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  • Explain solution RD Sharma class 12 chapter The Plane exercise 28.1 question 1 sub question (iv) maths

Explain solution RD Sharma class 12 chapter The Plane exercise 28.1 question 1 sub question (iv) maths

Answers (1)

Answer:-  The required equation of the plane is x+y-z-1=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:- (2,3,4), (-3,5,1) and (4,-1,2)

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 value

\begin{aligned} &\Rightarrow\left|\begin{array}{ccc} x-2 & y-3 & z-4 \\ -3-2 & 5-3 & 1-4 \\ 4-2 & -1-3 & 2-4 \end{array}\right|=0 \\\\ &\Rightarrow\left|\begin{array}{ccc} x-2 & y-3 & z-4 \\ -5 & 2 & -3 \\ 2 & -4 & -2 \end{array}\right|=0 \end{aligned}

\begin{gathered} (x-2)(-4-12)-(y-3)(10+6)+(z-4)(20-4)=0 \\\\ (x-2)(-16)-(y-3)(16)+(z-4)(16)=0 \\\\ -16 x+32-16 y+48+16 z-64=0 \end{gathered}

 

                        \begin{gathered} -16 x-16 y+16 z+16=0 \\\\ \therefore x+y-z-1=0 \end{gathered}

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