Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Provide solution for RD Sharma maths class 12 chapter The Plane exercise 28.8 question 10

Provide solution for RD Sharma maths class 12 chapter The Plane exercise 28.8 question 10

Answers (1)

Answer:-  The answer of the given question is \vec{r} \cdot(4 \hat{i}+2 \hat{j}-4 \hat{k})+6=0  or

\vec{r} \cdot(-2 \hat{i}+4 \hat{j}+4 \hat{k})+6=0

Hint:-  We know that, equation of a plane passing through the line of intersection of two planes a_{1} x+b_{1} y+c_{1} z+d_{1}=0 \text { and } a_{2} x+b_{2} y+c_{2} z+d_{2}=0 is given by

\left(a_{1} x+b_{1} y+c_{1} z+d_{1}\right)+k\left(a_{2} x+b_{2} y+c_{2} z+d_{2}\right)=0


Given:- \begin{aligned} &-\vec{r} \cdot(\hat{\imath}+3 \hat{\jmath})+6=0 \end{aligned}      

            \\\\ \vec{r} \cdot(3 \hat{\imath}-\hat{\jmath}-4 \hat{k})=0

Solution:-  The equation of a plane through the line of intersection of  the planesr \cdot(\hat{\imath}+3 \hat{j})+6=0   and \vec{r} \cdot(3 \hat{\imath}-\hat{\jmath}-4 k)=0

                                    \begin{gathered} \vec{r} \cdot(\hat{\imath}+3 \hat{\jmath})+6+\lambda[\vec{r} \cdot(3 \hat{\imath}-\hat{\jmath}-4 \hat{k})]=0 \\\\ \vec{r} \cdot[(\hat{\imath}+3 \hat{\jmath})]+6+3 \lambda \cdot \vec{r} \hat{\imath}-\vec{r} \lambda \hat{j}-4 \lambda \vec{r} \hat{k}=0 \end{gathered}

                                    r(\hat{\imath}+3 \hat{\jmath}+3 \lambda \hat{\imath}-\lambda \hat{\jmath}-4 \lambda \hat{k})=-6                ...........(i)

                \vec{r} \cdot \frac{[\hat{\imath}(1+3 \lambda)+\hat{j}(3-\lambda)+\hat{k}(-4 \lambda)]}{\sqrt{(1+3 \lambda)^{2}+(3-\lambda)^{2}+(-4 \lambda)^{2}}}=\frac{-6}{\sqrt{(1+3 \lambda)^{2}+(3-\lambda)^{2}+(-4 \lambda)^{2}}}

The perpendicular distance from the origin is unity

            \begin{aligned} &\frac{-6}{\sqrt{(1+3 \lambda)^{2}+(3-\lambda)^{2}+(-4 \lambda)^{2}}}=1 \\\\ &(1+3 \lambda)^{2}+(3-\lambda)^{2}+(-4 \lambda)^{2}=36 \end{aligned}

            \begin{gathered} 1+9 \lambda^{2}+6 \lambda+9+\lambda^{2}-6 \lambda+16 \lambda^{2}=36 \\\\ \lambda^{2}=1 \\\\ \lambda=\pm 1 \end{gathered}

Using equation (i) the required plane is

                \vec{r} \cdot(1 \pm 3) \hat{\imath}+(3 \mp 1) \hat{\jmath}+(\mp 4) \hat{k}=-6

                \vec{r} \cdot(4 \hat{\imath}+2 \hat{\jmath}-4 \hat{k})=-6 \text { or }\vec{r} \cdot(-2 \hat{\imath}+4 \hat{\jmath}+4 \hat{k})=-6

                \vec{r} \cdot(4 \hat{\imath}+2 \hat{\jmath}-4 \hat{k})+6=0 \text { orr. }(-2 \hat{\imath}+4 \hat{\jmath}+4 \hat{k})+6=0

Posted by

infoexpert26

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