Introduction to Three Dimensional Geometry Test - 20

Let the plane $$ax+by + cz + d =0$$ divides the lines joining $$(x_1, y_1, z_1)$$ and $$(x_2, y_2, z_2)$$ in the ratio $$k:1$$ as shown in figure.

$$\therefore $$Coordinates of $$\displaystyle P \left ( \dfrac{kx_2 + x_1}{k + 1}, \dfrac{ky_2 + y_1}{k+1}, \dfrac{kz_2 + z_1}{k+1} \right )$$ must satisfy $$ax + by + ca + d = 0$$

i.e., $$\displaystyle \left ( a \dfrac{kx_2 + x_1}{k + 1} \right ) + b\left ( \dfrac{ky_2 + y_1}{k + 1} \right) + c \left ( \dfrac{kz_2 + z_1}{k + 1} \right) + d = 0$$

$$\Rightarrow a(kx_2 + x_1) + b (ky_2 + y_1) + c (kz_2 + z_1) + c(kz_2 + z_1) + d (k + 1)= 0$$

$$\Rightarrow k (ax_2 + by_2 + cz_2 + d) + (ax_1 + by_1 + cz_1 + d)= 0$$

$$\Rightarrow \displaystyle k = - \dfrac{(ax_1 + by_1 + cz_1 + d)}{(ax_2 + by_2 + cz_2 + d)}$$

Hence, option A is correct.

$$2x + 3y + 5z = 1$$ divides $$(1, 0, -3)$$ and $$(1, -5, 7)$$ in the ratio of $$k : 1$$ at point $$P$$.

Then, $$\displaystyle P \left ( \displaystyle \dfrac{k + 1}{k + 1} , \displaystyle \dfrac{-5k}{k + 1}, \displaystyle \dfrac{7k - 3}{k + 1} \right )$$ which must satisfy $$2x + 3y + 5z = 1$$

$$\Rightarrow 2 \left ( \displaystyle \dfrac{k + 1}{k + 1} \right ) + 3 \left ( \displaystyle \dfrac{-5k}{k + 1} \right ) + 5 \left (\displaystyle \dfrac{7k - 3}{k + 1} \right ) = 1$$

$$\Rightarrow 2k + 2 - 15 k + 35 k - 15 = k + 1$$

$$\Rightarrow                21 k = 14$$

$$\Rightarrow                k = \displaystyle \dfrac{2}{3}$$

$$\therefore 2x + 3y + 5z = 1$$ divides $$(1, 0, -3)$$ and $$(1, -5, 7)$$ in the ratio of $$2 : 3$$.

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

Center of the circum circle of a right angle triangle is on the mid of the hypotenuse. 

The direction ratios of line joining the points $$(5,1,6)$$ and $$(3,4,1)$$ are $$2,-3,5$$

If the vertices of tetrahedron are,$$(x_1,y_1,z_1),(x_2,y_2,z_2),(x_3,y_3,z_3),(x_4,y_4,z_4)$$, then the centroid of tetrahedron is

If $$R$$ is a point of trisection of $$PQ$$, then it divides the line $$PQ$$ in the ratio of $$2:1$$ or $$1:2$$
Case 1: $$R$$ divides $$PQ$$ in $$2:1$$ ratio,
Then coordinates of $$R$$ are $$\left(\dfrac{2\times3+1\times0}{3}, \dfrac{2\times6+1\times0}{3}, \dfrac{2\times9+1\times0}{3}\right) = (2,4.6)$$
Case 2: $$R$$ divides $$PQ$$ in $$1:2$$ ratio,
Then coordinates of $$R$$ are $$\left(\dfrac{1\times3+2\times0}{3},

\dfrac{1\times6+2\times0}{3}, \dfrac{1\times9+2\times0}{3}\right) = (1,2,3)$$
$$\therefore R_y = 2$$ or $$4$$
Hence, option B is correct.

intercept form $$\dfrac{7x}{3003}+\dfrac{11y}{3003}+\dfrac{13z}{3003}=1$$

if the fourth vertex is $$(x_1,y_1,z_1)$$