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  • Need solution for RD Sharma maths class 12 chapter The Plane exercise 28.1 question 3 sub question (i)

Need solution for RD Sharma maths class 12 chapter The Plane exercise 28.1 question 3 sub question (i)

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Answer:-  Hence they are coplanar.

Hint:-  Use the 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,-1,0), (2,1,-1), (1,1,1) and (3,3,0)

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 plane passing through these four points and then let us substitute the fourth point in it. If it is 0 then the points lie 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}{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,-1, 0), (2,1,-1), (1,1,1) and find plane equation.

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

\begin{gathered} x(2+2)-(y+1)(2+1)+z(4-2)=0 \\\\ 4 x-3 y-3+2 z=0 \\\\ 4 x-3 y+2 z-3=0 \end{gathered}

Now, let us substitute fourth point (3,3,0) in plane equation 4x-3y+2z-3=0

                   4(3)-3(3)+2(0)-3=0

                             \begin{gathered} 12-9+0-3=0 \\\\ 12-12=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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