Correct Answer - A
We know that the angle between the planes `vecr.vecn_(1)=d_(1)` and`vecr.vecn_(2)=d_(2)` is given by
`cos theta=(vecn_(1).vecn_(2))/(|vecn_(1)||vecn_(2)|)`
Here `vecn_(1)=2hati-hatj+hatk` and `vecn_(2)=hati+hatj+2hatk`
`:.cos theta=((2hati-hatj+hatk).(hati+hatj+2hatk))/(|2hati-hatj+hatk||hati+hatj+2hatk|)=1/2`
`impliestheta=pi//3`
If `a_(1)x+b_(1)y+c_(1)z+d_(1)=0` and `a_(2)x+b_(2)y+c_(2)z+d_(2)=0` are Cartesian equations of two planes, then vectors normal to them are
`vecn_(1)=a_(1)hati+b_(1)hatj+c_(1)hatk` and `vecn_(2)=a_(2)hati+b_(2)hatj+c_(2)hatk` respectively.
Therefore, the angle `theta` between the planes is given by
`cos thet=(vecn_(1).vecn_(2))/(|vecn_(1)||vecn_(2)|)`
`impliescos theta=(a_(1)a_(2)+b_(1)b_(2)+c_(1)c_(2))/(sqrt(a_(1)^(2)+b_(1)^(2)+c_(1)^(2))sqrt(a_(1)^(2)+b_(2)^(2)+c_(2)^(2)))`
If the planes are perpendicular then `vecn_(1)` and `vecn_(2)` are perpendicular.
`:.vecn_(1).vecn_(2)=0impliesa_(1)a_(2)+b_(1)b_(2)+c_(1)c_(2)=0`
If the planes are parallel then `vecn_(1)` and `vecn_(2)` are parallel.
`:.(a_(1))/(a_(2))=(b_(1))/(b_(2))=(c_(1))/(c_(2))`
