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  • Explain solution RD Sharma class 12 chapter The Plane exercise 28.1 question 3 sub question (ii) maths

Explain solution RD Sharma class 12 chapter The Plane exercise 28.1 question 3 sub question (ii)  maths

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Answer:-  Hence they are coplanar.

Hint:-  Use equation of plane  \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

Given:-(0,4,3), (-1,-5,-3), (-2,-2,1) and (1,1,-1)

Solution:- Given that these four points are coplanar. So these four points lie on the same plane.

So, first let us take three points and find the equation of the plane passing through these four points and then let us substitute the fourth point in it. If it is 0 then the points lies on the plane formed by these three points then they are coplanar.

The equation of the plane passing through these three points is given by

\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, let us take (0, 4,3), (-1,-5,-3), (-2,-2,1) and find plane equation

\begin{aligned} &\Rightarrow\left|\begin{array}{ccc} x-0 & y-4 & z-3 \\ -1-0 & -5-4 & -3-3 \\ -2-0 & -2-4 & 1-3 \end{array}\right|=0 \\\\ &\Rightarrow\left|\begin{array}{ccc} x & y-4 & z-3 \\ -1 & -9 & -6 \\ -2 & -6 & -2 \end{array}\right|=0 \end{aligned}

\begin{gathered} x(18-36)-(y-4)(2-12)+(z-3)(6-18)=0 \\\\ x(-18)-(y-4)(-10)+(z-3)(-12)=0 \\\\ -18 x+10 y-40-12 z+36=0 \\\\ -18 x+10 y-12 z-4=0 \end{gathered}

Now, let us substitute (1, 1,-1) in plane equation

                \begin{gathered} -18 x+10 y-12 z-4=0 \\\\ -18(1)+10(1)-12(-1)-4=0 \\\\ -18+10+12-4=0 \end{gathered}

                                \begin{gathered} -22+22=0 \\\\ \therefore 0=0 \end{gathered}

                                \therefore \mathrm{LHS}=\mathrm{RHS}

So, this point also lies on the plane.

Hence they are coplanar.

Posted by

infoexpert26

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