Show that the points
Question:

Show that the points A(1, 1, 1), B(-2, 4, 1), C(1, -5, 5) and D(2, 2, 5) are the vertices of a square.

Solution:

To prove: Points A, B, C, D form square.

Formula: The distance between two points $\left(x_{1}, y_{1}, z_{1}\right)$ and $\left(x_{2}, y_{2}, z_{2}\right)$ is given by

$\mathrm{D}=\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}+\left(\mathrm{z}_{2}-\mathrm{z}_{1}\right)^{2}}$

Here

$\left(x_{1}, y_{1}, z_{1}\right)=(1,1,1)$

$\left(x_{2}, y_{2}, z_{2}\right)=(-2,4,1)$

$\left(x_{3}, y_{3}, z_{3}\right)=(-1,5,5)$

$\left(x_{4}, y_{4}, z_{4}\right)=(2,2,5)$

Length $\mathrm{AB}=\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}+\left(\mathrm{z}_{2}-\mathrm{z}_{1}\right)^{2}}$

$=\sqrt{(-2-1)^{2}+(4-1)^{2}+(1-1)^{2}}$

$=\sqrt{(-3)^{2}+(3)^{2}+(0)^{2}}$

$=\sqrt{9+9+0}$

$=\sqrt{18}$

Length $B C=\sqrt{\left(x_{3}-x_{2}\right)^{2}+\left(y_{3}-y_{2}\right)^{2}+\left(z_{3}-z_{2}\right)^{2}}$

$=\sqrt{(-1+2)^{2}+(5-4)^{2}+(5-1)^{2}}$

$=\sqrt{(1)^{2}+(1)^{2}+(4)^{2}}$

$=\sqrt{1+1+16}$

$=\sqrt{18}$

Length $C D=\sqrt{\left(x_{4}-x_{3}\right)^{2}+\left(y_{4}-y_{3}\right)^{2}+\left(z_{4}-z_{3}\right)^{2}}$

$=\sqrt{(2+1)^{2}+(2-5)^{2}+(5-5)^{2}}$

$=\sqrt{(3)^{2}+(3)^{2}+(0)^{2}}$

$=\sqrt{9+9+0}$

$=\sqrt{18}$

Length $A D=\sqrt{\left(x_{4}-x_{1}\right)^{2}+\left(y_{4}-y_{1}\right)^{2}+\left(z_{4}-z_{1}\right)^{2}}$

$=\sqrt{(2-1)^{2}+(2-1)^{2}+(5-1)^{2}}$

$=\sqrt{(1)^{2}+(1)^{2}+(4)^{2}}$

$=\sqrt{1+1+16}$

$=\sqrt{18}$

Length $A C=\sqrt{\left(x_{3}-x_{1}\right)^{2}+\left(y_{3}-y_{1}\right)^{2}+\left(z_{3}-z_{1}\right)^{2}}$

$=\sqrt{(-1-1)^{2}+(5-1)^{2}+(5-1)^{2}}$

$=\sqrt{(-2)^{2}+(4)^{2}+(4)^{2}}$

$=\sqrt{4+16+16}$

$=\sqrt{36}$

Length $B D=\sqrt{\left(x_{4}-x_{2}\right)^{2}+\left(y_{4}-y_{2}\right)^{2}+\left(z_{4}-z_{2}\right)^{2}}$

$=\sqrt{(2+2)^{2}+(2-4)^{2}+(5-1)^{2}}$

$=\sqrt{(4)^{2}+(-2)^{2}+(4)^{2}}$

$=\sqrt{16+4+16}$

$=\sqrt{36}$

Here, AB = BC = CD = AD

Also, AC = BD

This means all the sides are the same and diagonals are also equal.

Hence vertices A, B, C, D form a square.

 

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