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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

Answers (1)

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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