Get Answers to all your Questions

header-bg qa

Please solve RD Sharma class 12 chapter 28 The Plane exercise very short answer type question 21 maths textbook solution

Answers (1)

Answer:

 \left \{ \overrightarrow{r}-(a\widehat{i}+b\widehat{j}+c\widehat{k}) \right \}(\widehat{i}+\widehat{j}+\widehat{k})=0

Hint:

 we will use equation of plane as

\overrightarrow{r}.\overrightarrow{n}=\overrightarrow{d}

Given:

 Plane passing through (a,b,c) and parallel to plane

\overrightarrow{r}.(\widehat{i}+\widehat{j}+\widehat{k})=2

Solution:

\overrightarrow{r}.(\widehat{i}+\widehat{j}+\widehat{k})=2 \qquad \qquad \dots(i)

So the vector, normal to the plane is

\overrightarrow{n}=\widehat{i}+\widehat{j}+\widehat{k}

∴Eqn. of plane to (i) is

\overrightarrow{r}.\overrightarrow{n}=\overrightarrow{d}

\Rightarrow \overrightarrow{r}.(\widehat{i}+\widehat{j}+\widehat{k})=\overrightarrow{d} \qquad \qquad \dots(ii)

Plane (ii) passing through (a,b,c)

\Rightarrow (a\widehat{i}+b\widehat{j}+c\widehat{k}) .(\widehat{i}+\widehat{j}+\widehat{k})=\overrightarrow{d} \qquad \qquad \dots(iii)

Putting (iii) in (ii), we get

\begin{aligned} &\Rightarrow \overrightarrow{r}.(\widehat{i}+\widehat{j}+\widehat{k})=(a\widehat{i}+b\widehat{j}+c\widehat{k}).(\widehat{i}+\widehat{j}+\widehat{k})\\ \end{aligned}

\Rightarrow \left \{ \overrightarrow{r}-(a\widehat{i}+b\widehat{j}+c\widehat{k}) \right \}(\widehat{i}+\widehat{j}+\widehat{k})=0

Posted by

Gurleen Kaur

View full answer

Crack CUET with india's "Best Teachers"

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