P is a point on the plane lx + my + nz = p and a point Q is taken on the line such that OP. OQ= p^2 ; show that the locus of point Q is p(lx + my + nz)= x^2 + y^2 + z^2.
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Expert's answer
2019-09-02T09:19:23-0400
Let P point be (α,β,γ) and Q point be(x1,y1,z1)
direction ratios of OP are α,β,γ and direction ratios of OQ are x1,y1,z1.
Since O,Q,P are collinear, we have
x1α=y1β=z1γ=k......(1)
As P(α,β,γ) lies on the plane lx+my+nz=p,
So,
lα+mβ+nγ=p
or k(x1+my1+nz1)=p........(2)
Given, OP× OQ =p2
∴α2+β2+γ2(x1)2+(y1)2+(z12)=p2
or, k2(x12+y12+z12)(x1)2+(y1)2+(z12)=p2
or, k(x12+y12+z12)=p2......(3)
On dividing (2) by(3) ,we get
x12+y12+z12lx1+my1+nz1=p1
or, p(lx1+my1+nz1)=x12+y12+z12
Hence the locus of point Q is p(lx+my+nz)=x2+y2+z2.
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Kanha Sharma
02.09.19, 16:36
Thanks a lot for the solution.
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Dear Kanha Sharma, You are welcome. We are glad to be helpful. If you liked our service, please press a like-button beside the answer field. Thank you!
Thanks a lot for the solution.
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