MCQ
XOZ-plane divides the join of (2, 3, 1) and (6, 7, 1) in the ratio
Options
3 : 7
2 : 7
–3 : 7
–2 : 7
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Solution
−3:7
Let A\[\equiv\](2, 3, 1) and B\[\equiv\]Let the line joining A and B be divided by the xz-plane at point P in the ratio\[\lambda: 1\]
Then, we have,
P\[\equiv \left( \frac{6\lambda + 2}{\lambda + 1}, \frac{7\lambda + 3}{\lambda + 1}, \frac{\lambda + 1}{\lambda + 1} \right)\]
Since P lies on the xz-plane, the y-coordinate of P will be zero.
\[\therefore \frac{7\lambda + 3}{\lambda + 1} = 0\]
\[ \Rightarrow 7\lambda + 3 = 0\]
\[ \Rightarrow \lambda = \frac{- 3}{7}\]
Hence, the xz-plane divides AB in the ratio \[-\]3 : 7
Is there an error in this question or solution?
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