Class 12: The Plane – Exercise 29.13

$\displaystyle \textbf{Question 1: }~\text{Show that the lines }\overrightarrow{r}=(2\widehat{j}-3\widehat{k})+\lambda(\widehat{i}+2\widehat{j}+3\widehat{k})\text{ and } \\ \overrightarrow{r}=(2\widehat{i}+6\widehat{j}+3\widehat{k})+\mu(2\widehat{i}+3\widehat{j}+4\widehat{k})\text{ are coplanar. Also, find the equation of the} \\ \text{plane containing them.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We know that the lines } \overrightarrow{r}=\overrightarrow{a_1}+\lambda\overrightarrow{b_1} \text{ and } \overrightarrow{r}=\overrightarrow{a_2}+\mu\overrightarrow{b_2} \text{ are coplanar if}$
$\displaystyle \overrightarrow{a_1}\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})=\overrightarrow{a_2}\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})$
$\displaystyle \text{and the equation of the plane containing them is}$
$\displaystyle \overrightarrow{r}\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})=\overrightarrow{a_1}\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})$
$\displaystyle \text{Here,}$
$\displaystyle \overrightarrow{a_1}=0\hat{i}+2\hat{j}-3\hat{k},\;\overrightarrow{b_1}=\hat{i}+2\hat{j}+3\hat{k},\;\overrightarrow{a_2}=2\hat{i}+6\hat{j}+3\hat{k},\;\overrightarrow{b_2}=2\hat{i}+3\hat{j}+4\hat{k}$
$\displaystyle \overrightarrow{b_1}\times\overrightarrow{b_2}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&2&3\\2&3&4\end{vmatrix}$
$\displaystyle =\hat{i}(8-9)-\hat{j}(4-6)+\hat{k}(3-4)$
$\displaystyle =-\hat{i}+2\hat{j}-\hat{k}$
$\displaystyle \overrightarrow{a_1}\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})=(0\hat{i}+2\hat{j}-3\hat{k})\cdot(-\hat{i}+2\hat{j}-\hat{k})$
$\displaystyle =0+4+3=7$
$\displaystyle \overrightarrow{a_2}\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})=(2\hat{i}+6\hat{j}+3\hat{k})\cdot(-\hat{i}+2\hat{j}-\hat{k})$
$\displaystyle =-2+12-3=7$
$\displaystyle \text{Clearly, }\overrightarrow{a_1}\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})=\overrightarrow{a_2}\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})$
$\displaystyle \text{Hence, the given lines are coplanar.}$
$\displaystyle \text{The equation of the plane containing the given lines is}$
$\displaystyle \overrightarrow{r}\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})=\overrightarrow{a_1}\cdot(\overrightarrow{b_1}\times\overrightarrow{b_2})$
$\displaystyle \Rightarrow \overrightarrow{r}\cdot(-\hat{i}+2\hat{j}-\hat{k})=(0\hat{i}+2\hat{j}-3\hat{k})\cdot(-\hat{i}+2\hat{j}-\hat{k})$
$\displaystyle \Rightarrow \overrightarrow{r}\cdot(-\hat{i}+2\hat{j}-\hat{k})=7$
$\displaystyle \Rightarrow \overrightarrow{r}\cdot(\hat{i}-2\hat{j}+\hat{k})+7=0$

$\displaystyle \textbf{Question 2: }~\text{Show that the lines }\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{-1}\text{ and }\frac{x}{1}=\frac{y-7}{-3}=\frac{z+7}{2}\text{ are coplanar. Also, find the equation of the plane containing them.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We know that the lines}$
$\displaystyle \frac{x-x_1}{l_1}=\frac{y-y_1}{m_1}=\frac{z-z_1}{n_1} \text{ and } \frac{x-x_2}{l_2}=\frac{y-y_2}{m_2}=\frac{z-z_2}{n_2} \text{ are coplanar if}$
$\displaystyle \begin{vmatrix}x_2-x_1&y_2-y_1&z_2-z_1\\l_1&m_1&n_1\\l_2&m_2&n_2\end{vmatrix}=0$
$\displaystyle \text{and the equation of the plane containing these lines is}$
$\displaystyle \begin{vmatrix}x-x_1&y-y_1&z-z_1\\l_1&m_1&n_1\\l_2&m_2&n_2\end{vmatrix}=0$
$\displaystyle \text{Here,}$
$\displaystyle x_1=-1,\;y_1=3,\;z_1=-2,\;x_2=0,\;y_2=7,\;z_2=-7,\;l_1=-3,\;m_1=2,\;n_1=1,\;l_2=1,\;m_2=-3,\;n_2=2$
$\displaystyle \text{Now,}$
$\displaystyle \begin{vmatrix}0+1&7-3&-7+2\\-3&2&1\\1&-3&2\end{vmatrix}$
$\displaystyle =\begin{vmatrix}1&4&-5\\-3&2&1\\1&-3&2\end{vmatrix}$
$\displaystyle =1\begin{vmatrix}2&1\\-3&2\end{vmatrix}-4\begin{vmatrix}-3&1\\1&2\end{vmatrix}-5\begin{vmatrix}-3&2\\1&-3\end{vmatrix}$
$\displaystyle =1(4+3)-4(-6-1)-5(9-2)$
$\displaystyle =7+28-35$
$\displaystyle =0$
$\displaystyle \text{So, the given lines are coplanar.}$
$\displaystyle \text{The equation of the plane containing the given lines is}$
$\displaystyle \begin{vmatrix}x+1&y-3&z+2\\-3&2&1\\1&-3&2\end{vmatrix}=0$
$\displaystyle \Rightarrow (x+1)\begin{vmatrix}2&1\\-3&2\end{vmatrix}-(y-3)\begin{vmatrix}-3&1\\1&2\end{vmatrix}+(z+2)\begin{vmatrix}-3&2\\1&-3\end{vmatrix}=0$
$\displaystyle \Rightarrow 7(x+1)+7(y-3)+7(z+2)=0$
$\displaystyle \Rightarrow 7x+7y+7z=0$
$\displaystyle \Rightarrow x+y+z=0$

$\displaystyle \textbf{Question 3: }~\text{Find the equation of the plane containing the line }\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{-1}\text{ and the point }(0,7,-7)\text{ and show that the line }\frac{x}{1}=\frac{y-7}{-3}=\frac{z+7}{2} \\ \text{ also lies in the same plane.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the equation of the plane passing through }(0,7,-7)\text{ be}$
$\displaystyle a(x-0)+b(y-7)+c(z+7)=0\ldots(1)$
$\displaystyle \text{The line } \frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1} \text{ passes through }(-1,3,-2) \\ \text{ and its direction ratios are proportional to }-3,2,1.$
$\displaystyle \text{Since plane (1) contains this line, it must pass through the point }(-1,3,-2).$
$\displaystyle \Rightarrow a(-1-0)+b(3-7)+c(-2+7)=0$
$\displaystyle \Rightarrow -a-4b+5c=0$
$\displaystyle \Rightarrow a+4b-5c=0\ldots(2)$
$\displaystyle \text{Since plane (1) contains this line, it must be parallel to the line.}$
$\displaystyle \Rightarrow -3a+2b+c=0\ldots(3)$
$\displaystyle \text{Solving (1), (2) and (3), we get}$
$\displaystyle \begin{vmatrix}x-0&y-7&z+7\\1&4&-5\\-3&2&1\end{vmatrix}=0$
$\displaystyle \Rightarrow x\begin{vmatrix}4&-5\\2&1\end{vmatrix}-(y-7)\begin{vmatrix}1&-5\\-3&1\end{vmatrix}+(z+7)\begin{vmatrix}1&4\\-3&2\end{vmatrix}=0$
$\displaystyle \Rightarrow 14x+14(y-7)+14(z+7)=0$
$\displaystyle \Rightarrow 14x+14y+14z=0$
$\displaystyle \Rightarrow x+y+z=0$

$\displaystyle \textbf{Question 4: }~\text{Find the equation of the plane which contains two parallel lines }\frac{x-4}{1}=\frac{y-3}{-4}=\frac{z-2}{5}\text{ and }\frac{x-1}{1}=\frac{y+2}{-4}=\frac{z}{5}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We know that the equation of the plane containing two given parallel lines}$
$\displaystyle \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c} \text{ and } \frac{x-x_2}{a}=\frac{y-y_2}{b}=\frac{z-z_2}{c} \text{ is}$
$\displaystyle \begin{vmatrix}x-x_1&y-y_1&z-z_1\\x_2-x_1&y_2-y_1&z_2-z_1\\a&b&c\end{vmatrix}=0$
$\displaystyle \text{Here,}$
$\displaystyle x_1=4,\;y_1=3,\;z_1=2,\;x_2=3,\;y_2=-2,\;z_2=0,\;a=1,\;b=-4,\;c=5$
$\displaystyle \text{Now,}$
$\displaystyle \begin{vmatrix}x-4&y-3&z-2\\3-4&-2-3&0-2\\1&-4&5\end{vmatrix}=0$
$\displaystyle \Rightarrow \begin{vmatrix}x-4&y-3&z-2\\-1&-5&-2\\1&-4&5\end{vmatrix}=0$
$\displaystyle \Rightarrow (x-4)\begin{vmatrix}-5&-2\\-4&5\end{vmatrix}-(y-3)\begin{vmatrix}-1&-2\\1&5\end{vmatrix}+(z-2)\begin{vmatrix}-1&-5\\1&-4\end{vmatrix}=0$
$\displaystyle \Rightarrow (x-4)(-25-8)-(y-3)(-5+2)+(z-2)(4+5)=0$
$\displaystyle \Rightarrow -33(x-4)+3(y-3)+9(z-2)=0$
$\displaystyle \Rightarrow 11(x-4)-(y-3)-3(z-2)=0$
$\displaystyle \Rightarrow 11x-y-3z=35$

$\displaystyle \textbf{Question 5: }~\text{Show that the lines }\frac{x+4}{3}=\frac{y+6}{5}=\frac{z-1}{2}\text{ and }3x-2y+z+5=0=2x+3y+4z-4\text{ intersect. Find the equation of the plane in which} \\ \text{they lie and also their point of intersection.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The equation of the given line is}$
$\displaystyle \frac{x+4}{3}=\frac{y+6}{5}=\frac{z-1}{-2}$
$\displaystyle \text{The coordinates of any point on this line are of the form}$
$\displaystyle \frac{x+4}{3}=\frac{y+6}{5}=\frac{z-1}{-2}=\lambda$
$\displaystyle \Rightarrow x=3\lambda-4;\;y=5\lambda-6;\;z=-2\lambda+1$
$\displaystyle \text{So, the coordinates of the point on the given line are }(3\lambda-4,5\lambda-6,-2\lambda+1). \text{ Since this point lies on the plane }$
$\displaystyle 3x-2y+z+5=0,$
$\displaystyle 3(3\lambda-4)-2(5\lambda-6)+(-2\lambda+1)+5=0$
$\displaystyle \Rightarrow 9\lambda-12-10\lambda+12-2\lambda+1+5=0$
$\displaystyle \Rightarrow -3\lambda+6=0$
$\displaystyle \Rightarrow \lambda=2$
$\displaystyle \text{So, the coordinates of the point are}$
$\displaystyle (3\lambda-4,5\lambda-6,-2\lambda+1)$
$\displaystyle =(3(2)-4,5(2)-6,-2(2)+1)$
$\displaystyle =(2,4,-3)$
$\displaystyle \text{Substituting this point in another plane equation }2x+3y+4z-4=0,\text{ we get}$
$\displaystyle 2(2)+3(4)+4(-3)-4=0$
$\displaystyle \Rightarrow 4+12-12-4=0$
$\displaystyle \Rightarrow 0=0$
$\displaystyle \text{So, the point }(2,4,-3)\text{ lies on another plane too. So, this is the point of intersection of both the lines.}$
$\displaystyle \text{Finding the plane equation}$
$\displaystyle \text{Let the direction ratios be proportional to }a,b,c.$
$\displaystyle \text{Since the plane contains the line } \frac{x+4}{3}=\frac{y+6}{5}=\frac{z-1}{-2}, \\ \text{ it must pass through the point }(-4,-6,1)\text{ and is parallel to this line.}$
$\displaystyle \text{So, the equation of plane is}$
$\displaystyle a(x+4)+b(y+6)+c(z-1)=0\ldots(1)$
$\displaystyle \text{and}$
$\displaystyle 3a+5b-2c=0\ldots(2)$
$\displaystyle \text{Since the given plane contains the planes }3x-2y+z+5=0 \text{ and } 2x+3y+4z-4=0,$
$\displaystyle 3a-2b+c=0\ldots(3)$
$\displaystyle 2a+3b+4c=0\ldots(4)$
$\displaystyle \text{Solving (3) and (4) using cross-multiplication, we get}$
$\displaystyle \frac{a}{-11}=\frac{b}{-10}=\frac{c}{13}\ldots(5)$
$\displaystyle \text{Using (1), (2) and (5), the equation of plane is}$
$\displaystyle \begin{vmatrix}x+4&y+6&z-1\\3&5&-2\\11&10&-13\end{vmatrix}=0$
$\displaystyle \Rightarrow -45(x+4)+17(y+6)-25(z-1)=0$
$\displaystyle \Rightarrow 45(x+4)-17(y+6)+25(z-1)=0$
$\displaystyle \Rightarrow 45x-17y+25z+53=0$

$\displaystyle \textbf{Question 6: }~\text{Show that the plane whose vector equation is }\overrightarrow{r}\cdot(\widehat{i}+2\widehat{j}-\widehat{k})=3\text{ contains the line whose vector equation is }\overrightarrow{r}=\widehat{i}+\widehat{j}+\lambda(2\widehat{i}+\widehat{j}+4\widehat{k}).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The line } \overrightarrow{r}=(\hat{i}+\hat{j}+0\hat{k})+\lambda(2\hat{i}+\hat{j}+4\hat{k})\ldots(1)\text{ passes through a point whose position vector is } \\ \overrightarrow{a}=\hat{i}+\hat{j}+0\hat{k}\text{ and is parallel to the vector }\overrightarrow{b}=2\hat{i}+\hat{j}+4\hat{k}.$
$\displaystyle \text{If the plane } \overrightarrow{r}\cdot(\hat{i}+2\hat{j}-\hat{k})=3 \text{ contains the given line, then}$
$\displaystyle (1)\ \text{it should pass through the point }\hat{i}+\hat{j}+0\hat{k}$
$\displaystyle (2)\ \text{it should be parallel to the line}$
$\displaystyle \text{Now, }(\hat{i}+\hat{j}+0\hat{k})\cdot(\hat{i}+2\hat{j}-\hat{k})=1+2=3$
$\displaystyle \text{So, the plane passes through the point }\hat{i}+\hat{j}+0\hat{k}.$
$\displaystyle \text{The normal vector to the given plane is }\overrightarrow{n}=\hat{i}+2\hat{j}-\hat{k}$
$\displaystyle \text{We observe that}$
$\displaystyle \overrightarrow{b}\cdot\overrightarrow{n}=(2\hat{i}+\hat{j}+4\hat{k})\cdot(\hat{i}+2\hat{j}-\hat{k})=2+2-4=0$
$\displaystyle \text{Therefore, the plane is parallel to the line.}$
$\displaystyle \text{Hence, the given plane contains the given line.}$

$\displaystyle \textbf{Question 7: }~\text{Find the equation of the plane determined by the intersection of the lines } \\ \frac{x+3}{3}=\frac{y}{-2}=\frac{z-7}{6}\text{ and }\frac{x+6}{1}=\frac{y+5}{-3}=\frac{z-1}{2}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given equations of the lines are}$
$\displaystyle \frac{x+3}{3}=\frac{y}{-2}=\frac{z-7}{6}\ldots(1)$
$\displaystyle \frac{x+6}{1}=\frac{y+5}{-3}=\frac{z-1}{2}\ldots(2)$
$\displaystyle \text{Let the direction ratios of the plane be proportional to }a,b,c.$
$\displaystyle \text{Since the plane contains line (1), it should pass through }(-3,0,7)\text{ and is parallel to line (1).}$
$\displaystyle \text{Equation of the plane through (1) is}$
$\displaystyle a(x+3)+b(y)+c(z-7)=0\ldots(3)$
$\displaystyle \text{where }3a-2b+6c=0\ldots(4)$
$\displaystyle \text{Since the plane contains line (2), the plane is parallel to line (2) also.}$
$\displaystyle \Rightarrow a-3b+2c=0\ldots(5)$
$\displaystyle \text{Solving (4) and (5) using cross-multiplication, we get}$
$\displaystyle \frac{a}{14}=\frac{b}{0}=\frac{c}{-7}$
$\displaystyle \text{Substituting }a,b\text{ and }c\text{ in (3), we get}$
$\displaystyle 14(x+3)+0(y)-7(z-7)=0$
$\displaystyle \Rightarrow 2(x+3)+0(y)-1(z-7)=0$
$\displaystyle \Rightarrow 2x-z+13=0$

$\displaystyle \textbf{Question 8: }~\text{Find the vector equation of the plane passing through the points }(3,4,2) \\ \text{ and }(7,0,6)\text{ and perpendicular to the plane }2x-5y-15=0.\text{ Also, show that} \\ \text{the plane thus obtained contains the line }\overrightarrow{r}=\widehat{i}+3\widehat{j}-2\widehat{k}+\lambda(\widehat{i}-\widehat{j}+\widehat{k}).\ \text{[CBSE\ 2012]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The equation of any plane passing through }(3,4,2)\text{ is}$
$\displaystyle a(x-3)+b(y-4)+c(z-2)=0\ldots(1)$
$\displaystyle \text{It is given that (1) is passing through }(7,0,6).\text{ So,}$
$\displaystyle a(7-3)+b(0-4)+c(6-2)=0$
$\displaystyle \Rightarrow 4a-4b+4c=0$
$\displaystyle \Rightarrow a-b+c=0\ldots(2)$
$\displaystyle \text{It is given that (1) is perpendicular to the plane }2x-5y+0z+15z=0.\text{ So,}$
$\displaystyle 2a-5b+0c=0\ldots(3)$
$\displaystyle \text{Solving (1), (2) and (3), we get}$
$\displaystyle \begin{vmatrix}x-3&y-4&z-2\\1&-1&1\\2&-5&0\end{vmatrix}=0$
$\displaystyle \Rightarrow (x-3)\begin{vmatrix}-1&1\\-5&0\end{vmatrix}-(y-4)\begin{vmatrix}1&1\\2&0\end{vmatrix}+(z-2)\begin{vmatrix}1&-1\\2&-5\end{vmatrix}=0$
$\displaystyle \Rightarrow 5(x-3)+2(y-4)-3(z-2)=0$
$\displaystyle \Rightarrow 5x+2y-3z=17$
$\displaystyle \text{Or }\overrightarrow{r}\cdot(5\hat{i}+2\hat{j}-3\hat{k})=17$
$\displaystyle \text{Showing that the plane contains the line}$
$\displaystyle \text{The line }\overrightarrow{r}=(\hat{i}+3\hat{j}-2\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})\text{ passes through a point whose position vector is } \\ \overrightarrow{a}=\hat{i}+3\hat{j}-2\hat{k}\text{ and is parallel to the vector }\overrightarrow{b}=\hat{i}-\hat{j}+\hat{k}.$
$\displaystyle \text{If the plane }\overrightarrow{r}\cdot(5\hat{i}+2\hat{j}-3\hat{k})=17\text{ contains the given line, then}$
$\displaystyle (1)\ \text{it should pass through the point }\hat{i}+3\hat{j}-2\hat{k}$
$\displaystyle (2)\ \text{it should be parallel to the line}$
$\displaystyle \text{Now, }(\hat{i}+3\hat{j}-2\hat{k})\cdot(5\hat{i}+2\hat{j}-3\hat{k})=5+6+6=17$
$\displaystyle \text{So, the plane passes through the point }\hat{i}+3\hat{j}-2\hat{k}.$
$\displaystyle \text{The normal vector to the given plane is }\overrightarrow{n}=\hat{i}-\hat{j}+\hat{k}$
$\displaystyle \text{We observe that}$
$\displaystyle \overrightarrow{b}\cdot\overrightarrow{n}=(\hat{i}-\hat{j}+\hat{k})\cdot(5\hat{i}+2\hat{j}-3\hat{k})=5-2-3=0$
$\displaystyle \text{Therefore, the plane is parallel to the line.}$
$\displaystyle \text{Hence, the given plane contains the given line.}$

$\displaystyle \textbf{Question 9: }~\text{If the lines }\frac{x-1}{-3}=\frac{y-2}{-2k}=\frac{z-3}{2}\text{ and }\frac{x-1}{k}=\frac{y-2}{1}=\frac{z-3}{5}\text{ are perpendicular, find the value of }k\text{ and hence find the equation} \\ \text{of the plane containing these lines. \ [CBSE\ 2012]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We know that the lines}$
$\displaystyle \frac{x-x_1}{l_1}=\frac{y-y_1}{m_1}=\frac{z-z_1}{n_1} \text{ and } \frac{x-x_2}{l_2}=\frac{y-y_2}{m_2}=\frac{z-z_2}{n_2} \text{ are perpendicular if}$
$\displaystyle l_1l_2+m_1m_2+n_1n_2=0$
$\displaystyle \text{Here,}$
$\displaystyle l_1=-3,\;m_1=-2k,\;n_1=2,\;l_2=k,\;m_2=1,\;n_2=5$
$\displaystyle \text{It is given that the given lines are perpendicular.}$
$\displaystyle \Rightarrow l_1l_2+m_1m_2+n_1n_2=0$
$\displaystyle \Rightarrow (-3)(k)+(-2k)(1)+(2)(5)=0$
$\displaystyle \Rightarrow -3k-2k+10=0$
$\displaystyle \Rightarrow -5k=-10$
$\displaystyle \Rightarrow k=2$
$\displaystyle \text{Substituting this value in the given equations of the lines, we get}$
$\displaystyle \frac{x-1}{-3}=\frac{y-2}{-4}=\frac{z-3}{2}\ldots(1)$
$\displaystyle \frac{x-1}{2}=\frac{y-2}{1}=\frac{z-3}{5}\ldots(2)$
$\displaystyle \text{Finding the equation of the plane}$
$\displaystyle \text{Let the direction ratios of the required plane be proportional to }a,b,c.$
$\displaystyle \text{We know from (1) and (2) that lines (1) and (2) pass through the point } \\ (1,2,3)\text{ and the direction ratios of (1) and (2) are proportional to } \\ -3,-4,2 \text{ and } 2,1,5 \text{ respectively.}$
$\displaystyle \text{Since the plane contains the lines (1) and (2), the plane must pass through the point } \\ (1,2,3)\text{ and it must be parallel to the lines.}$
$\displaystyle \text{So, the equation of the plane is}$
$\displaystyle a(x-1)+b(y-2)+c(z-3)=0\ldots(3)$
$\displaystyle -3a-4b+2c=0\ldots(4)$
$\displaystyle 2a+b+5c=0\ldots(5)$
$\displaystyle \text{Solving (4) and (5), we get}$
$\displaystyle \begin{vmatrix}x-1&y-2&z-3\\-3&-4&2\\2&1&5\end{vmatrix}=0$
$\displaystyle \Rightarrow (x-1)\begin{vmatrix}-4&2\\1&5\end{vmatrix}-(y-2)\begin{vmatrix}-3&2\\2&5\end{vmatrix}+(z-3)\begin{vmatrix}-3&-4\\2&1\end{vmatrix}=0$
$\displaystyle \Rightarrow -22(x-1)+19(y-2)+5(z-3)=0$
$\displaystyle \Rightarrow -22x+19y+5z=31$

$\displaystyle \textbf{Question 10: }~\text{Find the coordinates of the point where the line }\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}\text{ intersects the plane }x-y+z-5=0.\text{ Also, find the angle between} \\ \text{the line and the plane. \ [CBSE\ 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The coordinates of any point on this line are of the form}$
$\displaystyle \frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}=\lambda$
$\displaystyle \Rightarrow x=3\lambda+2;\;y=4\lambda-1;\;z=2\lambda+2$
$\displaystyle \text{So, the coordinates of the point on the given line are }(3\lambda+2,4\lambda-1,2\lambda+2). \\ \text{ This point lies on the plane }x-y+z-5=0.$
$\displaystyle \Rightarrow 3\lambda+2-4\lambda+1+2\lambda+2-5=0$
$\displaystyle \Rightarrow \lambda=0$
$\displaystyle \text{So, the coordinates of the point are}$
$\displaystyle (3\lambda+2,4\lambda-1,2\lambda+2)$
$\displaystyle =(3(0)+2,4(0)-1,2(0)+2)$
$\displaystyle =(2,-1,2)$
$\displaystyle \text{Finding the angle between the line and the plane}$
$\displaystyle \text{The given line is parallel to the vector }\overrightarrow{b}=3\hat{i}+4\hat{j}+2\hat{k} \\ \text{ and the given plane is normal to the vector }\overrightarrow{n}=\hat{i}-\hat{j}+\hat{k}.$
$\displaystyle \text{We know that the angle }\theta\text{ between the line and the plane is given by}$
$\displaystyle \sin\theta=\frac{\overrightarrow{b}\cdot\overrightarrow{n}}{\left|\overrightarrow{b}\right|\left|\overrightarrow{n}\right|}$
$\displaystyle =\frac{(3\hat{i}+4\hat{j}+2\hat{k})\cdot(\hat{i}-\hat{j}+\hat{k})}{\sqrt{9+16+4}\sqrt{1+1+1}}$
$\displaystyle =\frac{3-4+2}{\sqrt{87}}=\frac{1}{\sqrt{87}}$
$\displaystyle \Rightarrow \theta=\sin^{-1}\!\left(\frac{1}{\sqrt{87}}\right)$

$\displaystyle \textbf{Question 11: }~\text{Find the vector equation of the plane passing through three points} \\ \text{with position vectors }\widehat{i}+\widehat{j}-2\widehat{k},\ 2\widehat{i}-\widehat{j}+\widehat{k}\text{ and }\widehat{i}+2\widehat{j}+\widehat{k}.\text{ Also, find} \\ \text{the coordinates of the point of intersection of this plane and the line }\overrightarrow{r}=3\widehat{i}-\widehat{j}-\widehat{k}+\lambda(2\widehat{i}-2\widehat{j}+\widehat{k}).\ \text{[CBSE\ 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A(1,1,-2),B(2,-1,1)\text{ and }C(1,2,1)\text{ be the points represented by the given position vectors.}$
$\displaystyle \text{The required plane passes through the point }A(1,1,-2)\text{ whose position vector is }\overrightarrow{a}=\hat{i}+\hat{j}-2\hat{k}\text{ and is normal to the vector }\overrightarrow{n}\text{ given by }$
$\displaystyle \overrightarrow{n}=\overrightarrow{AB}\times\overrightarrow{AC}$
$\displaystyle \text{Clearly, }\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=(2\hat{i}-\hat{j}+\hat{k})-(\hat{i}+\hat{j}-2\hat{k})=\hat{i}-2\hat{j}+3\hat{k}$
$\displaystyle \overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=(\hat{i}+2\hat{j}+\hat{k})-(\hat{i}+\hat{j}-2\hat{k})=0\hat{i}+\hat{j}+3\hat{k}$
$\displaystyle \overrightarrow{n}=\overrightarrow{AB}\times\overrightarrow{AC}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&-2&3\\0&1&3\end{vmatrix}$
$\displaystyle =\hat{i}(-6-3)-\hat{j}(3-0)+\hat{k}(1-0)$
$\displaystyle =-9\hat{i}-3\hat{j}+\hat{k}$
$\displaystyle \text{The vector equation of the required plane is}$
$\displaystyle \overrightarrow{r}\cdot\overrightarrow{n}=\overrightarrow{a}\cdot\overrightarrow{n}$
$\displaystyle \Rightarrow \overrightarrow{r}\cdot(-9\hat{i}-3\hat{j}+\hat{k})=(\hat{i}+\hat{j}-2\hat{k})\cdot(-9\hat{i}-3\hat{j}+\hat{k})$
$\displaystyle \Rightarrow \overrightarrow{r}\cdot(-9\hat{i}-3\hat{j}+\hat{k})=-9-3-2$
$\displaystyle \Rightarrow \overrightarrow{r}\cdot[-(9\hat{i}+3\hat{j}-\hat{k})]=-14$
$\displaystyle \Rightarrow \overrightarrow{r}\cdot(9\hat{i}+3\hat{j}-\hat{k})=14$
$\displaystyle \text{To find the point of intersection of this plane}$
$\displaystyle \text{The given equation of the line is}$
$\displaystyle \overrightarrow{r}=(3\hat{i}-\hat{j}-\hat{k})+\lambda(2\hat{i}-2\hat{j}+\hat{k})$
$\displaystyle \Rightarrow \overrightarrow{r}=(3+2\lambda)\hat{i}+(-1-2\lambda)\hat{j}+(-1+\lambda)\hat{k}$
$\displaystyle \text{The coordinates of any point on this line are in the form }(3+2\lambda,-1-2\lambda,-1+\lambda)$
$\displaystyle \text{Since this point lies on the plane }\overrightarrow{r}\cdot(9\hat{i}+3\hat{j}-\hat{k})=14,$
$\displaystyle [(3+2\lambda)\hat{i}+(-1-2\lambda)\hat{j}+(-1+\lambda)\hat{k}]\cdot(9\hat{i}+3\hat{j}-\hat{k})=14$
$\displaystyle \Rightarrow 27+18\lambda-3-6\lambda+1-\lambda=14$
$\displaystyle \Rightarrow 11\lambda=-11$
$\displaystyle \Rightarrow \lambda=-1$
$\displaystyle \text{So, the coordinates of the point are}$
$\displaystyle (3+2\lambda,-1-2\lambda,-1+\lambda)$
$\displaystyle =(3-2,-1+2,-1-1)$
$\displaystyle =(1,1,-2)$

$\displaystyle \textbf{Question 12: }~\text{Show that the lines }\frac{5-x}{-4}=\frac{y-7}{4}=\frac{z+3}{-5}\text{ and }\frac{x-7}{8}=\frac{2y-8}{2}=\frac{z-5}{3}\text{ are coplanar. \ [CBSE\ 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The equations of the given lines can be re-written as } \frac{x-5}{4}=\frac{y-7}{4}=\frac{z+3}{-5} \text{ and } \frac{x-8}{7}=\frac{y-4}{1}=\frac{z-5}{3}.$
$\displaystyle \text{We know that the lines}$
$\displaystyle \frac{x-x_1}{l_1}=\frac{y-y_1}{m_1}=\frac{z-z_1}{n_1} \text{ and } \frac{x-x_2}{l_2}=\frac{y-y_2}{m_2}=\frac{z-z_2}{n_2} \text{ are coplanar if}$
$\displaystyle \begin{vmatrix}x_2-x_1&y_2-y_1&z_2-z_1\\l_1&m_1&n_1\\l_2&m_2&n_2\end{vmatrix}=0$
$\displaystyle \text{Here,}$
$\displaystyle x_1=5,\;y_1=7,\;z_1=-3,\;x_2=8,\;y_2=4,\;z_2=5$
$\displaystyle l_1=4,\;m_1=4,\;n_1=-5,\;l_2=7,\;m_2=1,\;n_2=3$
$\displaystyle \therefore \begin{vmatrix}8-5&4-7&5-(-3)\\4&4&-5\\7&1&3\end{vmatrix}$
$\displaystyle =\begin{vmatrix}3&-3&8\\4&4&-5\\7&1&3\end{vmatrix}$
$\displaystyle =3\begin{vmatrix}4&-5\\1&3\end{vmatrix}-(-3)\begin{vmatrix}4&-5\\7&3\end{vmatrix}+8\begin{vmatrix}4&4\\7&1\end{vmatrix}$
$\displaystyle =3(12+5)+3(12+35)+8(4-28)$
$\displaystyle =51+141-192$
$\displaystyle =0$
$\displaystyle \text{So, the given lines are coplanar.}$

$\displaystyle \textbf{Question 13: }~\text{Find the equation of a plane which passes through the point } \\ (3,2,0) \text{ and contains the line }\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}. \text{[CBSE\ 2015]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the equation of the plane passing through }(3,2,0)\text{ be}$
$\displaystyle a(x-3)+b(y-2)+c(z-0)=0$
$\displaystyle \text{The line } \frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4} \text{ passes through the point }(3,6,4) \\ \text{ and its direction ratios are proportional to }1,5,4.$
$\displaystyle \text{If plane (1) contains this line, then it must pass through }(3,6,4)\text{ and must be parallel to the line.}$
$\displaystyle \therefore a(3-3)+b(6-2)+c(4-0)=0$
$\displaystyle \Rightarrow 4b+4c=0$
$\displaystyle \Rightarrow b+c=0\ldots(2)$
$\displaystyle \text{Also,}$
$\displaystyle 1\times a+5\times b+4\times c=0$
$\displaystyle \Rightarrow a+5b+4c=0\ldots(3)$
$\displaystyle \text{Solving (2) and (3), we get}$
$\displaystyle \frac{a}{4-5}=\frac{b}{1-0}=\frac{c}{0-1}$
$\displaystyle \Rightarrow \frac{a}{-1}=\frac{b}{1}=\frac{c}{-1}=\lambda$
$\displaystyle \Rightarrow a=-\lambda,\;b=\lambda,\;c=-\lambda$
$\displaystyle \text{Putting these values of }a,b,c\text{ in (1), we get}$
$\displaystyle -\lambda(x-3)+\lambda(y-2)-\lambda(z-0)=0$
$\displaystyle \Rightarrow -x+3+y-2-z=0$
$\displaystyle \Rightarrow -x+y-z+1=0$
$\displaystyle \Rightarrow x-y+z-1=0$
$\displaystyle \text{Thus, the equation of the required plane is }x-y+z-1=0.$

$\displaystyle \textbf{Question 14: }~\text{Show that the lines }\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}\text{ and } \\ \frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}\text{ are coplanar. Hence, find the equation of the plane} \\ \text{containing these lines. \ [CBSE\ 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The lines }\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}\text{ and }\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}\text{ are coplanar if}$
$\displaystyle \begin{vmatrix}x_2-x_1&y_2-y_1&z_2-z_1\\a_1&b_1&c_1\\a_2&b_2&c_2\end{vmatrix}=0$
$\displaystyle \text{The given lines are }\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}\text{ and }\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}$
$\displaystyle \text{Now,}$
$\displaystyle \begin{vmatrix}-1-(-3)&2-1&5-5\\-3&1&5\\-1&2&5\end{vmatrix}$
$\displaystyle =\begin{vmatrix}2&1&0\\-3&1&5\\-1&2&5\end{vmatrix}$
$\displaystyle =2\begin{vmatrix}1&5\\2&5\end{vmatrix}-1\begin{vmatrix}-3&5\\-1&5\end{vmatrix}+0\begin{vmatrix}-3&1\\-1&2\end{vmatrix}$
$\displaystyle =2(5-10)-1(-15+5)+0$
$\displaystyle =-10+10+0$
$\displaystyle =0$
$\displaystyle \text{So, the given lines are coplanar.}$
$\displaystyle \text{The equation of the plane containing the given lines is}$
$\displaystyle \begin{vmatrix}x+3&y-1&z-5\\-3&1&5\\-1&2&5\end{vmatrix}=0$
$\displaystyle \Rightarrow (x+3)\begin{vmatrix}1&5\\2&5\end{vmatrix}-(y-1)\begin{vmatrix}-3&5\\-1&5\end{vmatrix}+(z-5)\begin{vmatrix}-3&1\\-1&2\end{vmatrix}=0$
$\displaystyle \Rightarrow (x+3)(5-10)-(y-1)(-15+5)+(z-5)(-6+1)=0$
$\displaystyle \Rightarrow -5(x+3)+10(y-1)-5(z-5)=0$
$\displaystyle \Rightarrow x-2y+z=0$
$\displaystyle \text{Thus, the equation of the plane containing the given lines is }x-2y+z=0.$

$\displaystyle \textbf{Question 15: }~\text{If the line }\frac{x-3}{2}=\frac{y+2}{-1}=\frac{z+4}{-6}\text{ lies in the plane } \\ x+my-z=9,\text{ then find the value of }m.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The line }\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\text{ lies in the plane }Ax+By+Cz+D=0\text{ iff}$
$\displaystyle (i)\ Ax_1+By_1+Cz_1+D=0\text{ and }(ii)\ aA+bB+cC=0.$
$\displaystyle \text{It is given that the line }\frac{x-3}{2}=\frac{y+2}{-1}=\frac{z+4}{3}\text{ lies in the plane}$
$\displaystyle lx+my-z=9$
$\displaystyle \therefore l\times3+m\times(-2)-(-4)=9$
$\displaystyle \Rightarrow 3l-2m+4=9$
$\displaystyle \Rightarrow 3l-2m=5\ldots(1)$
$\displaystyle \text{Also,}$
$\displaystyle 2\times l+(-1)\times m+3\times(-1)=0$
$\displaystyle \Rightarrow 2l-m-3=0$
$\displaystyle \Rightarrow 2l-m=3\ldots(2)$
$\displaystyle \text{Solving (1) and (2), we get}$
$\displaystyle l=1\text{ and }m=-1$
$\displaystyle \therefore l^2+m^2=1^2+(-1)^2=1+1=2$
$\displaystyle \text{Thus, the value of }l^2+m^2\text{ is }2.$

$\displaystyle \textbf{Question 16: }~\text{Find the values of }\lambda\text{ for which the lines } \\ \frac{x-1}{1}=\frac{y-2}{2}=\frac{z+3}{\lambda^{2}} \text{ and }\frac{x-3}{1}=\frac{y-2}{\lambda^{2}}=\frac{z-1}{2}\text{ are coplanar.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The lines }\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}\text{ and }\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}\text{ are coplanar if}$
$\displaystyle \begin{vmatrix}x_2-x_1&y_2-y_1&z_2-z_1\\a_1&b_1&c_1\\a_2&b_2&c_2\end{vmatrix}=0$
$\displaystyle \text{The given lines } \frac{x-1}{1}=\frac{y-2}{2}=\frac{z+3}{\lambda^2} \text{ and } \frac{x-3}{1}=\frac{y-2}{\lambda^2}=\frac{z-1}{2}$
$\displaystyle \text{are coplanar.}$
$\displaystyle \therefore \begin{vmatrix}3-1&2-2&1-(-3)\\1&2&\lambda^2\\1&\lambda^2&2\end{vmatrix}=0$
$\displaystyle =\begin{vmatrix}2&0&4\\1&2&\lambda^2\\1&\lambda^2&2\end{vmatrix}$
$\displaystyle =2\begin{vmatrix}2&\lambda^2\\\lambda^2&2\end{vmatrix}-0\begin{vmatrix}1&\lambda^2\\1&2\end{vmatrix}+4\begin{vmatrix}1&2\\1&\lambda^2\end{vmatrix}$
$\displaystyle =2(4-\lambda^4)+4(\lambda^2-2)$
$\displaystyle =8-2\lambda^4+4\lambda^2-8$
$\displaystyle =-2\lambda^4+4\lambda^2$
$\displaystyle =\lambda^2(\lambda^2-2)$
$\displaystyle \Rightarrow \lambda^2=0 \text{ or }\lambda^2-2=0$
$\displaystyle \Rightarrow \lambda=0 \text{ or }\lambda=\pm\sqrt{2}$
$\displaystyle \text{Thus, the values of }\lambda\text{ are }0,-\sqrt{2}\text{ and }\sqrt{2}.$

$\displaystyle \textbf{Question 17: }~\text{If the lines }x=5,\ \frac{y}{3-\alpha}=\frac{z}{2}\text{ and }x=\alpha,\\ \frac{y}{-1}=\frac{z}{2-\alpha}\text{ are coplanar, find the values of }\alpha.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The lines }\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}\text{ and }\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}\text{ are coplanar if}$
$\displaystyle \begin{vmatrix}x_2-x_1&y_2-y_1&z_2-z_1\\a_1&b_1&c_1\\a_2&b_2&c_2\end{vmatrix}=0$
$\displaystyle \text{The given lines } \frac{x-5}{0}=\frac{y}{3-\alpha}=\frac{z}{-2} \text{ and } \frac{x-\alpha}{0}=\frac{y}{-1}=\frac{z}{2-\alpha} \text{ are coplanar.}$
$\displaystyle \therefore \begin{vmatrix}\alpha-5&0-0&0-0\\0&3-\alpha&-2\\0&-1&2-\alpha\end{vmatrix}=0$
$\displaystyle =\begin{vmatrix}\alpha-5&0&0\\0&3-\alpha&-2\\0&-1&2-\alpha\end{vmatrix}$
$\displaystyle =(\alpha-5)\begin{vmatrix}3-\alpha&-2\\-1&2-\alpha\end{vmatrix}-0+0$
$\displaystyle =(\alpha-5)\left[(3-\alpha)(2-\alpha)-2\right]$
$\displaystyle =(\alpha-5)(\alpha^2-5\alpha+4)$
$\displaystyle =(\alpha-5)(\alpha-1)(\alpha-4)$
$\displaystyle \Rightarrow \alpha-1=0 \text{ or }\alpha-4=0 \text{ or }\alpha-5=0$
$\displaystyle \Rightarrow \alpha=1 \text{ or }\alpha=4 \text{ or }\alpha=5$
$\displaystyle \text{Thus, the values of }\alpha\text{ are }1,4\text{ and }5.$

$\displaystyle \textbf{Question 18: }~\text{If the straight lines }\frac{x-1}{2}=\frac{y+1}{k}=\frac{z}{2}\text{ and } \\ \frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{k}\text{ are coplanar, find the equations of the planes containing them.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given lines }L_1:\ \frac{x-1}{2}=\frac{y+1}{k}=\frac{z}{2}\ \text{ and }\ L_2:\ \frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{k}$
$\displaystyle L_1\Rightarrow \text{point }A(1,-1,0),\ \text{direction ratios }\overrightarrow d_1=(2,\ k,\ 2)$
$\displaystyle L_2\Rightarrow \text{point }B(-1,-1,0),\ \text{direction ratios }\overrightarrow d_2=(5,\ 2,\ k)$
$\displaystyle \overrightarrow{AB}=B-A=(-1-1,\ -1+1,\ 0-0)=(-2,0,0)$
$\displaystyle \text{For coplanarity: }\overrightarrow{AB}\cdot(\overrightarrow d_1\times \overrightarrow d_2)=0$
$\displaystyle \overrightarrow d_1\times \overrightarrow d_2=\begin{vmatrix}\widehat i&\widehat j&\widehat k\\2&k&2\\5&2&k\end{vmatrix}$
$\displaystyle =(k^2-4)\widehat i+(10-2k)\widehat j+(4-5k)\widehat k$
$\displaystyle \Rightarrow (-2,0,0)\cdot(k^2-4,\ 10-2k,\ 4-5k)=0$
$\displaystyle \Rightarrow -2(k^2-4)=0$
$\displaystyle \Rightarrow k^2=4$
$\displaystyle \Rightarrow k=2\ \text{or}\ k=-2$
$\displaystyle \textbf{Case 1: }k=2$
$\displaystyle \overrightarrow d_1=(2,2,2),\ \overrightarrow{AB}=(-2,0,0)$
$\displaystyle \text{Normal to the plane } \overrightarrow n=\overrightarrow d_1\times\overrightarrow{AB}=(2,2,2)\times(-2,0,0)=(0,-4,4)$
$\displaystyle \Rightarrow \overrightarrow n\parallel(0,-1,1)$
$\displaystyle \text{Plane through }A(1,-1,0):\ 0(x-1)-1(y+1)+1(z-0)=0$
$\displaystyle \Rightarrow z-y-1=0$
$\displaystyle \textbf{Case 2: }k=-2$
$\displaystyle \overrightarrow d_1=(2,-2,2),\ \overrightarrow{AB}=(-2,0,0)$
$\displaystyle \text{Normal to the plane } \overrightarrow n=\overrightarrow d_1\times\overrightarrow{AB}=(2,-2,2)\times(-2,0,0)=(0,-4,-4)$
$\displaystyle \Rightarrow \overrightarrow n\parallel(0,1,1)$
$\displaystyle \text{Plane through }A(1,-1,0):\ 0(x-1)+1(y+1)+1(z-0)=0$
$\displaystyle \Rightarrow y+z+1=0$
$\displaystyle \therefore\ \text{If the lines are coplanar, then }k=2\text{ gives plane }z-y-1=0,\ \text{and }k=-2\text{ gives plane }y+z+1=0$