Question

If the point P which divides the line segment joining $A(1,1,1)$ and $B(2,2,2)$ in the ratio 1: m lies on the plane $x+2y+3z-1=0$, then $m=$

• A. $-\frac{3}{2}$
• B. $\frac{4}{3}$
• C. $-\frac{11}{5}$
• D. $-\frac{1}{2}$

Hint:

Shortcut: The ratio $\lambda$ in which a plane $ax+by+cz+d=0$ divides the segment $AB$ is given directly by $-\frac{ax_1+by_1+cz_1+d}{ax_2+by_2+cz_2+d}$.

Answer 1

The Correct Option is C

Step 1: Concept
The coordinates of a point dividing the line segment joining $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ in the ratio $\lambda : 1$ are given by $\left(\frac{\lambda x_2 + x_1}{\lambda + 1}, \frac{\lambda y_2 + y_1}{\lambda + 1}, \frac{\lambda z_2 + z_1}{\lambda + 1}\right)$.

Step 2: Meaning
Here, the ratio is $1 : m$, which is equivalent to $\frac{1}{m} : 1$. Let $\lambda = \frac{1}{m}$. The coordinates of P are: $x = \frac{\lambda(2) + 1}{\lambda + 1}, y = \frac{\lambda(2) + 1}{\lambda + 1}, z = \frac{\lambda(2) + 1}{\lambda + 1}$.

Step 3: Analysis
Since P lies on the plane $x+2y+3z-1=0$, substitute these coordinates into the plane equation: $\left(\frac{2\lambda+1}{\lambda+1}\right) + 2\left(\frac{2\lambda+1}{\lambda+1}\right) + 3\left(\frac{2\lambda+1}{\lambda+1}\right) - 1 = 0$ $\implies (2\lambda+1) + 2(2\lambda+1) + 3(2\lambda+1) - (\lambda+1) = 0$ $\implies 6(2\lambda+1) - \lambda - 1 = 0 \implies 12\lambda + 6 - \lambda - 1 = 0 \implies 11\lambda + 5 = 0 \implies \lambda = -\frac{5}{11}$.

Step 4: Conclusion
Since $\lambda = \frac{1}{m}$, we have $m = \frac{1}{\lambda} = -\frac{11}{5}$, which matches option (C).

Final Answer: (C)