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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 28 The Plane Exercise 28.12 Question 7 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 28 The Plane  Exercise 28.12 Question 7 Maths Textbook Solution.

Answers (1)

Answer: Therefore, the distance is 10\sqrt{3}units.

Hint:Use distance formula \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}+\left ( z_{2}-z_{1} \right )^{2}}

Given:

\left ( 1,-5,9 \right ) and x-y+z=5

x=y=z

Solution: The straight line is given by x=y=z

\because \frac{x}{1}=\frac{y}{1}=\frac{z}{1}

Direction\: ratios \: are \: 1,1,1

Again\: the\: given\: point \: is (1,-5,9)

\therefore \frac{x-1}{1}=\frac{y+5}{1}=\frac{z-9}{1}=t( say )

\Rightarrow x=t+1, y=t-5, z=t+9

\begin{aligned} &\text { This point satisfies the plane equation }\\ &x-y+z=5\\ &t+1-t+5+t+9=5\\ &t+15=5\\ &t=-10 \end{aligned}

\begin{aligned} &x=-9, y=-15, z=-1\\ &\text { The distance between the coordinates }(-9,-15,-1) \text { and }(1,-5,9)\\ &=\sqrt{(1+9)^{2}+(-5+15)^{2}+(9+1)^{2}}\\ &=\sqrt{10^{2}+10^{2}+10^{2}}\\ &=10 \sqrt{3} \end{aligned}

 

 

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