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प्रश्न
The plane ax + by = 0 is rotated about its line of intersection with the plane z = 0 through an angle α. Prove that the equation of the plane in its new position is `"a"x + "b"y +- (sqrt("a"^2 + "b"^2) tan alpha)`z = 0.
उत्तर
Given planes are:
ax + by = 0 ......(i)
z = 0 ......(ii)
Equation of any plane passing through the line of intersection of plane (i) and (ii) is
(ax + by) + kz = 0
⇒ ax + by + kz = 0 ......(iii)
Dividing both sides by `sqrt("a"^2 + "b"^2 + "k"^2)`, we get
`"a"/sqrt("a"^2 + "b"^2 + "k"^2) x + "b"/sqrt("a"^2 + "b"^2 + "k"^2) y + "k"/sqrt("a"^2 + "b"^2 + "k"^2) z` = 0
∴ Direction cosines of the normal to the plane are
`"a"/sqrt("a"^2 + "b"^2 + "k"^2)`
`"b"/sqrt("a"^2 + "b"^2 + "k"^2)`
`"k"/sqrt("a"^2 + "b"^2 + "k"^2)`
And the direction cosines of the plane (i) are
`"a"/sqrt("a"^2 + "b"^2), "b"/sqrt("a"^2 + "b"^2)`, 0
Since, a is the angle between the planes (i) and (iii), we get
`cos alpha = ("a"."a" + "b"."b" + "k".0)/(sqrt("a"^2 + "b"^2 + "k"^2)*sqrt("a"^2 + "b"^2)`
⇒ `cos alpha = ("a"^2 + "b"^2)/(sqrt("a"^2 + "b"^2 + "k"^2)*sqrt("a"^2 + "b"^2)`
⇒ `cos alpha = sqrt("a"^2 + "b"^2)/sqrt("a"^2 + "b"^2 + "k"^2)`
⇒ `cos^2alpha = ("a"^2 + "b"^2)/sqrt("a"^2 + "b"^2 + "k"^2)`
⇒ (a2 + b2 + k2) cos2α = a2 + b2
⇒ a2 cos2α + b2 cos2α + k2 cos2α = a2 + b2
⇒ k2 cos2α = a2 – a2 cos2α + b2 – b2 cos2α
⇒ k2 cos2α = a2(1 – cos2α) + b2(1 – cos2α)
⇒ k2 cos2α = a2 sin2α + b2 sin2α
⇒ k2 cos2α = (a2 + b2) sin2α
⇒ `"k"^2 = ("a"^2 + "b"^2) (sin^2 alpha)/(cos^2 alpha)`
⇒ k = `+- sqrt("a"^2 + "b"^2) . tan alpha`
Putting the value of k in equation (iii) we get
`"a"x + "b"y +- (sqrt("a"^2 + "b"^2) tan alpha)`z = 0 which is the required equation of plane.
Hence proved.
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