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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Three Dimensional Geometry

NDA 2024 · Mathematics · Three Dimensional Geometry

If p is the perpendicular distance from origin to the plane passing...

If pp is the perpendicular distance from origin to the plane passing through (1,0,0),(0,1,0)(1,0,0),(0,1,0) and (0,0,1)(0,0,1), then what is 3p23 p^2 equal to?

Options

  1. A4
  2. B3
  3. C2
  4. D1

Correct answer

D. 1

Step-by-step solution

∵\because Plane passing through (1,0,0),(0,1,0)(1,0,0),(0,1,0) and (0,0,1)a⃗=i^,b⃗=j^,c⃗=k^(0,0,1)\vec{a}=\hat{i}, \vec{b}=\hat{j}, \vec{c}=\hat{k} Let, r^=xi^+yj^+zk^\hat{r}=x \hat{i}+y \hat{j}+z \hat{k}\begin{aligned} & (\vec{r}-\vec{a}) \cdot[(\vec{b}-\vec{a}) \times(\vec{c}-\vec{a})]=0 \\ & \begin{aligned} \vec{b}-\vec{a} & =\hat{j}-\hat{i} \text{ and } \vec{c}-\vec{a}=\hat{k}-\hat{i} \\ \therefore \quad(\vec{b}-\vec{a}) \times(\vec{c}-\vec{a}) & =\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right|=\hat{i} \hat{j} \hat{k} \\ \therefore \quad(\vec{r}-\vec{a}) & =(x \hat{i}+y \hat{j}+z \hat{k})-(\hat{i}) \\ & =(x-1) \hat{i}+y \hat{j}+z \hat{k} \end{aligned}\end{aligned} We have, [(x−1)i^+yj^+zk^]⋅(i^+j^+k^)=0(x−1)+y+z=0\begin{aligned} {[(x-1) \hat{i}+y \hat{j}+z \hat{k}] \cdot(\hat{i}+\hat{j}+\hat{k}) } & =0 \\ \quad & (x-1)+y+z=0 \end{aligned}\thereforeEquatingofplaneEquating of planex+y+z-1=0\perpdistancefromorigintotheplanedistance from origin to the plane=\frac{|-1|}{\sqrt{(1)^2+(1)^2+(1)^2}}=\frac{1}{\sqrt{3}} \text{ unit } Here, p=13⇒p2=13∴3p2=1\begin{aligned} \text{ Here, } & & p & =\frac{1}{\sqrt{3}} \Rightarrow p^2=\frac{1}{3} \\ & \therefore & 3 p^2 & =1\end{aligned}$
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