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.
Show that the point $A(1,-1,3), B(2,-4,5)$ and $C(5,-13,11)$ are collinear.
$$\begin{aligned} &\text { Given points, } A(1,-1,3), B(2,-4,5) \text { and } C(5,-13,11) \text {. }\\ &\begin{aligned} A B & =\sqrt{(1-2)^2+(-1+4)^2+(3-5)^2} \\ & =\sqrt{1+9+4}=\sqrt{14} \\ B C & =\sqrt{(2-5)^2+(-4+13)^2+(5-11)^2} \\ & =\sqrt{9+81+36}=\sqrt{126} \\ A C & =\sqrt{(1-5)^2+(-1+13)^2+(3-11)^2} \\ & =\sqrt{16+144+64}=\sqrt{224} \end{aligned} \end{aligned}$$
$$\begin{aligned} &\begin{array}{ll} \because & A B+B C=A C \\ \Rightarrow & \sqrt{14}+\sqrt{126}=\sqrt{224} \\ \Rightarrow & \sqrt{14}+3 \sqrt{14}=4 \sqrt{14} \end{array}\\ &\text { So, the points } A, B \text { and } C \text { are collinear. } \end{aligned}$$