Question

समतल 3x+y+5z = 0 और शंकु 6yz - 2zx + 5xy = 0 क प्रतिच्छेद रेखाओं के बीच का
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Answer 1

Find the angles between the line of section of the plane 3x+y+5z=0 and the cone 6yz-2zx+5xy = 0.

θ = cos-1 (1/6)

Given

the plane 3x+y+5z=0

the cone 6yz-2zx+5xy = 0

To find

The angles between the line of section of the plane 3x+y+5z=0 and the cone 6yz-2zx+5xy = 0

Solution:

Vertex of the given cone is is the origin O(0, 0,0) .

Also the given plane passes through the vertex.

Hence, the lines of section also passes through the origin.

Let one of the lines be,

• x/l = y/m = z/n .... (1)

The general point on (1) is (kl, km, kn)

This lines on the plane 3x+y+5z=0 if 3l+m+5n=0 gives

• m = -(3l+5n)

Also, the point lies on 6yz-2zx+5xy = 0 gives

• 6mn-2nl+5lm=0
• -6(3l+5n)n-2nl-5l(3l+5n) = 0
• -18ln-30n²-2nl-15l²-25ln = 0
• 15l²+45ln+30n²=0
• l²+3ln+2n² = 0
• (l+2n) (l+n) = 0
• l = -2n Or l = -n

Therefore,

• m = n Or m = -2n

dr's of the line are,

• (-2n, n, n) & ( -n, -2n, n) or (-2,1,1) & (-1,-2,1)

The angle between them is given by

• cos θ = (-2)(-1) + 1(-2)+1(1) / (√(-2)²+1²+1² ) (√(-1)²+(-2)²+1²)
• cos θ = 1/6
• θ = cos-1 (1/6)

Hence, the angles between the line of section of the plane 3x+y+5z=0 and the cone 6yz-2zx+5xy = 0 is θ = cos-1 (1/6)

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