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

Answers (1)

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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