Question

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

Answer 1

$\left(a\right)$

since, distance of a point $P(a,b,c)$ from $y-axis=\sqrt{a^2+z^2}$

so , here distance of point $P(-3,-1,6)$ from $y-axis=\sqrt{(-3{)}^2+(6{)}^2}$

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

$\left(b\right)$

given point is $P(-3,-1,6)$

If we draw a perpendicular from the point $P(-3,-1,6)$ on $xz-plane$ then the foot of the perpendicular is $(-3,0,6)$

Now the distance of the point $P(-3,-1,6)$ from the $xz-plane$ 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

$\left(c\right)$

If a point has coordinates $(x,y,z)$, then the image of this point in the $XY-plane$ is given by $(x,y,-z)$

so , here Image of $P(-3,-1,6)$ with respect to $XY-plane$ is given by $(-3,-1,-6)$

$\left(d\right)$

The $x-coordinate$, $y-coordinate$ and $z-coordinate$ of point $(-3,-1,6)$ are negative

negative and positive respectively. Therefore this point lies in octant $III$