MHT CET · Maths · Three Dimensional Geometry
The distance of a point \((1,2,-1)\) from the plane \(x-2 y+4 z+10=0\) is
- A \(\frac{3}{\sqrt{7}}\) units
- B \(\frac{\sqrt{3}}{7}\) units
- C \(\sqrt{\frac{7}{3}}\) units
- D \(\sqrt{\frac{3}{7}}\) units
Answer & Solution
Correct Answer
(D) \(\sqrt{\frac{3}{7}}\) units
Step-by-step Solution
Detailed explanation
(D)
\(\begin{aligned} \text { Required distance } &=\frac{|1(1)+-2(2)+4(-1)+10|}{\sqrt{1^{2}+(-2)^{2}+(4)^{2}}} \\ &=\frac{3}{\sqrt{21}}=\frac{\sqrt{3}}{\sqrt{7}} \text { units } \end{aligned}\)
\(\begin{aligned} \text { Required distance } &=\frac{|1(1)+-2(2)+4(-1)+10|}{\sqrt{1^{2}+(-2)^{2}+(4)^{2}}} \\ &=\frac{3}{\sqrt{21}}=\frac{\sqrt{3}}{\sqrt{7}} \text { units } \end{aligned}\)
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