If P \left(\right. - 3 , - 1 , 6 \left.\right) be any point in space, then find the(a ) distance of P from Y-axis. (b ) distance of P from X Z-plane.(c ) Image of P with respect to X Y-plane.(d ) Octant in which P lies.

Engineering

Mathematics

Line of Greatest Slope

Question

P (- 3, - 1, 6)

be any point in space, then find the
(a ) distance of

P

from

Y

-axis.
(b ) distance of

P

from

X Z

-plane.
(c ) Image of

P

with respect to

X Y

-plane.
(d ) Octant in which

P

lies.

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(a)

since, distance of a point

P (a, b, c)

from

y - a x i s = \sqrt{a^{2} + z^{2}}

so , here distance of point

P (- 3, - 1, 6)

from

y - a x i s = \sqrt{(- 3)^{2} + (6)^{2}}

= \sqrt{9 + 36} = \sqrt{45} = 3 \sqrt{5}

units

(b)

given point is

P (- 3, - 1, 6)

If we draw a perpendicular from the point

P (- 3, - 1, 6)

x z - p l a n e

then the foot of the perpendicular is

(- 3, 0, 6)

Now the distance of the point

P (- 3, - 1, 6)

from the

x z - p l a n e

is the distance between the ponts

P (- 3, - 1, 6)

and

(- 3, 0, 6)

and that is

\sqrt{(- 3 + 3)^{2} + (- 1 - 0)^{2} + (6 - 6)^{2}} = \sqrt{0 + 1 + 0} = 1

unit

(c)

If a point has coordinates

(x, y, z)

, then the image of this point in the

X Y - p l a n e

is given by

(x, y, - z)

so , here Image of

P (- 3, - 1, 6)

with respect to

X Y - p l a n e

is given by

(- 3, - 1, - 6)

(d)

The

x - c o o r d i n a t e

y - c o o r d i n a t e

and

z - c o o r d i n a t e

of point

(- 3, - 1, 6)

are negative

negative and positive respectively. Therefore this point lies in octant

I I I

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