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

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Answer:-  The required equation of plane is x-3y-6z+8=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:- (1,1,1), (1,-1,2) and (-2,-2,2)

Solution:- We know that, the equation of the plane passing through given three 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}{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

Now, substitute the given values

\begin{aligned} &\Rightarrow\left|\begin{array}{ccc} x-1 & y-1 & z-1 \\ 1-1 & -1-1 & 2-1 \\ -2-1 & -2-1 & 2-1 \end{array}\right|=0 \\\\ &\Rightarrow\left|\begin{array}{ccc} x-1 & y-1 & z-1 \\ 0 & -2 & 1 \\ -3 & -3 & 1 \end{array}\right|=0 \end{aligned}

 

\begin{gathered} (x-1)(-2+3)-(y-1)(0+3)+(z-1)(0-6)=0 \\\\ (x-1)(1)-(y-1) 3+(z-1)(-6)=0 \\\\ \therefore x-3 y-6 z+8=0 \end{gathered}

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