Question

Find the angle between the lines of intersection of the plane x-3y +z=0and the cone x^-5y^+z^=0

Answer 1

Answer:

The angle between the lines of intersection of the plane x-3y + z = 0 and a quadric cone x^2 - 5y^2 + 2 = 0 is 360°.

Answer 2

$\sf \purple{by \: vector \: method}$

$\blue{ \boxed{ \sf \theta = \cos ^{ - 1} ( \frac{n_ 1 .n_2}{ | n_1 || n_2 | } ) }}$

$\sf n_1 = i - 3j + k$

$\sf n_2 = i - 5j + k$

$\sf | n_1 | = \sqrt{ {1}^{2} + {( - 3)}^{2} + {1}^{2} } = \sqrt{11}$

$\sf | n_1 | = \sqrt{ {1}^{2} + {( -5)}^{2} + {1}^{2} } = \sqrt{27}$

$\implies \sf \theta = \cos ^{ - 1} ( \frac{(1)(1) + ( - 3)( - 5) + (1)(1)}{ \sqrt{11} \sqrt{27} })$

$\orange{ \therefore\sf \theta = \cos ^{ - 1} ( \frac{17}{ \sqrt{297} })}$

HOPE IT HELPS!!