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Provide solution for RD Sharma maths class 12 chapter The Plane exercise 28.1 question 2

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Answer:-  The required equation of the plane is 5x-7y+11z+4=0 and hence they are coplanar.

Hint:-  Use equation of the 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,-1,-1), (4,5,1), (3,9,4) and (-4,4,4)

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 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

\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, let us take (0,-1,-1), (4, 5,1), (3,9,4) and find plane equation

\begin{aligned} &\Rightarrow\left|\begin{array}{ccc} x-0 & y+1 & z+1 \\ 4-0 & 5+1 & 1+1 \\ 3-0 & 9+1 & 4+1 \end{array}\right|=0 \\\\ &\Rightarrow\left|\begin{array}{ccc} x & y+1 & z+1 \\ 4 & 6 & 2 \\ 3 & 10 & 5 \end{array}\right|=0 \end{aligned}

 

\begin{gathered} x(30-20)-(y+1)(20-6)+(z+1)(40-18)=0 \\\\ 10 x-14 y+22 z+8=0 \\\\ 5 x-7 y+11 z+4=0 \end{gathered}

Now let us substitute fourth point (-4, 4, 4) we get

\begin{gathered} 5(-4)-7(4)+11(4)+4=0 \\\\ -20-28+44+4=0 \\\\ 48+48=0 \\\\ \therefore 0=0 \end{gathered}

 

                ∴LHS=RHS

 

So, as said above this fourth point satisfies. So, this point also lies on the same plane.

Hence they are coplanar.

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