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  • Explain solution RD Sharma class 12 chapter The Plane exercise 28.3 question 20 maths

Explain solution RD Sharma class 12 chapter The Plane exercise 28.3  question 20 maths

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Answer:

The answer of given question is a x+b y+c z=a^{2}+b^{2}+c^{2}

Hint:

\vec{A} \cdot \vec{B}=A_{x} B_{x}+A_{y} B_{y}+A_{z} B_{z}

Given:

The required plane is passing through A(a,b,c) and perpendicular to OA.

Solution:

Let the position vector of this point A be

\vec{a}=a \hat{\imath}+b \hat{\jmath}+c \hat{k} \ldots(i)

And it is also given the planes is normal to the line joining the points O(0,0,0) and position vector of point A(a,b,c)

Then

\vec{n}=\overrightarrow{O A}

\Rightarrow \vec{n} = position vector of \vec{A} - Position vector of \vec{O}

\begin{aligned} &\Rightarrow \vec{n}=(a \hat{\imath}+b \hat{\jmath}+c \hat{k})-(0 \hat{\imath}+0 \hat{\jmath}+0 \hat{k}) \\ &\Rightarrow \vec{n}=a \hat{\imath}+b \hat{\jmath}+c \hat{k} \ldots .(i i) \end{aligned}

The direction ratio of OA are proportional to a,b,c.

Hence, the direction cosines are

\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}}, \frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}}, \frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}

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} &{[\vec{r}-(a \hat{\imath}+b \hat{\jmath}+c \hat{k})] \cdot(a \hat{\imath}+b \hat{\jmath}+c \hat{k})=0} \\ &\Rightarrow \vec{r} \cdot(a \hat{\imath}+b \hat{\jmath}+c \hat{k})-(a \hat{\imath}+b \hat{\jmath}+c \hat{k}) \cdot(a \hat{\imath}+b \hat{\jmath}+c \hat{k})=0 \end{aligned}

\Rightarrow \vec{r} \cdot(a \hat{\imath}+b \hat{\jmath}+c \hat{k})-\left[a^{2}+b^{2}+c^{2}\right]=0

By 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 \vec{r} \cdot(a \hat{\imath}+b \hat{\jmath}+c \hat{k})=a^{2}+b^{2}+c^{2} \\ &\text { Let } \vec{r}=(x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \end{aligned}

Then, the above vector equation of the plane becomes

(x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot(a \hat{\imath}+b \hat{\jmath}+c \hat{k})=a^{2}+b^{2}+c^{2}

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)(a)+(y)(b)+(z)(c)=a^{2}+b^{2}+c^{2} \\ &\Rightarrow a x+b y+c z=a^{2}+b^{2}+c^{2} \end{aligned}

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

Posted by

infoexpert26

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