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

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

Answers (1)

best_answer

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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