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Please Solve R.D.Sharma class 12 Chapter 28 The Plane Exercise 28.10 Question 1 Maths textbook Solution.

Answers (1)

Answer: Distance is $\frac{25}{3\sqrt{14}}$

Hint: Use Formula $P=\frac{a x_{1}+b y_{1}+c_{z_{1}}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}$

Given: $-2x-y+3z-4=0$ and $6x-3y+9z+13=0$

Solution:Let $P\left ( x_{1},y_{1},z_{1} \right )$ be any point on $2x-y+3z-4=0$ Then

$2x_{1}-y_{1}+3z_{1}=4$ .......(1)

Let, P = Distance between $\left ( x_{1},y_{1},z_{1} \right )$ and the plane $6x-3y+9z+13=0.$

We know the distance of point $\left ( x_{1},y_{1},z_{1} \right )$ from the plane $ax+by+cz+d=0$ is given by

$\begin{aligned} &\mathrm{P}=\left|\frac{a x_{1}+b y_{1}+c_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right| \\ \end{aligned}$

Putting the values

$\begin{aligned} &\mathrm{P}=\left|\frac{6 x_{1}+(-3) y_{1}+9 z_{1}+13}{\sqrt{6^{2}+(-3)^{2}+9^{2}}}\right| \\ \end{aligned}$

$\begin{aligned} &\mathrm{P}=\left|\frac{3\left(2 x_{1}-y_{1}+3 z_{1}\right)+13}{\sqrt{36+9+81}}\right| \\ \end{aligned}$

$\begin{aligned} &\mathrm{P}=\left|\frac{3(4)+13}{3 \sqrt{14}}\right| \\ \end{aligned}$ (From equation (1))

$\begin{aligned} &\mathrm{P}=\frac{25}{3 \sqrt{14}} \end{aligned}$

Therefore, the distance between the parallel planes $2x-y+3z-4=0$ and $6x-3y+3z+13=0$ is $\frac{25}{3\sqrt{14}}$ units.

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