+B_{1}y+C_{1}z=0$ & $A_{2}x+B_{2}y+C_{2}z=0$ are hom ogeneous linear factor