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  • Explain Solution R.D.Sharma Class 12 Chapter 28 The Plane Exercise 28.10 Question 4 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 28 The Plane Exercise 28.10 Question 4 Maths Textbook Solution.

Answers (1)

Answer: -  The required distance is \frac{7}{\sqrt{56}}unit.

Hint: - Use formula \mathrm{P}=\left|\frac{\vec{a} \cdot \vec{n}-d}{|\vec{n}|}\right|

Given: -  r(\hat{i}+2 \hat{j}+3 \hat{k})+7=0 \text { and } \vec{r} \cdot(2 \hat{i}+4 \hat{j}+6 \hat{k})+7=0

Solution: - Let \bar{a}  be the Position vector of any point on plane \bar{r}.(\hat{i}+2 \hat{j}+3 \hat{k})+7=0 \: \: \:

  \vec{a} \cdot( \hat{i}+2 \hat{j}+3 \hat{k})+7=0       ----- (1)

We know the distance of \vec{a} from the plane \vec{r}.\vec{n}-d=0

\begin{aligned} &\mathrm{P}=\left|\frac{\vec{a} \cdot{n}-d}{|\vec{n}|}\right| \\ \end{aligned}

Putting the value of \vec{a} and \vec{n}.

\begin{aligned} &\mathrm{P}=\left|\frac{\vec{a} \cdot(2 \hat{i}+4 \hat{j}+6 \hat{k})+7}{|(2 \hat{i}+4 \hat{j}+6 \hat{k})|}\right| \\ &\mathrm{P}=\left|\frac{2 \vec{a} \cdot(\hat{i}+2 \hat{j}+3 \hat{k})+7}{\sqrt{2^{2}+4^{2}+6^{2}}}\right| \\ &\mathrm{P}=\left|\frac{2(-7)+7}{\sqrt{56}}\right| \\ &\mathrm{P}=\left|\frac{-7}{\sqrt{56}}\right| \\ &\mathrm{P}=\frac{7}{\sqrt{56}} \end{aligned}

Therefore, the distance between the plane \bar{r}.(\hat{i}+2 \hat{j}+3 \hat{k})+7=0 \: \: \:and \vec{r} \cdot( 2\hat{i}+4 \hat{j}+6 \hat{k})+7=0 is \frac{7}{\sqrt{56}} units.

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