Prove that the points $(0,-1,-7),(2,1,-9)$ and $(6,5,-13)$ are collinear. Find the ratio in which the first point divides the join of the other two.
Given points are $A(0,-1,-7), B(2,1,-9)$ and $C(6,5,-13)$
$$\begin{aligned} & A B=\sqrt{(0-2)^2+(-1-1)^2+(-7+9)^2}=\sqrt{4+4+4}=2 \sqrt{3} \\ & B C=\sqrt{(2-6)^2+(1-5)^2+(-9+13)^2}=\sqrt{16+16+16}=4 \sqrt{3} \\ & A C=\sqrt{(0-6)^2+(-1-5)^2+(-7+13)^2}=\sqrt{36+36+36}=6 \sqrt{3} \end{aligned}$$
$$\begin{aligned} &\begin{array}{ll} \because & A B+B C=2 \sqrt{3}+4 \sqrt{3}=6 \sqrt{3} \\ \text { So, } & A B+B C=A C \end{array}\\ &\text { Hence, the points } A, B \text { and } C \text { are collinear. } \end{aligned}$$
$$\begin{aligned} &A B: A C=2 \sqrt{3}: 6 \sqrt{3}=1: 3\\ &\text { So, point } A \text { divide } B \text { and } C \text { in 1: 3 externally. } \end{aligned}$$
What are the coordinates of the vertices of a cube whose edge is 2 units, one of whose vertices coincides with the origin and the three edges passing through the origin, coincides with the positive direction of the axes through the origin?
The coordinates of the cube which edge is 2 units, are $(2,0,0),(2,2,0),(0,2,0)$, $(0,2,2),(0,0,2),(2,0,2),(0,0,0)$ and $(2,2,2)$.
The distance of point $P(3,4,5)$ from the $Y Z$-plane is
What is the length of foot of perpendicular drawn from the point $P$ $(3,4,5)$ on $Y$-axis?
Distance of the point $(3,4,5)$ from the origin $(0,0,0)$ is