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 \(\mathrm{p}=\)
- A \(\frac{4}{5}\)
- B \(\frac{5}{6}\)
- C 6
- D 5
Answer & Solution
Correct Answer
(D) 5
Step-by-step Solution
Detailed explanation
\(6 = \frac{|(-\overline{\mathrm{i}}+\mathrm{p} \overline{\mathrm{j}}-3 \overline{\mathrm{k}}) \cdot (2 \overline{\mathrm{i}}-3 \overline{\mathrm{j}}+6 \overline{\mathrm{k}}) - 7|}{ |2 \overline{\mathrm{i}}-3 \overline{\mathrm{j}}+6 \overline{\mathrm{k}}| }\)…
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