Question

Assertion :Distance of point $D(1,0,-1)$ from the plane of points $A(1,-2,0),B(3,1,2)$ and $C(-1,1,-1)$ is $\frac{8}{\sqrt{229}}$ Reason: Volume of tetrahedron formed by the points $A,B,C$ and $D$ is $\frac{\sqrt{229}}{2}$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is D Assertion is incorrect but Reason is correct

$\overrightarrow{AD}=2\hat{j}-\hat{k},\overrightarrow{BD}=-2\hat{i}-\hat{j}-3\hat{k}$ and $\overrightarrow{CD}=2\hat{i}-\hat{j}$
Volume of tetrahedron $=\frac{1}{6}$ $\left[\overrightarrow{AD}\overrightarrow{BD}\overrightarrow{CD}\right]$
$=\frac{1}{6}\Vert \begin{array}{ccc}0& 2& -1\\ -2& -1& -3\\ 2& -1& 0\end{array}\Vert =\frac{8}{3}$

Also, area of the triangle $ABC=\frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|$
$=\frac{1}{2}\Vert \begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& 2\\ -2& 3& -1\end{array}\Vert$
$=\frac{1}{2}|-9\hat{i}-2\hat{j}+12\hat{k}|$

$=\frac{\sqrt{229}}{2}$

Then $\frac{8}{3}=\frac{1}{3}\times ($Distance of $D$ from base $ABC)\times ($Area of triangle $ABC)$
Distance of $D$ from base $ABC=\frac{16}{\sqrt{229}}$