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प्रश्न
Find the vector equation of the plane that contains the lines `vecr = (hat"i" + hat"j") + λ (hat"i" + 2hat"j" - hat"k")` and the point (–1, 3, –4). Also, find the length of the perpendicular drawn from the point (2, 1, 4) to the plane thus obtained.
उत्तर
Let the vector equation of the required plane be `vec"r" . vec"n" = d`
The plane contains the line `vec"r" = hat"i" + hat"j" + λ (hat"i" + 2hat"j" - hat"k")`
Since the plane passes through point A and B. So `vec"n"` will be parallel to vector `vec"AB" xx (hat"i" + 2hat"j" - hat"k")`
`vec"AB" = vec"OB" -vec"OA"`
= `(-hat"i" + 3hat"j" - 4hat"k") (hat"i" + hat"j")`
= `-2hat"i" + 2hat"j" - 4hat"k"`
`vec"AB" xx (hat"i" + 2hat"j" - hat"k") = |(hat"i",hat"j",hat"k"),(1,2,-1),(-2,2, -4)|`
= `hat"i" (-8 + 2) - hat"j" (-4-2) + hat"k" (2+4)`
= `-6hat"i" + 6hat"j" + 6hat"k"`
which is a normal vector to the plane.
So the equation of plane will be `vec"r" . (-6hat"i" + 6hat"j" + 6hat"k") = d`
∴ it passes through (1, 1, 0) so `(hat"i" + hat"j"). (-6hat"i" + 6hat"j" + 6hat"k") = d or, d = 0`
equation of plane is `vecr . (-6hat"i" + 6hat"j" + 6hat"k") = 0`
`vecr (hat"i" - hat"j" -hat"k") = 0`
in Cartesian plane,
`(xhat"i" + yhat"j" + zhat"k") . (hat"i" - hat"j" -hat"k") = 0`
x -y - z = 0
So, the perpendicular distance of the plane from the point (2, 1, 4) is ` = |(2 -1-4)/sqrt(1^2 + (-1)^2 + (-1)^2)| = |(-3)/sqrt(3)| = sqrt(3) "unit"`.
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