If $A, B, C$ be the feet of perpendiculars from a point $P$ on the $X, Y$ and $Z$-axes respectively, then find the coordinates of $A, B$ and $C$ in each of the following where the point $P$ is
(i) $\mathrm{A}(3,4,2)$
(ii) $B(-5,3,7)$
(iii) $C(4,-3,-5)$
The coordinates of $A, B$ and $C$ are the following
(i) $A(3,0,0), B(0,4,0), C(0,0,2)$
(ii) $A(-5,0,0), B(0,3,0), C(0,0,7)$
(iii) $A(4,0,0), B(0,-3,0), C(0,0,-5)$
If $A, B$, and $C$ be the feet of perpendiculars from a point $P$ on the $X Y, Y Z$ and $Z X$-planes respectively, then find the coordinates of $A, B$ and $C$ in each of the following where the point $P$ is
(i) $(3,4,5)$
(ii) $(-5,3,7)$
(iii) $(4,-3,-5)$
We know that, on $X Y$-plane $z=0$, on $Y Z$-plane, $x=0$ and on $Z X$-plane, $y=0$. Thus, the coordinate of $A, B$ and $C$ are following
(i) $A(3,4,0), B(0,4,5), C(3,0,5)$
(ii) $A(-5,3,0), B(0,3,7), C(-5,0,7)$
(iii) $A(4,-3,0), B(0,-3,-5), C(4,0,-5)$
How far apart are the points $(2,0,0)$ and $(-3,0,0)$ ?
$$ \begin{aligned} &\text { Given points, } A(2,0,0) \text { and } B(-3,0,0)\\ &A B=\sqrt{(2+3)^2+0^2+0^2}=5 \end{aligned}$$
6 Find the distance from the origin to $(6,6,7)$.
Distance from origin to the point $(6,6,7)$
$$\begin{aligned} & =\sqrt{(0-6)^2+(0-6)^2+(0-7)^2} \quad\left[\because d=\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2+\left(z_1-z_2\right)^2}\right] \\ & =\sqrt{36+36+49} \\ & =\sqrt{121}=11 \end{aligned}$$
Show that, if $x^2+y^2=1$, then the point $\left(x, y, \sqrt{1-x^2-y^2}\right)$ is at a distance 1 unit form the origin.
Given that, $x^2+y^2=1$
$\therefore$ Distance of the point $\left(x, y, \sqrt{1-x^2-y^2}\right)$ from origin is given as
$$\begin{aligned} d & =\left|\sqrt{x^2+y^2+\left(\sqrt{1-x^2-y^2}\right)^2}\right| \\ & =\left|\sqrt{x^2+y^2+1-x^2-y^2}\right|=1 \end{aligned}$$
Hence proved.