Question

Let A($\alpha$,4,7) and B(3,$\beta$,8) be two points in space. If YZ plane and ZX plane respectively divide the line segment joining the points A and B in the ratio 2:3 and 4:5, then the point C which divides AB in the ratio $\alpha:\beta$ externally is

• A. $(\frac{16}{3}, 10, 3)$
• B. $(-\frac{16}{3}, \frac{28}{3}, \frac{19}{3})$
• C. $(-\frac{16}{3}, -\frac{28}{3}, -\frac{19}{3})$
• D. $(-\frac{16}{3}, 10, \frac{19}{3})$

Hint:

A useful shortcut: The ratio in which the xy-plane divides the line segment joining $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is $-z_1:z_2$. Similarly, for the yz-plane the ratio is $-x_1:x_2$, and for the xz-plane the ratio is $-y_1:y_2$.

Answer 1

The Correct Option is D

Step 1: Find the value of $\alpha$.
The YZ-plane (where $x=0$) divides the line segment joining $A(\alpha,4,7)$ and $B(3,\beta,8)$ in the ratio $m:n=2:3$. Using the section formula for the x-coordinate, the point of division has an x-coordinate of 0. \[ x = \frac{m x_B + n x_A}{m+n} = 0 \implies \frac{2(3) + 3(\alpha)}{2+3} = 0. \] \[ 6 + 3\alpha = 0 \implies 3\alpha = -6 \implies \alpha = -2. \]

Step 2: Find the value of $\beta$.
The ZX-plane (where $y=0$) divides the same segment in the ratio $m:n=4:5$. Using the section formula for the y-coordinate, the point of division has a y-coordinate of 0. \[ y = \frac{m y_B + n y_A}{m+n} = 0 \implies \frac{4(\beta) + 5(4)}{4+5} = 0. \] \[ 4\beta + 20 = 0 \implies 4\beta = -20 \implies \beta = -5. \]

Step 3: Apply the external section formula.
We need to find the point C that divides the segment AB externally in the ratio $\alpha:\beta$, which is $-2:-5$ or $2:5$. The coordinates are $A(-2,4,7)$ and $B(3,-5,8)$. The ratio is $m:n=2:5$. The formula for external division is $C = \left(\frac{mx_2 - nx_1}{m-n}, \frac{my_2 - ny_1}{m-n}, \frac{mz_2 - nz_1}{m-n}\right)$. \[ x_C = \frac{2(3) - 5(-2)}{2-5} = \frac{6+10}{-3} = -\frac{16}{3}. \] \[ y_C = \frac{2(-5) - 5(4)}{2-5} = \frac{-10-20}{-3} = \frac{-30}{-3} = 10. \] \[ z_C = \frac{2(8) - 5(7)}{2-5} = \frac{16-35}{-3} = \frac{-19}{-3} = \frac{19}{3}. \] The coordinates of point C are $(-\frac{16}{3}, 10, \frac{19}{3})$.