AP EAMCET · Maths · Three Dimensional Geometry
A plane \(\pi\) is passing through the points \(\mathrm{A}(1,-2,3)\) and \(\mathrm{B}(6,4,5)\). If the plane \(\pi\) is perpendicular the plane \(3 x-y+z=2\), then the perpendicular distance from \((0,0,0)\) to the plane \(\pi\) is
- A \(\frac{63}{\sqrt{594}}\)
- B \(\frac{32}{\sqrt{594}}\)
- C \(\frac{72}{\sqrt{435}}\)
- D \(\frac{23}{\sqrt{135}}\)
Answer & Solution
Correct Answer
(A) \(\frac{63}{\sqrt{594}}\)
Step-by-step Solution
Detailed explanation
\(\vec{AB} = (6-1, 4-(-2), 5-3) = (5, 6, 2)\) Normal vector \(\vec{n} = (a, b, c)\). \(\vec{n} \cdot \vec{AB} = 0 \Rightarrow 5a + 6b + 2c = 0\) \(\vec{n} \cdot (3, -1, 1) = 0 \Rightarrow 3a - b + c = 0\) Solving equations: \((a, b, c) = (8k, k, -23k)\). Let \(k=1\), so…
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