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  • Need solution for RD Sharma maths class 12 chapter The Plane exercise 28.5 question 3

Need solution for RD Sharma maths class 12 chapter The Plane exercise 28.5 question 3

Answers (1)

Answer:

 The required vector equation of plane is:

\vec{r}=(bc\hat{i}+ac\hat{j}+ab\hat{k})=abc

Hint:

 Using

\vec{r}.\vec{n}=\vec{a}.\vec{n}

Given:

Vector equation of the plane passing through the point A (a, 0, 0), B (0, b, 0) and

C (0, 0, c). Prove that

\frac{1}{p^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}

Solution:

The required plane passing through the point A (a, 0, 0) whose position vector is 

\vec{a}=(a\hat{i}+0\hat{j}+0\hat{k})

and is normal to the vector given by

The vector equation of required plane is

\begin{aligned} &\vec{r}.(bc\hat{i}+ac\hat{j}+ab\hat{k})=(a\hat{i}+0\hat{j}+0\hat{k}).(bc\hat{i}+ac\hat{j}+ab\hat{k})\\ &\vec{r}.(bc\hat{i}+ac\hat{j}+ab\hat{k})=abc+0+0\\ &\vec{r}.(bc\hat{i}+ac\hat{j}+ab\hat{k})=abc \qquad \qquad \dots(i)\\ &\left | \vec{n} \right |=\sqrt{(bc)^2+(ac)^2+(ab)^2}=\sqrt{b^2c^2+a^2c^2+a^2b^2} \end{aligned}

For reducing (i) to normal form, we need to divide both sides of (i) by

\begin{aligned} &\sqrt{b^2c^2+a^2c^2+a^2b^2} \end{aligned}

Then, we get

\begin{aligned} &\vec{r}.\left ( \frac{bc\hat{i}+ac\hat{j}+ab\hat{k}}{\sqrt{b^2c^2+a^2c^2+a^2b^2}} \right )=\frac{abc}{\sqrt{b^2c^2+a^2c^2+a^2b^2}} \end{aligned}

Therefore, the normal form of plane (i) is

\begin{aligned} &\frac{abc}{\sqrt{b^2c^2+a^2c^2+a^2b^2}} \end{aligned}

So, the distance of plane (i) from the origin

\begin{aligned} &P=\frac{abc}{\sqrt{b^2c^2+a^2c^2+a^2b^2}}\\ &\frac{1}{P}=\frac{\sqrt{b^2c^2+a^2c^2+a^2b^2}}{abc}\\ &\frac{1}{P^2}=\frac{b^2c^2+a^2c^2+a^2b^2}{a^2b^2c^2}\\ &\frac{1}{P^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \end{aligned}

Hence proved

Posted by

Gurleen Kaur

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