MHT CET · Maths · Three Dimensional Geometry
If the line passing through the points \((a, 1,6)\) and \((3,4, b)\) crosses the yz-plane at the point \(\left(0, \frac{17}{2}, \frac{-13}{2}\right)\), then
- A \(a=-5, n=1\)
- B \(a=5, b=1\)
- C \(a=-5, b=-1\)
- D \(a=5, b=-1\)
Answer & Solution
Correct Answer
(B) \(a=5, b=1\)
Step-by-step Solution
Detailed explanation
Let \(\mathrm{y}-\mathrm{z}\) plane divides the join in the ratio \(\lambda: 1\)
\(\begin{aligned} & \Rightarrow\left(\frac{3 \lambda+a}{\lambda+1}, \frac{4 \lambda+1}{\lambda+1}, \frac{b \lambda+6}{\lambda+1}\right) \equiv\left(0 ; \frac{17}{2}, \frac{-13}{2}\right) \\ & \Rightarrow 3 \lambda+a=0, \frac{4 \lambda+1}{\lambda+1}=\frac{17}{2}, \frac{b \lambda+6}{\lambda+1}=\frac{-13}{2} \\ & \Rightarrow \lambda=\frac{-5}{3}, a=5, b=1\end{aligned}\)
\(\begin{aligned} & \Rightarrow\left(\frac{3 \lambda+a}{\lambda+1}, \frac{4 \lambda+1}{\lambda+1}, \frac{b \lambda+6}{\lambda+1}\right) \equiv\left(0 ; \frac{17}{2}, \frac{-13}{2}\right) \\ & \Rightarrow 3 \lambda+a=0, \frac{4 \lambda+1}{\lambda+1}=\frac{17}{2}, \frac{b \lambda+6}{\lambda+1}=\frac{-13}{2} \\ & \Rightarrow \lambda=\frac{-5}{3}, a=5, b=1\end{aligned}\)
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