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  • Please solve RD Sharma class 12 chapter The Plane exercise 28.3 question 13 sub question (i) maths textbook solution

Please solve RD Sharma class 12 chapter The Plane exercise 28.3  question 13 sub question (i)  maths textbook solution

Answers (1)

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Answer:

The normal to the given pairs of planes are perpendicular to each other.

Hint:

\overrightarrow{n_{1}} \cdot \overrightarrow{n_{2}}=0

Given:
\begin{aligned} &x-y+z-2=0 \\ &3 x+2 y-z+4=0 \end{aligned}

Solution:

The vector equation of the planes x-y+z-2=0 can be written as
\begin{aligned} &(x \hat{\imath}+y \hat{\jmath}+z \hat{k})(\hat{\imath}-\hat{\jmath}+\hat{k})=2 \\ &\Rightarrow \vec{r} \cdot(\hat{\imath}-\hat{\jmath}+\hat{k})=2 \end{aligned}

The normal to this plane is \overrightarrow{n_{1}}=\hat{\imath}-\hat{\jmath}+\hat{k} \ldots(\mathrm{i})

The vector equation of the plane 3 x+2 y-z+4=0 can be written as
\begin{aligned} &(x \hat{\imath}+y \hat{\jmath}+z \hat{k})(3 \hat{\imath}+2 \hat{\jmath}-\hat{k})=-4 \\ &\Rightarrow \vec{r} \cdot(3 \hat{\imath}+2 \hat{\jmath}-\hat{k})=-4 \end{aligned}

The normal to this plane is

\overrightarrow{n_{2}}=3 \hat{\imath}+2 \hat{\jmath}-\hat{k} \ldots(ii)

Now,

\begin{aligned} &\overrightarrow{n_{1}} \cdot \overrightarrow{n_{2}}=(\hat{\imath}-\hat{\jmath}+\hat{k}) \cdot(3 \hat{\imath}+2 \hat{\jmath}-\hat{k}) \\ &\Rightarrow \overrightarrow{n_{1}} \cdot \overrightarrow{n_{2}}=(1)(3)+(-1)(2)+(1)(-1)=0 \end{aligned}

Hence \overrightarrow{n_{1}}  is perpendicular to \overrightarrow{n_{2}}

Therefore, the normal to the given pair of planes are perpendicular to each other.

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