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  • Please solve RD Sharma Class 12 Chapter 28 The Plane Exercise 28 .14 Question 3 maths textbook solution.

Please solve RD Sharma Class 12 Chapter 28 The Plane Exercise 28.14 Question 3 maths textbook solution.

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Answer \frac{8}{\sqrt{14}}

\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|

Given – Find the shortest distance between the lines  \frac{x-1}{2}=\frac{y-3}{4}=\frac{z+2}{1} and 3 x-y-2 z+4=0=2 x+y+z+1

\frac{x-1}{2}=\frac{y-3}{4}=\frac{z+2}{1}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\frac{x-x_{1}}{\alpha}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\right]

x_{1}=1, y_{1}=3, z_{1}=-2

length of perpendicular from (1,3,-2) to the plane 3x-y-2z+4=0

\begin{aligned} &D=\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right| \\ &D=\left|\frac{3(1)-1(3)-2(-2)+4}{\sqrt{(3)^{2}+(-1)^{2}+(-2)^{2}}}\right| \end{aligned}

\begin{aligned} &D=\left|\frac{8}{\sqrt{14}}\right| \\ &D=\frac{8}{\sqrt{14}} \text { (answer) } \end{aligned}

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