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AP EAMCET · Maths · Three Dimensional Geometry

For a positive real number \(p\), if the perpendicular distance from a point \(-\overline{\mathrm{i}}+\mathrm{p} \overline{\mathrm{j}}-3 \overline{\mathrm{k}}\) to the plane \(\overline{\mathrm{r}} \cdot(2 \overline{\mathrm{i}}-3 \overline{\mathrm{j}}+6 \overline{\mathrm{k}})=7\) is 6 units, then \(\mathrm{p}=\)

  1. A \(\frac{4}{5}\)
  2. B \(\frac{5}{6}\)
  3. C 6
  4. D 5
Verified Solution

Answer & Solution

Correct Answer

(D) 5

Step-by-step Solution

Detailed explanation

\(6 = \frac{|(-\overline{\mathrm{i}}+\mathrm{p} \overline{\mathrm{j}}-3 \overline{\mathrm{k}}) \cdot (2 \overline{\mathrm{i}}-3 \overline{\mathrm{j}}+6 \overline{\mathrm{k}}) - 7|}{ |2 \overline{\mathrm{i}}-3 \overline{\mathrm{j}}+6 \overline{\mathrm{k}}| }\)…