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

Board Exam Time Table 2025 Live: UP Board exams from February 24; CBSE, BSEB date sheet soon
anam-khan

CBSE Date Sheet 2025 Live: Class 10, 12 board exam dates soon; syllabus, sample papers
anam-khan

Extra time in CBSE exams for students with type 1 diabetes: Kerala SHRC seeks report on plea

Latest Question