TS EAMCET · Maths · Three Dimensional Geometry
The perpendicular distance of the point \((1,-1,2)\) from the plane \(x+2 y+z=4\), is
- A \(\sqrt{17}\)
- B \(\sqrt{6}\)
- C \(\sqrt{\frac{3}{2}}\)
- D \(\sqrt{\frac{2}{3}}\)
Answer & Solution
Correct Answer
(C) \(\sqrt{\frac{3}{2}}\)
Step-by-step Solution
Detailed explanation
Perpendicular distance of point \(\left(x_1, y_1, z_1\right)\) from the plane \(a x+b y+c z+d=0\) is given by \(\left|\frac{a x_1+b y_1+c z_1+d}{\sqrt{a^2+b^2+c^2}}\right|\)…
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