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Need solution for RD Sharma maths class 12 chapter 28 The Plane exercise Fill in the blank question 19

Answers (1)

Answer:

 3 units is the required distance.

Hint:

 use vector dot product to find the distance

Given:

 2x+y-2z-6=0 \text { and } 4x+2y-4z=0

Solution:

Let

\begin{aligned} &P_{1}\rightarrow 2x+y-2z-6=0\\ &ax+by+cz+d_{1}=0\\ &P_{2}\rightarrow 4x+2y-4z=0\\ &ax+by+cz+d_{2}=0\\ \end{aligned}

\begin{aligned} &\because P_{1}\left | \right |P_{2} \end{aligned}

\begin{aligned} &\text { Distance between } P_{1} \text { and } P_{2}=\left | \frac{d_{2}-d_{1}}{\sqrt{a^{2}+b^{2}+c^{2}}} \right |\\ &=\left | \frac{0-6}{\sqrt{2^{2}+1^{2}+(-2)^{2}}} \right |=\left | \frac{-6}{4+1+4} \right |\\ &=3\: units \end{aligned}

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Gurleen Kaur

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