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Find the foot of perpendicular from the point (2,3,–8) to the line 4-x2=y6=1-z3. Also, find the perpendicular distance from the given point to the line. - Mathematics

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प्रश्न

Find the foot of perpendicular from the point (2,3,–8) to the line `(4 - x)/2 = y/6 = (1 - z)/3`. Also, find the perpendicular distance from the given point to the line.

योग

उत्तर

Let the given equation be  `(4 - x)/2 = y/6 = (1 - z)/3 = lambda`

This can be written as `(x - 4)/(-2) = y/6 = (z - 1)/(-3) = lambda`  .......(1)

∴ The coordinates of any point on the line is x = 4 − 2λ, y = 6λ, z = 1 − 3λ

Let Q(4 − 2λ, 6λ, 1 − 3λ) be the foot of perpendicular from the point P(2, 3, −8) on line  ......(1)

We know the direction ratios of any line segement PQ is given by (x2 ​− x1​, y2 ​− y1​, z2 ​− z1​)

The direction cosines of PQ is given by

= (−2λ + 4 − 2, 6λ − 3, −3λ + 1 + 8)

= (−2λ + 2, 6λ − 3, −3λ + 9)

Now Q is the foot of the perpendicular of the line (1)

`vec(PQ)` ​is the perpendicular to the line (1)

Hence the sum of the product of this direction ratios is 0

= (−2λ + 2)(−2) + (6λ − 3) . 6 + (−3λ + 9)(−3) = 0

⇒ 4λ − 4 + 36λ − 18 + 9λ − 27 = 0

⇒ 49λ − 49 = 0
∴ λ = 1
Substituting λ = 1 in Q we get
Q(0, 3, 6)
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अध्याय 12: Introduction to Three Dimensional Geometry - Exercise [पृष्ठ २३६]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
अध्याय 12 Introduction to Three Dimensional Geometry
Exercise | Q 16 | पृष्ठ २३६

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