Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Please help! The distance of the point (1, −2, 4) from the plane passing through the point (1, 2, 2) and perpendicular to the planes x − y + 2z = 3 and 2x −2y + z + 12=0, is :

 The distance of the point (1, −2, 4) from the plane passing through the point (1, 2, 2) and perpendicular to the planes x − y + 2z = 3 and  2x −2y + z + 12=0, is :

 

  • Option 1)

    2\sqrt{2}

  • Option 2)

    2

  • Option 3)

    \sqrt{2}

  • Option 4)

    \frac{1}{\sqrt{2}}

 

Answers (1)

As we learnt in 

Distance of a point from plane (Cartesian form) -

The length of perpendicular from P(x_{1},y_{1},z_{1}) to the plane

ax+by+cz+d= 0 is given by  \frac{\left [ ax_{1}+by_{1} +cz_{1}+d\right ]}{\left | \sqrt{a^{2}+b^{2}+c^{2}} \right |}

 

-

 

 Normal vector of plane is \begin{vmatrix} \hat{i} &\hat{j} & \hat{k}\\ 1 & -1 & 2\\ 2& -2& 1 \end{vmatrix}=3\hat{i}+3\hat{j}

So equation of plane is

3x+3y=C

passes through (1,2,2)

3+6=C

3x+3y=9    =>    x+y=3

Distance =\left | \frac{1-2-3}{\sqrt{2}} \right |=\frac{4}{\sqrt{2}}=2\sqrt{2}


Option 1)

2\sqrt{2}

This option is correct

Option 2)

2

This option is incorrect

Option 3)

\sqrt{2}

This option is incorrect

Option 4)

\frac{1}{\sqrt{2}}

This option is incorrect

Posted by

Sabhrant Ambastha

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 Mains 2025 session 1 registration deadline on November 22; apply on jeemain.nta.nic.in
anam-khan

BTech Admission 2025 Live: JEE Mains, SRMJEEE, MET registrations underway; MHT CET syllabus updates
anam-khan

Centre issues coaching guidelines, prohibits centres from using false claims like '100% selection'

Latest Question