question_answer

Vertices of a parallelogram taken in order are \[A(2,\,\,-1,\,\,4),\]\[B(1,\,\,0,\,\,-1),\]\[C(1,\,\,2,\,\,3)\] and \[D\]. Distance of the point \[P(8,\,\,2,\,\,-12)\] from the plane of the parallelogram is-

A) \[\frac{4\sqrt{6}}{9}\]                       

B) \[\frac{32\sqrt{6}}{9}\]

C) \[\frac{16\sqrt{6}}{9}\]                     

D)  None of these

Correct Answer: B

Solution :

\[\overrightarrow{n}=7\widehat{i}+2\widehat{j}-\widehat{k}\]is normal to plane (Assuming \[n=a\widehat{i}+b\widehat{j}+c\widehat{k}\] and using\[\overrightarrow{n}.\overrightarrow{AB}=0,\,\,\overrightarrow{n}.\overrightarrow{BC}=\overrightarrow{n}.\overrightarrow{AC}=0)\] \[P=(8,\,\,2,\,\,-12)\] \[\overrightarrow{AP}=6\widehat{i}+3\widehat{j}-16\widehat{k}\] \[\therefore \]Distance \[d=\left| \frac{\overrightarrow{AP}.\overrightarrow{n}}{|\overrightarrow{n}|} \right|=\left| \frac{42+6+16}{\sqrt{49+4+1}} \right|\]\[=\frac{64}{\sqrt{54}}=\frac{64}{3\sqrt{6}}=\frac{64\sqrt{6}}{18}=\frac{32\sqrt{6}}{9}\]