JEE Mains · Maths · STD 12 - 11. three dimension geometry
The distance of the point \((1, 0, 2)\) from the point of intersection of the line \(\frac{{x - 2}}{3} = \frac{{y + 1}}{4} = \frac{{z - 2}}{{12}}\) and the plane \(x - y + z = 16\) is
- A \(13\)
- B \(2\sqrt {14} \)
- C \(8\)
- D \(3\sqrt {21} \)
Answer & Solution
Correct Answer
(A) \(13\)
Step-by-step Solution
Detailed explanation
Let \(\frac{{x - 2}}{3} = \frac{{y + 1}}{4} - \frac{{z - 2}}{{12}} = \lambda \) \( \Rightarrow x = 3\lambda + 2,y = 4\lambda - 1,z = 12\lambda + 2\) The coorodinates of any point on ve lies are given by \(\left( {3\lambda + 2,4\lambda - 1,12\lambda + 2} \right).\) The ponit of…
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