MHT CET · Maths · Three Dimensional Geometry
The length of the perpendicular to the plane \(\bar{r} \cdot(\hat{\imath}-2 \hat{\jmath}+3 \hat{k})=14\) from the origin is
- A \(\sqrt{7}\) units
- B 7 units
- C 14 units
- D \(\sqrt{14}\) units
Answer & Solution
Correct Answer
(D) \(\sqrt{14}\) units
Step-by-step Solution
Detailed explanation
Length of \(\perp\) er from the point \(A(\bar{a})\) to the plane \(\bar{r} \cdot \bar{n}=p\) is \(\frac{|\bar{a} \cdot \bar{n}-p|}{|\bar{n}|}\)
Here \(\overline{\mathrm{a}}=0 \hat{\mathrm{i}}+0 \hat{\mathrm{j}}+0 \hat{\mathrm{k}}\) and \(\overline{\mathrm{n}}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\)
\(\therefore \overline{\mathrm{a}} \cdot \overline{\mathrm{n}}=0 \quad\) and \(|\overline{\mathrm{n}}|=\sqrt{1+4+9}=\sqrt{14}\)
Hence required distance is \(\frac{|0-14|}{\sqrt{14}}=\sqrt{14}\)
Here \(\overline{\mathrm{a}}=0 \hat{\mathrm{i}}+0 \hat{\mathrm{j}}+0 \hat{\mathrm{k}}\) and \(\overline{\mathrm{n}}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\)
\(\therefore \overline{\mathrm{a}} \cdot \overline{\mathrm{n}}=0 \quad\) and \(|\overline{\mathrm{n}}|=\sqrt{1+4+9}=\sqrt{14}\)
Hence required distance is \(\frac{|0-14|}{\sqrt{14}}=\sqrt{14}\)
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