Question

The angle between the planes represented by $2x^2-6y^2-12z^2+18yz+2zx+xy=0$ is

• A. ${\cos }^{-1}\frac{16}{21}$
• B. ${\cos }^{-1}\frac{17}{21}$
• C. ${\cos }^{-1}\frac{19}{21}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is A ${\cos }^{-1}\frac{16}{21}$

Let $2x^2-6y^2-12z^2+18yz+2zx+xy=(a_{1}x+b_{1}y+c_{1}z)(a_{2}x+b_{2}y+c_{2}z)=0.$
Here $a_1{a}_2=2,b_1{b}_2=-6,c_1{c}_2=12$and $a_1{b}_2+a_2{b}_1=1,a_1{c}_2+a_2{c}_1=2,b_1{c}_2+b_2{c}_1=18$
On solving, the planes represented by $2x^2-6y^2-12z^2+18yz+2zx+xy=0$ are
$2x-3y+6z=0\cdots \left(i\right)$
$x+2y-2z=0\cdots \left(ii\right)$
If $\theta$ is the angle between the planes $\left(i\right)$ and $\left(ii\right)$ then,
$\cos \theta =\frac{|\left(2\right)\left(1\right)+(-3)\left(2\right)+\left(6\right)(-2)|}{\sqrt{2^2+(-3{)}^2+6^2}\sqrt{1^2+2^2+(-2{)}^2}}=\frac{16}{21}$
$\therefore \theta ={\cos }^{-1}\frac{16}{21}$