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
- A \(2 x-3 y+4 z=6\)
- B \(2 x+3 y-4 z=6\)
- C \(-2 x-3 y+4 z=6\)
- D \(2 x+3 y+4 z=-6\)
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…
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