Introduction to Three Dimensional Geometry Test - 36

$$OABC$$ is a tetrahedron such that  $$OA^{2}+BC^{2}=OB^{2}+CA^{2}=OC^{2}+AB^{2}$$
Let $$O(0,0,0),A(x_{1},y_{1},z_{1}),B(x_{2},y_{2},z_{2}),C(x_{3},y_{3},z_{3})$$ are coordinates of vertices.
$$OA^{2}+BC^{2}$$ = $$(x_{1}^{2}+y_{1}^{2}+z_{1}^{2})+((x_{3}-x_{2})^{2}+(y_{3}-y_{2})^{2}+(z_{3}-z_{2})^{2})$$
                  = $$(x_{1}^{2}+y_{1}^{2}+z_{1}^{2})+(x_{2}^{2}+y_{2}^{2}+z_{2}^{2})+(x_{3}^{2}+y_{3}^{2}+z_{3}^{2})-2(x_{2}x_{3}+y_{2}y_{3}+z_{2}z_{3})$$
Similarly,
$$OB^{2}+CA^{2}$$ = $$(x_{1}^{2}+y_{1}^{2}+z_{1}^{2})+(x_{2}^{2}+y_{2}^{2}+z_{2}^{2})+(x_{3}^{2}+y_{3}^{2}+z_{3}^{2})-2(x_{1}x_{3}+y_{1}y_{3}+z_{1}z_{3})$$
$$OC^{2}+AB^{2}$$ = $$(x_{1}^{2}+y_{1}^{2}+z_{1}^{2})+(x_{2}^{2}+y_{2}^{2}+z_{2}^{2})+(x_{3}^{2}+y_{3}^{2}+z_{3}^{2})-2(x_{1}x_{2}+y_{1}y_{2}+z_{1}z_{2})$$
$$\therefore$$ $$OA^{2}+BC^{2}=OB^{2}+CA^{2}\Rightarrow x_{2}x_{3}+y_{2}y_{3}+z_{2}z_{3} = x_{1}x_{3}+y_{1}y_{3}+z_{1}z_{3}$$
$$\Rightarrow (x_{1}-x_{2})x_{3}+(y_{1}-y_{2})y_{3}+(z_{1}-z_{2})z_{3} = 0$$

For finding the coordinates of the point where the plane $$ax+by+cz-3=0$$ cuts the $$x$$ axis, 

Given, $$BL:LC=1:2$$ and $$DM:MC=1:2$$

 the centroid of the triangle is

Let $$P(x,y,z)$$ be the required point.

Given points are, $$A= \left ( 0,0,2 \right ),B= \left (\sqrt{2},\sqrt{2},2 \right ),C=

\left ( \sqrt{2},\sqrt{2},0 \right )$$ and $$D= \left ( \displaystyle

\frac{8\sqrt{2}-20}{17},\frac{12\sqrt{2}+4}{17},\frac{20-8\sqrt{2}}{17}

\right )$$

Length $$AB = \sqrt{\sqrt{2}^2 + \sqrt{2}^2 + (2-2)^2} = 2$$
Length $$BC = \sqrt{(\sqrt{2}-\sqrt{2})^2 + (\sqrt{2}-\sqrt{2})^2 + 2^2} = 2$$
Length $$CD = \sqrt{\left(

\dfrac{8\sqrt{2}-20}{17}-\sqrt{2}\right)^2 + \left(\dfrac{12\sqrt{2}+4}{17}-\sqrt{2}\right)^2 + \left(\dfrac{20-8\sqrt{2}}{17}\right)^2} = 2$$
Length $$AD = \sqrt{\left(

\dfrac{8\sqrt{2}-20}{17}\right)^2 + \left(\dfrac{12\sqrt{2}+4}{17}\right)^2 + \left(\dfrac{20-8\sqrt{2}}{17} - 2\right)^2} = 2$$

Angle between two vectors $$\overline{AB} = p_{1}\hat{i}+q_{1}\hat{j}+r_{1}\hat{k} = \sqrt{2}\hat{i} + \sqrt{2}\hat{j} + 0\hat{k}$$ and $$\overline{BC} = p_{2}\hat{i}+q_{2}\hat{j}+r_{2}\hat{k} = 0\hat{i} + 0\hat{j} + 2 \hat{k}$$ is $$cos\theta = 0 \Rightarrow \theta = 90^o$$

All sides are equal and angle between $$AB$$ and $$BC$$ is $$90^o$$, Hence, its a square.

Eliminating t from the given equation, we get the equation of the path $$\dfrac{x}{2}=\dfrac{y}{-4}=\dfrac{z}{4}=t $$

The distance between the points $$(1,0,5)$$ and $$(-3,6,3)$$ is given below:

Equation of the plane containing the triangle 
$$ABC$$ is $$\displaystyle \frac{x}{\alpha }+\frac{y}{\beta }+\frac{z}{\gamma }=1$$
which meets the axes in $$\displaystyle \left ( \alpha ,0,0 \right ),\left ( 0,\beta ,0 \right )$$ and $$\displaystyle \left ( 0,0,y \right )$$.
Let the coordinates of $$A$$ be $$\displaystyle \left (x_1,y_1,z_1 \right )$$
Since the middle point of $$AB$$ lies on the $$z$$-axis it is $$\displaystyle \left (0,0,\gamma \right )$$ and thus the coordinates of $$B$$ are $$\displaystyle \left (-x_1,-y_1,2\gamma -z_1 \right )$$
Similarly the coordinates of $$C$$ are $$\displaystyle \left (-x_1,2\beta -y_1,-z_1 \right )$$
So that the middle point of $$\displaystyle BC=\left ( -x_1,\beta -y_1,\gamma -z_1 \right ) \displaystyle =\left ( \alpha ,0,0 \right )$$
$$\displaystyle \Rightarrow x_1=-\alpha ,y_1=\beta ,z_1=\gamma$$.
And thus the coordinates of $$A$$ are $$\displaystyle \left ( -\alpha ,\beta ,\gamma  \right )$$
Similarly the coordinates of $$B$$ are $$\displaystyle \left ( \alpha ,-\beta ,\gamma  \right )$$ and those of $$C$$ are $$\displaystyle \left ( \alpha ,\beta ,-\gamma  \right )$$.
Hence, the coordinates of the centroid of the triangle $$ABC$$ are $$\displaystyle \left ( \dfrac {\alpha }{3},\dfrac {\beta }{3}, \dfrac {\gamma }{3} \right )$$

Let us consider x be the position vector of B,