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Provide solution for RD Sharma maths class 12 chapter The Plane exercise 28.1 question 1 sub question (ii)

Answers (1)

Answer:-  The required equation of the plane is 2x-y-2z-2=0

Hint:-  Use equation of the plane \left(x_{1}, y_{1}, z_{1}\right),\left(x_{2}, y_{2}, z_{2}\right) and \left(x_{3}, y_{3}, z_{3}\right)

Given:- (-5,0,-6), (-3,10,-9) and (-2,6,-6)

Solution:- We know that, the equation of the plane passing through three non-collinear points \left(x_{1}, y_{1}, z_{1}\right),\left(x_{2}, y_{2}, z_{2}\right) and \left(x_{3}, y_{3}, z_{3}\right) is

\left|\begin{array}{ccc} 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

Now, substitute the given value

\begin{aligned} &\Rightarrow\left|\begin{array}{ccc} x+5 & y-0 & z+6 \\ -3+5 & 10-0 & -9+6 \\ -2+5 & 6-0 & -6+6 \end{array}\right|=0 \\\\ &\Rightarrow\left|\begin{array}{ccc} x+5 & y & z+6 \\ 2 & 10 & -3 \\ 3 & 6 & 0 \end{array}\right|=0 \end{aligned}

 

\begin{gathered} (x+5)(18)-y(9)+(z+6)(-18)=0 \\\\ 18 x+90-9 y-18 z-108=0 \\\\ \therefore 2 x-y-2 z-2=0 \end{gathered}

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