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  • Please solve RD Sharma Class 12 Chapter 28 The Plane Exercise 28.9 Question 11 maths textbook solution.

Please solve RD Sharma Class 12 Chapter 28 The Plane Exercise 28.9 Question 11 maths textbook solution.

Answers (1)

Answer :  The answer of the given question is \sqrt{29} Units

Hint :

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

Given :

         3 Point (7,2,4) and the plane determine by the point

      A(2,5,-3), B(-2,-3,5) \&(5,3,-3)

Solution :

The equation of the plane passing through \left(x_{1}, y_{1}, z_{1}\right),\left(x_{2}, y_{2}, z_{2}\right) \&\left(x_{3}, y_{3}, z_{3}\right) is given by the following equation

\left|\begin{array}{lll} x-x_{1} & y-y_{1} & z-z_{1} \\ x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1} \end{array}\right|=0

According to questions

\begin{aligned} &\left(x_{1}, y_{1}, z_{1}\right)=(2,5,-3) \\ &\left(x_{2}, y_{2}, z_{2}\right)=(-2,-3,5) \\ &\left(x_{3}, y_{3}, z_{3}\right)=(5,3,-3) \end{aligned}

Putting these values

\left|\begin{array}{ccc} x-2 & y-5 & z-(-3) \\ -4 & -8 & 8 \\ 3 & -2 & 0 \end{array}\right|=0

\begin{aligned} &(x-2)(16)+(y-5)(24)+(z+3)(32)=0 \\ &2 x+3 y+4 z-7=0 \end{aligned}

Distance of 2 x+3 y+4 z-7=0 from (7,2,4)

We know, the distance of point \left(x_{1}, y_{1}, z_{1}\right) from the plane

\pi: a x+b y+c z+d=0 Is given by

\begin{aligned} &P=\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right| \\ &P=\left|\frac{2 \times 7+3 \times 2+4 \times 4-7}{\sqrt{2^{2}+3^{2}+4^{2}}}\right| \end{aligned}

P= \sqrt{29} Units

     

Posted by

infoexpert23

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