Question
The angle between the pair of planes represented by equation $$2{x^2} - 2{y^2} + 4{z^2} + 6xz + 2yz + 3xy = 0{\text{ is :}}$$
A.
$${\text{co}}{{\text{s}}^{ - 1}}\left( {\frac{1}{3}} \right)$$
B.
$${\text{co}}{{\text{s}}^{ - 1}}\left( {\frac{4}{{21}}} \right)$$
C.
$${\text{co}}{{\text{s}}^{ - 1}}\left( {\frac{4}{9}} \right)$$
D.
$${\text{co}}{{\text{s}}^{ - 1}}\left( {\frac{7}{{\sqrt {84} }}} \right)$$
Answer :
$${\text{co}}{{\text{s}}^{ - 1}}\left( {\frac{4}{9}} \right)$$
Solution :
$$\eqalign{
& 2{x^2} - 2{y^2} + 4{z^2} + 6xz + 2yz + 3xy = 0 \cr
& {\text{or, }}2{x^2} + x\left( {6z + 3y} \right) - 2{y^2} + 4{z^2} + 2yz = 0 \cr
& \Rightarrow x = \frac{{ - \left( {6z + 3y} \right) \pm \sqrt {36{z^2} + 9{y^2} + 36yz - 8\left( { - 2{y^2} + 4{z^2} + 2yz} \right)} }}{4} \cr
& \Rightarrow x = \frac{{ - \left( {6z + 3y} \right) \pm \sqrt {{{\left( {2z + 5y} \right)}^2}} }}{4} \cr
& \Rightarrow x = \frac{{ - \left( {6z + 3y} \right) \pm \left( {2z + 5y} \right)}}{4} \cr
& {\text{or, }}2x - y + 2z = 0,\,x + 2y + 2z = 0 \cr
& \therefore {\text{ Angle between planes is }}{\cos ^{ - 1}}\left( {\frac{4}{9}} \right). \cr} $$