Question

find the equation of the lines in which the plane 3x+4y+z=0 cuts the cone 15x^2-32y^2-7z^2=0​

Answer 1

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MATHS

An angle between the plane, x+y+z=5 and the line of intersection of the planes, 3x+4y+z−1=0 and 5x+8y+2z+14=0, is

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VIDEO EXPLANATION

ANSWER

Normal to 3x+4y+z=1 is 3

i

^

+4

j

^

+

k

^

.

Normal to 5x+8y+2z=−14 is 5

i

^

+8

j

^

+2

k

^

The line at which these planes intersect is perpendicular to both normals, hence its direction ratios are directly proportional to the cross product vector of the normals

So, the direction ratios of the line can be chosen as −

j

^

+4

k

^

So, the angle between the plane x+y+z+5=0 and the line obtained is given by sin

−1

17

3

−1+4

=sin

−1

17

3

So, option D is the correct answer.