MHT CET · Maths · Three Dimensional Geometry
The distance of the point \((2,4,0)\) from the point of intersection of the lines \(\frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}\) and \(\frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}\) is
- A 3 units
- B \(3 \sqrt{3}\) units
- C 2 units
- D \(2 \sqrt{3}\) units
Answer & Solution
Correct Answer
(A) 3 units
Step-by-step Solution
Detailed explanation
Line 1: \(\frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}=k_1 \implies x=3k_1-6, y=2k_1, z=k_1-1\) Line 2: \(\frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}=k_2 \implies x=4k_2+7, y=3k_2+9, z=2k_2+4\)
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