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.
$$\begin{aligned} &\text { Given points, } A(1,-1,3), B(2,-4,5) \text { and } C(5,-13,11) \text {. }\\ &\begin{aligned} A B & =\sqrt{(1-2)^2+(-1+4)^2+(3-5)^2} \\ & =\sqrt{1+9+4}=\sqrt{14} \\ B C & =\sqrt{(2-5)^2+(-4+13)^2+(5-11)^2} \\ & =\sqrt{9+81+36}=\sqrt{126} \\ A C & =\sqrt{(1-5)^2+(-1+13)^2+(3-11)^2} \\ & =\sqrt{16+144+64}=\sqrt{224} \end{aligned} \end{aligned}$$
$$\begin{aligned} &\begin{array}{ll} \because & A B+B C=A C \\ \Rightarrow & \sqrt{14}+\sqrt{126}=\sqrt{224} \\ \Rightarrow & \sqrt{14}+3 \sqrt{14}=4 \sqrt{14} \end{array}\\ &\text { So, the points } A, B \text { and } C \text { are collinear. } \end{aligned}$$
Three consecutive vertices of a parallelogram $A B C D$ are $A(6,-2,4)$, $B(2,4,-8)$ and $C(-2,2,4)$. Find the coordinates of the fourth vertex.
Let the coordinates of the fourth vertices $D(x, y, z)$.
$$\begin{aligned} \text{Mid-points of diagonal }AC, \quad & x=\frac{x_1+x_2}{2}, y=\frac{y_1+y_2}{2}, z=\frac{z_1+z_2}{2} \\ \text{and}\quad & x=\frac{6-2}{2}=2, y=\frac{-2+2}{2}=0, z=\frac{4+4}{2}=4 \end{aligned}$$
Since, the mid-point of $A C$ are $(2,0,4)$.
Now, mid-point of $B D, 2=\frac{x+2}{2} \Rightarrow x=2$
$$\begin{array}{ll} \Rightarrow & 0=\frac{y+4}{2} \Rightarrow y=-4 \\ \Rightarrow & 4=\frac{z-8}{2} \Rightarrow z=16 \end{array}$$
So, the coordinates of fourth vertex $D$ is $(2,-4,16)$.
Show that the $\triangle A B C$ with vertices $A(0,4,1), B(2,3,-1)$ and $C(4,5,0)$ is right angled.
Given that, the vertices of the $\triangle A B C$ are $A(0,4,1), B(2,3,-1)$ and $C(4,5,0)$.
Now,
$$\begin{aligned} A B & =\sqrt{(0-2)^2+(4-3)^3+(1+1)^2} \\ & =\sqrt{4+1+4}=3 \\ B C & =\sqrt{(2-4)^2+(3-5)^2+(-1-0)^2} \\ & =\sqrt{4+4+1}=3 \\ A C & =\sqrt{(0-4)^2+(4-5)^2+(1-0)^2} \\ & =\sqrt{16+1+1}=\sqrt{18} \end{aligned}$$
$$\begin{aligned} & \because \quad A C^2=A B^2+B C^2 \\ & \Rightarrow \quad 18=9+9 \end{aligned}$$
Hence, vertices $\triangle A B C$ is a right angled triangle.
Find the third vertex of triangle whose centroid is origin and two vertices are $(2,4,6)$ and $(0,-2,5)$.
Let third vertex of $\Delta ABC$ i.e., is $A(x,y,z)$.
Given that, the coordinate of centroid $G$ are $(0,0,0)$.
$$\begin{aligned} \because \quad 0 & =\frac{x+2+0}{3} \Rightarrow x=-2 \\ 0 & =\frac{y+4-2}{3} \Rightarrow y=-2 \\ 0 & =\frac{z+6-5}{2} \Rightarrow z=-1 \end{aligned}$$
Hence, the third vertex of triangle is $(-2,-2,-1)$.