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Please solve RD Sharma class 12 chapter The Plane exercise 28.3  question 17 maths textbook solution

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Answer:

\vec{r} \cdot(4 \hat{\imath}-7 \hat{\jmath}+3 \hat{k})=28 is the vector equation of a required plane  &
4 x-7 y+3 z=28 is the cartesian form of equation of the required plane.

Hint:

By using formula \vec{A} . \vec{B}=A_{x} B_{x}+A_{y} B_{y}+A_{z} B_{z}

Given:

(-1,2,3)\; and \; (3,-5,6).

Solution:

It means that the plane is passing through the midpoint of the line AB.

\begin{aligned} &\vec{a}=\frac{\text { Position vector of } A+\text { Position vector of } B}{2} \\ &\Rightarrow \vec{a}=\frac{(-\hat{\imath}+2 \hat{\jmath}+3 \hat{k})+(3 \hat{\imath}-5 \hat{\jmath}+6 \hat{k})}{2} \\ &\Rightarrow \vec{a}=\frac{2 \hat{\imath}-3 \hat{\jmath}+9 \hat{k}}{2} \\ &\Rightarrow \vec{a}=\hat{\imath}-\frac{3}{2} \hat{\jmath}+\frac{9}{2} \hat{k} \ldots(i) \end{aligned}

And also given the plane B normal to the line joining the points A(-1,2,3) \; and \; B(3,-5,6)

Then \vec{n}=\overrightarrow{A B}

? \vec{n} = position vector of \vec{B} - Position vector of \vec{A}
\begin{aligned} &\vec{n}=(3 \hat{\imath}-5 \hat{\jmath}+6 \hat{k})-(-\hat{\imath}+2 \hat{\jmath}+3 \hat{k}) \\ &\vec{n}=4 \hat{\imath}-7 \hat{\jmath}+3 \hat{k} \ldots(i i) \end{aligned}

We know that

(\vec{r}-\vec{a}) \cdot \vec{n}=0

Substituting the values from equation (i) and equation (ii) in the above equation we get\begin{aligned} &{\left[\vec{r}-\left(\hat{\imath}-\frac{3}{2} \hat{\jmath}+\frac{9}{2} \hat{k}\right)\right] \cdot(4 \hat{\imath}-7 \hat{\jmath}+3 \hat{k})=0} \\ &\Rightarrow \vec{r} \cdot(4 \hat{\imath}-7 \hat{\jmath}+3 \hat{k})-\left(\hat{\imath}-\frac{3}{2} \hat{\jmath}+\frac{9}{2} \hat{k}\right) \cdot(4 \hat{\imath}-7 \hat{\jmath}+3 \hat{k})=0 \\ &\Rightarrow \vec{r} \cdot(4 \hat{\imath}-7 \hat{\jmath}+3 \hat{k})-\left[(1)(4)+\left(-\frac{3}{2}\right)(-7)+\left(\frac{9}{2}\right)(3)\right]=0 \end{aligned}

By multiplying the two vectors using the formula

\vec{A} . \vec{B}=A_{x} B_{x}+A_{y} B_{y}+A_{z} B_{z}
\begin{aligned} &\Rightarrow \vec{r} \cdot(4 \hat{\imath}-7 \hat{\jmath}+3 \hat{k})-\left[4+\frac{21}{2}+\frac{27}{2}\right]=0 \\ &\Rightarrow \vec{r} \cdot(4 \hat{\imath}-7 \hat{\jmath}+3 \hat{k})-\left[\frac{8+21+27}{2}\right]=0 \\ &\Rightarrow \vec{r} \cdot(4 \hat{\imath}-7 \hat{\jmath}+3 \hat{k})-28=0 \end{aligned}

\Rightarrow \vec{r} \cdot(4 \hat{\imath}-7 \hat{\jmath}+3 \hat{k})=28  is the vector equation of a required plane.

\vec{r}=(x \hat{\imath}+y \hat{\jmath}+z \hat{k})

Then the above vector equation of a plane becomes,

(x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot(4 \hat{\imath}-7 \hat{\jmath}+3 \hat{k})=28

Now multiplying the two vectors using the formula
\begin{aligned} &\vec{A} \cdot \vec{B}=A_{x} B_{x}+A_{y} B_{y}+A_{z} B_{z} \\ &\Rightarrow(x)(4)+(y)(-7)+(z)(3)=28 \\ &\Rightarrow 4 x-7 y+3 z=28 \end{aligned}

This is the cartesian form of equation of the required plane.

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