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TS EAMCET · Maths · Three Dimensional Geometry

The equation of the plane in cartesian form, which is at a distance of \(\frac{6}{\sqrt{29}}\) from the origin and its normal vector drawn from the origin being \(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}\), is

  1. A \(2 x-3 y+4 z=6\)
  2. B \(2 x+3 y-4 z=6\)
  3. C \(-2 x-3 y+4 z=6\)
  4. D \(2 x+3 y+4 z=-6\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(2 x-3 y+4 z=6\)

Step-by-step Solution

Detailed explanation

Equation of plane, which is distance of \(\frac{6}{\sqrt{29}}\) from the origin and its normal vector \(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}\) is…