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: Set up the section point.
Point $P$ divides $A(1,1,1)$ and $B(2,2,2)$ in the ratio $1:m$. Since $A$ and $B$ have equal coordinates in all three axes, $P$ also has all three coordinates equal to one value $t$.
Step 2: Find that common value.
Using the section formula with ratio $1:m$, \[ t=\frac{1\cdot2+m\cdot1}{1+m}=\frac{m+2}{m+1}. \]
Step 3: Put $P$ on the plane.
$P=(t,t,t)$ lies on $x+2y+3z-1=0$, so $t+2t+3t-1=0$, that is $6t-1=0$.
Step 4: Solve for $t$.
\[ 6\cdot\frac{m+2}{m+1}=1. \]
Step 5: Clear and simplify.
$6(m+2)=m+1\Rightarrow 6m+12=m+1\Rightarrow 5m=-11$.
Step 6: Final value of $m$.
\[ m=-\frac{11}{5}. \] \[ \boxed{-\dfrac{11}{5}} \]