MHT CET · Maths · Three Dimensional Geometry
The line joining points \((3,5,-7)\) and \((-2,1,8)\) meets yz-plane at point
- A \(\left(0, \frac{13}{5}, 2\right)\)
- B \((0,13,2)\)
- C \(\left(0, \frac{13}{5},-3\right)\)
- D \(\left(0, \frac{-13}{5}, 2\right)\)
Answer & Solution
Correct Answer
(A) \(\left(0, \frac{13}{5}, 2\right)\)
Step-by-step Solution
Detailed explanation
Let \(\mathrm{y}_{\mathrm{z}}\) plane divides the join in the ratio \(\lambda: 1\) then
\(\frac{\lambda \times(-2)+1 \times 3}{\lambda+1}=0 \Rightarrow \lambda=\frac{3}{2}\)
So \(\mathrm{y}_{\mathrm{z}}\) plane divides the join in the ratio \(3: 2\).
Hence, the required point is
\( \left(\frac{3 \times(-2)+2 \times 3}{3+2}, \frac{3 \times 1+2 \times 5}{3+2},\ \frac{3 \times 8+2 \times(-7)}{3+2}\right)\)
\(=\left(0, \frac{13}{5}, 2\right)\)
\(\frac{\lambda \times(-2)+1 \times 3}{\lambda+1}=0 \Rightarrow \lambda=\frac{3}{2}\)
So \(\mathrm{y}_{\mathrm{z}}\) plane divides the join in the ratio \(3: 2\).
Hence, the required point is
\( \left(\frac{3 \times(-2)+2 \times 3}{3+2}, \frac{3 \times 1+2 \times 5}{3+2},\ \frac{3 \times 8+2 \times(-7)}{3+2}\right)\)
\(=\left(0, \frac{13}{5}, 2\right)\)
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