Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A plane is perpendicular to the two planes <img alt="\mathrm{2 x-2 y+z=0 \: and \: x-y+2 z=4}" src="/latex-image/?%5Cmathrm%7B2%20x-2%2

A plane \mathrm{E} is perpendicular to the two planes \mathrm{2 x-2 y+z=0 \: and \: x-y+2 z=4}, and passes through the point \mathrm{P(1,-1,1)}. If the distance of the plane \mathrm{ E} from the point \mathrm{ Q(a, a, 2) \: is\: 3 \sqrt{2},\: then \: (P Q)^{2}} is equal to

Option: 1

9


Option: 2

12


Option: 3

21


Option: 4

33


Answers (1)

best_answer

\mathrm{Normal \;\vec{n} \;to \;plane \;is\; \vec{n}_{1} \times \overrightarrow{n_{2}}}

\mathrm{=\left|\begin{array}{rrr} i & j & k \\ 2 & -2 & 1 \\ 1 & -1 & 2 \end{array}\right| }\\

\mathrm{=(-4+1) i-(4-1) j+(-2+2) k} \\

\mathrm{=-3 i-3 j}

Equation of required plane through \mathrm{P(1,-1,1)} is

\mathrm{-3(x-1)-3(y+1)+0(z-1)=0} \\

\mathrm{\Rightarrow \quad x-1+y+1=0 }\\

\mathrm{\Rightarrow x+y=0}

\mathrm{Distance\; from\; Q(a, a, 2)}

\mathrm{\frac{|a+a|}{\sqrt{2}}=3 \sqrt{2} }

\mathrm{\Rightarrow(a)=3}\\

\mathrm{\Rightarrow a=\pm 3}\\

\mathrm{\therefore \quad Q \text { is }(3,3,2) \text { or }(-3,-3,2)}\\

\mathrm{\therefore P Q^{2}=(3-1)^{2}+(3+1)^{2}+(2-1)^{2}}\\

=4+16+1\\

=21

Hence correct option is 3

Posted by

avinash.dongre

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025: 11 from Kota coaching centres among 24 students with 100 percentile
anam-khan

Low JEE Main rank? Explore alternative paths for engineering admission
anam-khan

Gender-gap in JEE Main 2025: Female participation remains around 31%, overall pool drops by 7.09%

Latest Question