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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 28 The Plane Exercise 28.10 Question 2 Maths Textbook Solution.

Provide Solution For  R.D.Sharma Maths Class 12 Chapter 28 The Plane Exercise 28.10 Question 2 Maths Textbook Solution.

Answers (1)

Answer: - Therefore, equation of the plane is 2x-3y+5z+11=0 and distance of the plane is  \frac{4}{\sqrt{38}}units.

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

Given: - Point (3, 4, -1) and 2x-3y+5z+7=0

Solution: - Since the plane is parallel to  2x-3y+5z+7=0 it must be of the form:

 2x-3y+5z+\theta =0                                                        …… (i)

According to question:

The plane passes through (3, 4, - 1)

2\left ( 3 \right )-3\left ( 4 \right )+5\left ( -1 \right )+\theta =0

\theta =1

Putting the value of \theta  in eqn (i)

So, we get the equation of the plane is 2x-3y+5z+11=0

Distance of the plane 2x-3y+5z+11=0 from (3, 4, -1)

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 z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right| \\ \end{aligned}

Putting the values,

\begin{aligned} &\mathrm{P}=\left|\frac{2(3)+(-3)(4)+5(-1)+7}{\sqrt{2^{2}+(-3)^{2}+5^{2}}}\right| \\ &\mathrm{P}=\left|\frac{-4}{\sqrt{38}}\right| \\ &\mathrm{P}=\frac{4}{\sqrt{38}} \end{aligned}

    Therefore, the distance of the plane 2x-3y+5z+7=0 from (3, 4, - 1) is \frac{4}{\sqrt{38}}units.

 

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