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Question
Find the perpendicular distance of the point (3, −1, 11) from the line \[\frac{x}{2} = \frac{y - 2}{- 3} = \frac{z - 3}{4} .\]
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Solution
Given equation of line AB is
`x/2=("y"-2)/3=("z"-3)/4=lambda`[say]
or `x/2=lambda,("y"-2)/3=lambda`
and `("z"-3)/4=lambda`
or x = 2λ, y = 3λ + 2
and z = 4λ + 3
∴ Any point P on the given line
= (2λ, 3λ + 2, 4λ + 3)
Let P be the foot of Perpendicular drawn from point
Q(3, -1, 11) on line AB. Now, DR's of line
QP = (2λ - 3, 3λ + 2 + 1, 4λ + 3 - 11)
= (2λ - 3, 3λ + 3, 4λ - 8)
Here, a1 = 2λ -3, b1 = 3λ + 3, c1 = 4λ - 8,
and a2 = 2, b2 = 3, c2 = 4
Since, QP ⊥ AB
∴ We have, a1a2 + b1b2 + c1c2 = 0 ...(i)
or 2(2λ - 3) + 3 (3λ + 3) + 4 (4λ - 8) = 0
or 4λ - 6 + 9λ + 9 + 16λ - 32 = 0
or 29λ - 29 = 0 or 29λ = 29 or λ = 1
∴ Foot of Perpendicular
P = (2, 3 + 2, 4 + 3)
= (2, 5, 7)
Now, equation of Perpendicular QP, Where Q (3, -1, 11) and P (2, 5, 7) is
`(x-3)/(2-3)=("y"+1)/(5+1)=("z"-11)/(7-11)`
`[("using two points form of equation of line"), ("i"."e". (x-x_1)/(x_2-x_1)=("y"-"y"_1)/("y"_2-"y"_1)=("z"-"z"_1)/("z"_2-"z"_1))]`
or `(x-3)/-1=("y"+1)/6=("z"-11)/-4`
Now, length of Perpendicular QP = distance between points Q(3, -1, 11) and P (2, 5, 7)
= `sqrt((2-3)^2+(5+1)^2+(7-11)^2)`
`[(∵ "distance"), (=sqrt((x_2-x_1)^2+("y"_2-"y"_1)^2+("z"_2-"z"_1)^2))]`
= `sqrt(1+36+16)`
= `sqrt53`
Hence, length of Perpendicular is `sqrt53`.
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