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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 28 The Plane Exercise 28 .10 Question 3 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 28 The Plane  Exercise 28 .10 Question 3 Maths Textbook Solution.

Answers (1)

Answer: - Therefore, equation of the mid- parallel plane is 2x-2y+z+6=0

Hint: - Use formula \mathrm{P}=\left|\frac{a x_{1}+b y_{1}+c_{-1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|

Given:-  2x-2y+z+3=0 and 2x-2y+z+9=0

             Let \pi _{1}-2x-2y+z+3=0 and 2x-2y+z+9=0

             And \pi _{2}=2x-2y+z+9=0

Solution: - Let the equation of the plane mid- parallel to these plane be

 \pi _{3}=2x-2y+z+\theta =0

Now, letP\left ( x_{1},y_{1},z_{1} \right )  be any point on this plane.

2x_{1}-2y_{1}+z_{1}+\theta =0----- (1)

We know, the distance of point from the plane ax +by +cz + d = 0

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

Distance of P from \pi _{1}

\begin{aligned} &\mathrm{P}=\left|\frac{2 x_{1}-2 y_{1}+z_{1}+3}{\sqrt{2^{2}+(-2)^{2}+1^{2}}}\right|\\ &\mathrm{P}=\left|\frac{(-\theta)+3}{3}\right| \text { (using equation (1)) }\\ \end{aligned}

Similarly, Distance of Q from \pi _{2}

\begin{aligned} &\mathrm{Q}=\left|\frac{2 x_{1}-2 y_{1}+z_{1}+9}{\sqrt{2^{2}+(-2)^{2}+1^{2}}}\right|\\ &\mathrm{Q}=\left|\frac{(-\theta)+9}{3}\right| \text { (using equation (1)) }\\ \end{aligned}

As \pi _{3} is mid- parallel to \pi _{1} and  \pi _{2}

\begin{aligned} &\mathrm{P}=\mathrm{Q}\\ \end{aligned}

\begin{aligned} &\left|\frac{(-\theta)+3}{3}\right|=\left|\frac{(-\theta)+9}{3}\right|\\ \end{aligned}

Squaring both sides

\begin{aligned} &\left(\frac{(-\theta)+3}{3}\right)^{2}=\left(\frac{(-\theta)+9}{3}\right)^{2}\\ \end{aligned}

\begin{aligned} &(3-\theta)^{2}=(9-\theta)^{2}\\ \end{aligned}

\begin{aligned} &9-6 \theta+\theta^{2}=81-18 \theta+\theta^{2}\\ &12 \theta=72 \end{aligned}

\theta =6

 Therefore, equation of the mid-parallel plane is2x-2y+z+6=0

 

Posted by

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