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  • Explain solution RD Sharma class 12 chapter 28 The Plane exercise very short answer type question 20 maths

Explain solution RD Sharma class 12 chapter 28 The Plane exercise very short answer type question 20 maths

Answers (1)

Answer:

 (\widehat{i}-2\widehat{j}-3\widehat{k})+(2\widehat{i}+\widehat{j}+2\widehat{k})

Hint:

 we will use vector equation of line as

\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}

Given:

 Line passing through (1,-2,-3) and normal the plane

\overrightarrow{r}.(2\widehat{i}+\widehat{j}+2\widehat{k})=5

Solution:

Position vector of point (1,-2,-3).

\overrightarrow{a}=(\widehat{i}-2\widehat{j}-3\widehat{k})

Given equation of plane

\overrightarrow{r}.(2\widehat{i}+\widehat{j}+2\widehat{k})=5

So the vector, normal to the plane is

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

∴Eqn. of line

\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}

\overrightarrow{r}=(\widehat{i}-2\widehat{j}-3\widehat{k})+\lambda (2\widehat{i}+\widehat{j}+2\widehat{k})

Where λ is parameter.

Posted by

Gurleen Kaur

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