Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Explain solution RD Sharma class 12 chapter 28 The Plane exercise multiple choice question 16 maths

Explain solution RD Sharma class 12 chapter 28 The Plane exercise multiple choice question 16 maths

Answers (1)

Answer:

 Option (b)

Hint:

 Use simultaneous equation.

Given:

 (-1, -5, -10)

Solution:

Let the point of intersection of the line

\overrightarrow{r}.(2\widehat{i}-\widehat{j}+2\widehat{k})+\lambda (3\widehat{i}+4\widehat{j}+12\widehat{k}) -1, -5

And the plane

\overrightarrow{r}.(\widehat{i}-\widehat{j}+\widehat{k})=5 \text { be } \left (x_0,y_0,z_0 \right )

As x0, y0, z0 is the point of intersection of the line and the plane, so the position vector of this point i.e.,

\overrightarrow{r_0}.(x_0\widehat{i}+y_0\widehat{j}+z_0\widehat{k})

Must satisfy both equations of line and the equation of plane.

Substituting \overrightarrow{r_0} in plane of \overrightarrow{r} in both the equations, we get,

(x_0\widehat{i}+y_0\widehat{j}+z_0\widehat{k})=(2\widehat{i}-\widehat{j}+2\widehat{k})+\lambda (3\widehat{i}+4\widehat{j}+12\widehat{k})

and

(x_0\widehat{i}+y_0\widehat{j}+z_0\widehat{k}).(\widehat{i}-\widehat{j}+\widehat{k})=5 \qquad \qquad \dots (2)

Substituting, these values in equation (2)

\begin{aligned} &\left [ ((2+3\lambda )\times 1)-(1\times (-1+4\lambda )) \right ]+\left [ 1\times (2+12\lambda ) \right ]=5\\ &2+3\lambda +1-4\lambda +2+12\lambda =5\\ &11\lambda =0\\ &\lambda =0 \end{aligned}

Therefore,

\begin{aligned} &x_0=2+3\lambda \\ &=2,\\ &y_0=-1+4\lambda \\ &=-1\\ &z_0=2+12\lambda \\ &=2 \end{aligned}

Hence, the point of intersection is (2, -1, 2)

Now, the distance between the point (-1, -5, -10) and (2, -1, 2) is

\begin{aligned} &=\sqrt{[2-(-1)]^2+[(-1)-(-5)]^2+[2-(-10)]^2}\\ &=\sqrt{3^2+4^2+12^2}\\ &=\sqrt{169}\\ &=13 \end{aligned}

Hence, the required distance between the point (-1, -5, -10) and the point where the line

\begin{aligned} &\overrightarrow{r}.(2\widehat{i}-\widehat{j}+2\widehat{k})+\lambda (3\widehat{i}+4\widehat{j}+12\widehat{k}) \end{aligned}

And the plane

\begin{aligned} &\overrightarrow{r}.(\widehat{i}-\widehat{j}+\widehat{k})=5 \end{aligned}

intersects is 13 units

Posted by

Gurleen Kaur

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Class 12 result verification 2024 application out at cbse.gov.in; last date, fees
anam-khan

CBSE results 2024 later for 102 students over fake certificates; board issues show cause notice
anam-khan

What after CBSE Class 12 results 2024? From CUET UG to JEE Advanced, all you need to know

Latest Question