Class 12: The Plane – Exercise 29.1

$\displaystyle \textbf{Question 1: }~\text{Find the equation of the plane passing through the following points:}$
$\displaystyle \textbf{(i) }\ (2,1,0),\ (3,-2,-2)\text{ and }(3,1,7)$
$\displaystyle \textbf{(ii) }\ (-5,0,-6),\ (-3,10,-9)\text{ and }(-2,6,-6)$
$\displaystyle \textbf{(iii) }\ (1,1,1),\ (1,-1,2)\text{ and }(-2,-2,2)$
$\displaystyle \textbf{(iv) }\ (2,3,4),\ (-3,5,1)\text{ and }(4,-1,2)\text{ \ [CBSE\ 2004]}$
$\displaystyle \textbf{(v) }\ (0,-1,0),\ (3,3,0)\text{ and }(1,1,1)$
$\displaystyle \text{Answer:}$
$\displaystyle \textbf{(i) }$
$\displaystyle \text{The equation of the plane passing through points }(2,1,0),(3,-2,-2) \\ \text{and }(3,1,7)\text{ is given by}$
$\displaystyle \begin{vmatrix}x-2&y-1&z-0\\3-2&-2-1&-2-0\\3-2&1-1&7-0\end{vmatrix}=0$
$\displaystyle \Rightarrow \begin{vmatrix}x-2&y-1&z-0\\1&-3&-2\\1&0&7\end{vmatrix}=0$
$\displaystyle \Rightarrow -21(x-2)-9(y-1)+3z=0$
$\displaystyle \Rightarrow -21x+42-9y+9+3z=0$
$\displaystyle \Rightarrow -21x-9y+3z+51=0$
$\displaystyle \Rightarrow 21x+9y-3z=51$
$\displaystyle \Rightarrow 7x+3y-z=17$

$\displaystyle \textbf{(ii) }$
$\displaystyle \text{The equation of the plane passing through points }(-5,0,-6),(-3,10,-9) \\ \text{and }(-2,6,-6)\text{ is given by}$
$\displaystyle \begin{vmatrix}x+5&y-0&z+6\\-3+5&10-0&-9+6\\-2+5&6-0&-6+6\end{vmatrix}=0$
$\displaystyle \Rightarrow \begin{vmatrix}x+5&y&z+6\\2&10&-3\\3&6&0\end{vmatrix}=0$
$\displaystyle \Rightarrow 18(x+5)-9y-18(z+6)=0$
$\displaystyle \Rightarrow 2(x+5)-y-2(z+6)=0$
$\displaystyle \Rightarrow 2x+10-y-2z-12=0$
$\displaystyle \Rightarrow 2x-y-2z-2=0$

$\displaystyle \textbf{(iii) }$
$\displaystyle \text{The equation of the plane passing through points }(1,1,1),(1,-1,2) \\ \text{and }(-2,-2,2)\text{ is given by}$
$\displaystyle \begin{vmatrix}x-1&y-1&z-1\\1-1&-1-1&2-1\\-2-1&-2-1&2-1\end{vmatrix}=0$
$\displaystyle \Rightarrow \begin{vmatrix}x-1&y-1&z-1\\0&-2&1\\-3&-3&1\end{vmatrix}=0$
$\displaystyle \Rightarrow 1(x-1)-3(y-1)-6(z-1)=0$
$\displaystyle \Rightarrow x-1-3y+3-6z+6=0$
$\displaystyle \Rightarrow x-3y-6z+8=0$

$\displaystyle \textbf{(iv) }$
$\displaystyle \text{The equation of the plane passing through points }(2,3,4),(-3,5,1) \\ \text{and }(4,-1,2)\text{ is given by}$
$\displaystyle \begin{vmatrix}x-2&y-3&z-4\\-3-2&5-3&1-4\\4-2&-1-3&2-4\end{vmatrix}=0$
$\displaystyle \Rightarrow \begin{vmatrix}x-2&y-3&z-4\\-5&2&-3\\2&-4&-2\end{vmatrix}=0$
$\displaystyle \Rightarrow -16(x-2)-16(y-3)+16(z-4)=0$
$\displaystyle \Rightarrow (x-2)+(y-3)-(z-4)=0$
$\displaystyle \Rightarrow x+y-z=1$

$\displaystyle \textbf{(v) }$
$\displaystyle \text{The equation of the plane passing through points }(0,-1,0),(3,3,0) \\ \text{and }(1,1,1)\text{ is given by}$
$\displaystyle \begin{vmatrix}x-0&y+1&z-0\\3-0&3+1&0-0\\1-0&1+1&1-0\end{vmatrix}=0$
$\displaystyle \Rightarrow \begin{vmatrix}x&y+1&z\\3&4&0\\1&2&1\end{vmatrix}=0$
$\displaystyle \Rightarrow 4x-3(y+1)+2z=0$
$\displaystyle \Rightarrow 4x-3y+2z=3$

$\displaystyle \textbf{Question 2: }~\text{Show that the four points }(0,-1,-1),\ (4,5,1),\ (3,9,4) \\ \text{and }(-4,4,4)\text{ are coplanar and find the equation of the common plane.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The equation of the plane passing through the points }(0,-1,-1),(4,5,1)\text{ and }(3,9,4)\text{ is given by}$
$\displaystyle \begin{vmatrix}x-0&y+1&z+1\\4-0&5+1&1+1\\3-0&9+1&4+1\end{vmatrix}=0$
$\displaystyle \Rightarrow \begin{vmatrix}x&y+1&z+1\\4&6&2\\3&10&5\end{vmatrix}=0$
$\displaystyle \Rightarrow 10x-14(y+1)+22(z+1)=0$
$\displaystyle \Rightarrow 5x-7(y+1)+11(z+1)=0$
$\displaystyle \Rightarrow 5x-7y+11z+4=0$
$\displaystyle \text{Substituting the last point }(-4,4,4)\text{ (it means }x=-4;y=4;z=4\text{) in this plane equation, we get}$
$\displaystyle 5(-4)-7(4)+11(4)+4=0$
$\displaystyle \Rightarrow -48+48=0$
$\displaystyle \Rightarrow 0=0$
$\displaystyle \text{So, the plane equation is satisfied by the point }(-4,4,4).$
$\displaystyle \text{So, the given points are coplanar and the equation of the common plane (as we already found) is}$
$\displaystyle 5x-7y+11z+4=0$

$\displaystyle \textbf{Question 3: }~\text{Show that the following points are coplanar:}$
$\displaystyle \textbf{(i) }\ (0,-1,0),\ (2,1,-1),\ (1,1,1)\text{ and }(3,3,0)$
$\displaystyle \textbf{(ii) }\ (0,4,3),\ (-1,-5,-3),\ (-2,-2,1)\text{ and }(1,1,-1)$
$\displaystyle \text{Answer:}$
$\displaystyle \textbf{(i) }$
$\displaystyle \text{The equation of the plane passing through points }(0,-1,0),(2,1,-1)\text{ and }(1,1,1)\text{ is given by}$
$\displaystyle \begin{vmatrix}x-0&y+1&z-0\\2-0&1+1&-1-0\\1-0&1+1&1-0\end{vmatrix}=0$
$\displaystyle \Rightarrow \begin{vmatrix}x&y+1&z\\2&2&-1\\1&2&1\end{vmatrix}=0$
$\displaystyle \Rightarrow 4x-3(y+1)+2z=0$
$\displaystyle \Rightarrow 4x-3y+2z-3=0$
$\displaystyle \text{Substituting the last point }(3,3,0)\text{ (it means }x=3;y=3;z=0\text{) in this plane equation, we get}$
$\displaystyle 4(3)-3(3)+2(0)-3=0$
$\displaystyle \Rightarrow 12-12=0$
$\displaystyle \Rightarrow 0=0$
$\displaystyle \text{So, the plane equation is satisfied by the point }(3,3,0).$
$\displaystyle \text{So, the given points are coplanar.}$

$\displaystyle \textbf{(ii) }$
$\displaystyle \text{The equation of the plane passing through }(0,4,3),(-1,-5,-3)\text{ and }(-2,-2,1)\text{ is}$
$\displaystyle \begin{vmatrix}x-0&y-4&z-3\\-1-0&-5-4&-3-3\\-2-0&-2-4&1-3\end{vmatrix}=0$
$\displaystyle \Rightarrow \begin{vmatrix}x&y-4&z-3\\-1&-9&-6\\-2&-6&-2\end{vmatrix}=0$
$\displaystyle \Rightarrow -18x+10(y-4)-12(z-3)=0$
$\displaystyle \Rightarrow 9x-5(y-4)+6(z-3)=0$
$\displaystyle \Rightarrow 9x-5y+6z+2=0$
$\displaystyle \text{Substituting the last point }(1,1,-1)\text{ (it means }x=1;y=1;z=-1\text{) in this plane equation, we get}$
$\displaystyle 9(1)-5(1)+6(-1)+2=0$
$\displaystyle \Rightarrow 4-4=0$
$\displaystyle \Rightarrow 0=0$
$\displaystyle \text{So, the plane equation is satisfied by the point }(1,1,-1).$
$\displaystyle \text{So, the given points are coplanar.}$

$\displaystyle \textbf{Question 4: }~\text{Find the coordinates of the point }P\text{ where the line through } \\ A(3,-4,-5) \text{ and }B(2,-3,1)\text{ crosses the plane passing through three points } \\ L(2,2,1), M(3,0,1)\text{ and }N(4,-1,0). \text{Also, find the ratio in which }P\text{ divides the} \\ \text{line segment }AB.\text{ \ [CBSE\ 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Equation of the plane passing through the points }L(2,2,1),M(3,0,1)\text{ and }N(4,-1,0)\text{ is}$
$\displaystyle [\overrightarrow{r}-(2\widehat{i}+2\widehat{j}+\widehat{k})]\cdot[(\widehat{i}-2\widehat{j})\times(\widehat{i}-\widehat{j}-\widehat{k})]=0$
$\displaystyle \Rightarrow [\overrightarrow{r}-(2\widehat{i}+2\widehat{j}+\widehat{k})]\cdot(2\widehat{i}+\widehat{j}+\widehat{k})=0$
$\displaystyle \Rightarrow \overrightarrow{r}\cdot(2\widehat{i}+\widehat{j}+\widehat{k})=(2\widehat{i}+2\widehat{j}+\widehat{k})\cdot(2\widehat{i}+\widehat{j}+\widehat{k})$
$\displaystyle \Rightarrow \overrightarrow{r}\cdot(2\widehat{i}+\widehat{j}+\widehat{k})=4+2+1=7\quad ...(1)$
$\displaystyle \text{The equation of line segment through }A(3,-4,-5)\text{ and }B(2,-3,1)\text{ is}$
$\displaystyle \frac{x-3}{2-3}=\frac{y+4}{-3+4}=\frac{z+5}{1+5}$
$\displaystyle \text{i.e. }\frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}$
$\displaystyle \text{Any point on this line is of the form }(-\lambda+3,\lambda-4,6\lambda-5)$
$\displaystyle \text{This point lies on the plane }(1).$
$\displaystyle [(-\lambda+3)\widehat{i}+(\lambda-4)\widehat{j}+(6\lambda-5)\widehat{k}]\cdot(2\widehat{i}+\widehat{j}+\widehat{k})=7$
$\displaystyle \Rightarrow 2(-\lambda+3)+(\lambda-4)+(6\lambda-5)=7$
$\displaystyle \Rightarrow 5\lambda=10$
$\displaystyle \Rightarrow \lambda=2$
$\displaystyle \text{Thus, the coordinates of the point }P\text{ are }(-2+3,2-4,6\times2-5)\text{ i.e. }(1,-2,7).$
$\displaystyle \text{Suppose }P\text{ divides the line segment }AB\text{ in the ratio }\mu:1.$
$\displaystyle \therefore (1,-2,7)=\left(\frac{2\mu+3}{\mu+1},\frac{-3\mu-4}{\mu+1},\frac{\mu-5}{\mu+1}\right)$
$\displaystyle \Rightarrow \frac{2\mu+3}{\mu+1}=1,\frac{-3\mu-4}{\mu+1}=-2,\frac{\mu-5}{\mu+1}=7$
$\displaystyle \Rightarrow 2\mu+3=\mu+1,-3\mu-4=-2\mu-2,\mu-5=7\mu+7$
$\displaystyle \Rightarrow \mu=-2$
$\displaystyle \text{Thus, the point }P\text{ divides the line segment }AB\text{ externally in the ratio }2:1.$