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Question -

In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.
(a) 2x + 3y + 4z тАУ 12 = 0
(b) 3y + 4z тАУ 6 = 0
(c) x + y + z = 1
(d) 5y + 8 = 0



Answer -

(a)┬а2x + 3y + 4z тАУ 12 = 0

Let the coordinate ofthe foot of┬атКе┬аP from theorigin to the given plane be P(x, y, z).

2x + 3y + 4z = 12 тАж.(1)

Direction ratio are(2, 3, 4)

тИЪ[(2)2┬а+(3)2┬а+ (4)2] =┬атИЪ(4 + 9 + 16)

=┬атИЪ29

Now,

Divide both the sidesof equation (1) by┬атИЪ29, weget

2x/(тИЪ29) + 3y/(тИЪ29) + 4z/(тИЪ29) =12/тИЪ29

So this is of the formlx + my + nz = d

Where, l, m, n are thedirection cosines and d is the distance

тИ┤ The direction cosinesare 2/тИЪ29, 3/тИЪ29, 4/тИЪ29

Coordinate of the foot(ld, md, nd) =

= [(2/тИЪ29) (12/тИЪ29), (3/тИЪ29) (12/тИЪ29), (4/тИЪ29) (12/тИЪ29)]

= 24/29, 36/29, 48/29

(b)┬а3y + 4z тАУ 6 = 0

Let the coordinate ofthe foot of┬атКе┬аP from theorigin to the given plane be P(x, y, z).

0x + 3y + 4z = 6 тАж.(1)

Direction ratio are(0, 3, 4)

тИЪ[(0)2┬а+(3)2┬а+ (4)2] =┬атИЪ(0 + 9 + 16)

=┬атИЪ25

= 5

Now,

Divide both the sidesof equation (1) by 5, we get

0x/(5) + 3y/(5) +4z/(5) = 6/5

So this is of the formlx + my + nz = d

Where, l, m, n are thedirection cosines and d is the distance

тИ┤ The direction cosinesare 0/5, 3/5, 4/5

Coordinate of the foot(ld, md, nd) =

= [(0/5) (6/5), (3/5)(6/5), (4/5) (6/5)]

= 0, 18/25, 24/25

(c)┬аx + y + z = 1

Let the coordinate ofthe foot of┬атКе┬аP from theorigin to the given plane be P(x, y, z).

x + y + z = 1 тАж. (1)

Direction ratio are(1, 1, 1)

тИЪ[(1)2┬а+(1)2┬а+ (1)2] =┬атИЪ(1 + 1 + 1)

=┬атИЪ3

Now,

Divide both the sidesof equation (1) by┬атИЪ3, weget

1x/(тИЪ3) + 1y/(тИЪ3) + 1z/(тИЪ3) = 1/тИЪ3

So this is of the formlx + my + nz = d

Where, l, m, n are thedirection cosines and d is the distance

тИ┤ The direction cosinesare 1/тИЪ3, 1/тИЪ3, 1/тИЪ3

Coordinate of the foot(ld, md, nd) =

= [(1/тИЪ3) (1/тИЪ3), (1/тИЪ3) (1/тИЪ3), (1/тИЪ3) (1/тИЪ3)]

= 1/3, 1/3, 1/3

(d)┬а5y + 8 = 0

Let the coordinate ofthe foot of┬атКе┬аP from theorigin to the given plane be P(x, y, z).

0x тАУ 5y + 0z = 8 тАж.(1)

Direction ratio are(0, -5, 0)

тИЪ[(0)2┬а+(-5)2┬а+ (0)2] =┬атИЪ(0 + 25 + 0)

=┬атИЪ25

= 5

Now,

Divide both the sidesof equation (1) by 5, we get

0x/(5) тАУ 5y/(5) +0z/(5) = 8/5

So this is of the formlx + my + nz = d

Where, l, m, n are thedirection cosines and d is the distance

тИ┤ The direction cosinesare 0, -1, 0

Coordinate of the foot(ld, md, nd) =

= [(0/5) (8/5), (-5/5)(8/5), (0/5) (8/5)]

= 0, -8/5, 0

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