\displaystyle \textbf{Question 1: }~\text{Find the coordinates of the point where the line through }(5,1,6) \\ \text{ and }(3,4,1)\text{ crosses the}
\displaystyle \text{(i) }\ yz\text{-plane}\qquad \text{(ii) }\ zx\text{-plane.}
\displaystyle \text{Also, find the angle which this line makes with these planes. \ [ CBSE\ 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Line passes through }(5,1,6)\text{ and }(3,4,1)
\displaystyle \text{Direction ratios }=(3-5,\ 4-1,\ 1-6)=(-2,\ 3,\ -5)
\displaystyle \text{Equation of the line: }\frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}=\lambda
\displaystyle \Rightarrow x=-2\lambda+5,\ y=3\lambda+1,\ z=-5\lambda+6

\displaystyle \text{(i) Point where the line crosses the yz-plane}
\displaystyle \text{On yz-plane, }x=0
\displaystyle \Rightarrow -2\lambda+5=0
\displaystyle \Rightarrow \lambda=\frac{5}{2}
\displaystyle \Rightarrow y=3\left(\frac{5}{2}\right)+1=\frac{15}{2}+\frac{2}{2}=\frac{17}{2}
\displaystyle \Rightarrow z=-5\left(\frac{5}{2}\right)+6=-\frac{25}{2}+\frac{12}{2}=-\frac{13}{2}
\displaystyle \therefore\ \text{Point on yz-plane }=\left(0,\frac{17}{2},-\frac{13}{2}\right)
\displaystyle \text{Angle made by the line with the yz-plane}
\displaystyle \text{Normal to yz-plane (}x=0\text{) is }(1,0,0)
\displaystyle \text{If }\theta_1\text{ is the angle between the line and the yz-plane, then}
\displaystyle \sin\theta_1=\frac{|( -2,\ 3,\ -5)\cdot(1,0,0)|}{\sqrt{(-2)^2+3^2+(-5)^2}\ \sqrt{1^2+0^2+0^2}}
\displaystyle =\frac{|-2|}{\sqrt{4+9+25}}=\frac{2}{\sqrt{38}}
\displaystyle \therefore\ \theta_1=\sin^{-1}\left(\frac{2}{\sqrt{38}}\right)

\displaystyle \text{(ii) Point where the line crosses the zx-plane}
\displaystyle \text{On zx-plane, }y=0
\displaystyle \Rightarrow 3\lambda+1=0
\displaystyle \Rightarrow \lambda=-\frac{1}{3}
\displaystyle \Rightarrow x=-2\left(-\frac{1}{3}\right)+5=\frac{2}{3}+\frac{15}{3}=\frac{17}{3}
\displaystyle \Rightarrow z=-5\left(-\frac{1}{3}\right)+6=\frac{5}{3}+\frac{18}{3}=\frac{23}{3}
\displaystyle \therefore\ \text{Point on zx-plane }=\left(\frac{17}{3},0,\frac{23}{3}\right)
\displaystyle \text{Angle made by the line with the zx-plane}
\displaystyle \text{Normal to zx-plane (}y=0\text{) is }(0,1,0)
\displaystyle \text{If }\theta_2\text{ is the angle between the line and the zx-plane, then}
\displaystyle \sin\theta_2=\frac{|( -2,\ 3,\ -5)\cdot(0,1,0)|}{\sqrt{(-2)^2+3^2+(-5)^2}\ \sqrt{0^2+1^2+0^2}}
\displaystyle =\frac{|3|}{\sqrt{38}}=\frac{3}{\sqrt{38}}
\displaystyle \therefore\ \theta_2=\sin^{-1}\left(\frac{3}{\sqrt{38}}\right)

\displaystyle \textbf{Question 2: }~\text{Find the coordinates of the point where the line through }(3,-4,-5) \\ \text{ and }(2,-3,1)\text{ crosses the plane }2x+y+z=7.\ \text{[\ CBSE\ 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the line through the points }(3,-4,-5)\text{ and }(2,-3,1)\text{ is}
\displaystyle \frac{x-3}{2-3}=\frac{y+4}{-3+4}=\frac{z+5}{1+5}
\displaystyle \Rightarrow \frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}
\displaystyle \text{The coordinates of any point on this line are of the form}
\displaystyle \frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}=\lambda
\displaystyle \Rightarrow x=-\lambda+3,\ y=\lambda-4,\ z=6\lambda-5
\displaystyle \text{So, the coordinates of the point on the given line are }(-\lambda+3,\ \lambda-4,\ 6\lambda-5)
\displaystyle \text{Since this point lies on the plane }2x+y+z=7,
\displaystyle 2(-\lambda+3)+\lambda-4+6\lambda-5=7
\displaystyle \Rightarrow -2\lambda+6+\lambda-4+6\lambda-5=7
\displaystyle \Rightarrow 5\lambda-3=7
\displaystyle \Rightarrow 5\lambda=10
\displaystyle \Rightarrow \lambda=2
\displaystyle \text{So, the coordinates of the point are}
\displaystyle (-2+3,\ 2-4,\ 6(2)-5)
\displaystyle =(1,\ -2,\ 7)

\displaystyle \textbf{Question 3: }~\text{Find the distance of the point }(-1,-5,-10)\text{ from the point of} \\ \text{intersection of the line }\overrightarrow{r}=(2\widehat{i}-\widehat{j}+2\widehat{k})+\lambda(3\widehat{i}+4\widehat{j}+2\widehat{k})\text{ and the plane }\overrightarrow{r}\cdot(\widehat{i}-\widehat{j}+\widehat{k})=5.\ \text{[\ CBSE\ 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{The given equation of the line is}
\displaystyle \overrightarrow r=(2\widehat i-\widehat j+2\widehat k)+\lambda(3\widehat i+4\widehat j+2\widehat k)
\displaystyle \Rightarrow \overrightarrow r=(2+3\lambda)\widehat i+(-1+4\lambda)\widehat j+(2+2\lambda)\widehat k
\displaystyle \text{The coordinates of any point on this line are }(2+3\lambda,\,-1+4\lambda,\,2+2\lambda)
\displaystyle \text{Since this point lies on the plane }\overrightarrow r\cdot(\widehat i-\widehat j+\widehat k)=5,
\displaystyle [(2+3\lambda)\widehat i+(-1+4\lambda)\widehat j+(2+2\lambda)\widehat k]\cdot(\widehat i-\widehat j+\widehat k)=5
\displaystyle \Rightarrow 2+3\lambda+1-4\lambda+2+2\lambda-5=0
\displaystyle \Rightarrow \lambda=0
\displaystyle \text{So, the coordinates of the point are}
\displaystyle (2+3\lambda,\,-1+4\lambda,\,2+2\lambda)
\displaystyle =(2,\,-1,\,2)
\displaystyle \text{Distance between }(2,-1,2)\text{ and }(-1,-5,-10)
\displaystyle =\sqrt{(-1-2)^2+(-5+1)^2+(-10-2)^2}
\displaystyle =\sqrt{9+16+144}
\displaystyle =13\text{ units}

\displaystyle \textbf{Question 4: }~\text{Find the distance of the point }(2,12,5)\text{ from the point of} \\ \text{intersection of the line }\overrightarrow{r}=2\widehat{i}-4\widehat{j}+2\widehat{k}+\lambda(3\widehat{i}+4\widehat{j}+2\widehat{k})\text{ and the plane }\overrightarrow{r}\cdot(\widehat{i}-2\widehat{j}+\widehat{k})=0.\ \text{[CBSE\ 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the given line is}
\displaystyle \overrightarrow r=(2\widehat i-4\widehat j+2\widehat k)+\lambda(3\widehat i+4\widehat j+2\widehat k)
\displaystyle \text{The position vector of any point on the given line is}
\displaystyle \overrightarrow r=(2+3\lambda)\widehat i+(-4+4\lambda)\widehat j+(2+2\lambda)\widehat k\ \ \ \ \ \ (1)
\displaystyle \text{If this lies on the plane }\overrightarrow r\cdot(\widehat i-2\widehat j+\widehat k)=0\text{ then}
\displaystyle [(2+3\lambda)\widehat i+(-4+4\lambda)\widehat j+(2+2\lambda)\widehat k]\cdot(\widehat i-2\widehat j+\widehat k)=0
\displaystyle \Rightarrow (2+3\lambda)-2(-4+4\lambda)+(2+2\lambda)=0
\displaystyle \Rightarrow 2+3\lambda+8-8\lambda+2+2\lambda=0
\displaystyle \Rightarrow 12-3\lambda=0
\displaystyle \Rightarrow 3\lambda=12
\displaystyle \Rightarrow \lambda=4
\displaystyle \text{Putting }\lambda=4\text{ in (1), we get}
\displaystyle \overrightarrow r=(2+3\times4)\widehat i+(-4+4\times4)\widehat j+(2+2\times4)\widehat k
\displaystyle =14\widehat i+12\widehat j+10\widehat k
\displaystyle \text{So, the position vector of the point of intersection is }14\widehat i+12\widehat j+10\widehat k
\displaystyle \text{The position vector of the given point is }2\widehat i+12\widehat j+5\widehat k
\displaystyle \text{Required distance }=\left|(14\widehat i+12\widehat j+10\widehat k)-(2\widehat i+12\widehat j+5\widehat k)\right|
\displaystyle =|12\widehat i+5\widehat k|
\displaystyle =\sqrt{12^2+0^2+5^2}
\displaystyle =\sqrt{169}
\displaystyle =13\ \text{units}

\displaystyle \textbf{Question 5: }~\text{Find the distance of the point }P(-1,-5,-10)\text{ from the point of} \\ \text{intersection of the line joining the points }A(2,-1,2)\text{ and }B(5,3,4)\text{ with the plane }x-y+z=5.\ \text{[CBSE\ 2014,\ 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the line passing through the points }A(2,-1,2)\text{ and }B(5,3,4)\text{ is given by}
\displaystyle \frac{x-2}{5-2}=\frac{y-(-1)}{3-(-1)}=\frac{z-2}{4-2}
\displaystyle \text{Or }\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}
\displaystyle \text{The coordinates of any point on the line}
\displaystyle \frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}=\lambda\ (\text{say})\text{ are}
\displaystyle (3\lambda+2,\ 4\lambda-1,\ 2\lambda+2)\ \ \ \ \ \ (1)
\displaystyle \text{If it lies on the plane }x-y+z=5,\text{ then}
\displaystyle 3\lambda+2-(4\lambda-1)+2\lambda+2=5
\displaystyle \Rightarrow \lambda+5=5
\displaystyle \Rightarrow \lambda=0
\displaystyle \text{Putting }\lambda=0\text{ in (1), we get }(2,-1,2)\text{ as the coordinates of the point of intersection of the given line and plane.}
\displaystyle \therefore\ \text{Required distance }=\text{Distance between points }(-1,-5,-10)\text{ and }(2,-1,2)
\displaystyle =\sqrt{(2+1)^2+(-1+5)^2+(2+10)^2}
\displaystyle =\sqrt{9+16+144}
\displaystyle =\sqrt{169}
\displaystyle =13\ \text{units}

\displaystyle \textbf{Question 6: }~\text{Find the distance of the point }P(3,4,4)\text{ from the point,} \\ \text{where the line joining the points }A(3,-4,-5)\text{ and }B(2,-3,1)\text{ intersects the} \\ \text{plane }2x+y+z=7.\ \text{[CBSE\ 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the line passing through the points }A(3,-4,-5)\text{ and }B(2,-3,1)\text{ is given by}
\displaystyle \frac{x-3}{2-3}=\frac{y-(-4)}{-3-(-4)}=\frac{z-(-5)}{1-(-5)}
\displaystyle \text{Or }\frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}
\displaystyle \text{The coordinates of any point on the line}
\displaystyle \frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}=\lambda\ (\text{say})\text{ are }(-\lambda+3,\ \lambda-4,\ 6\lambda-5)\ \ \ \ \ \ (1)
\displaystyle \text{If it lies on the plane }2x+y+z=7,\text{ then}
\displaystyle 2(-\lambda+3)+(\lambda-4)+(6\lambda-5)=7
\displaystyle \Rightarrow 5\lambda-3=7
\displaystyle \Rightarrow 5\lambda=10
\displaystyle \Rightarrow \lambda=2
\displaystyle \text{Putting }\lambda=2\text{ in (1), we get }(1,-2,7)\text{ as the coordinates of the point of intersection of the given line and plane.}
\displaystyle \therefore\ \text{Required distance }=\text{Distance between points }(3,4,4)\text{ and }(1,-2,7)
\displaystyle =\sqrt{(3-1)^2+(4+2)^2+(4-7)^2}
\displaystyle =\sqrt{4+36+9}
\displaystyle =\sqrt{49}
\displaystyle =7\ \text{units}

\displaystyle \textbf{Question 7: }~\text{Find the distance of the point }(1,-5,9)\text{ from the plane }x-y+z=5\text{ measured along the line }x=y=z.\ \text{[CBSE\ 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the line parallel to }x=y=z\text{ and passing through the point }(1,-5,9)\text{ is}
\displaystyle \frac{x-1}{1}=\frac{y+5}{1}=\frac{z-9}{1}\ \ \ \ \ \ (1)
\displaystyle \text{Any point on this line is of the form }(k+1,\ k-5,\ k+9)
\displaystyle \text{If }(k+1,\ k-5,\ k+9)\text{ is the point of intersection of line (1) and the given plane, then}
\displaystyle (k+1)-(k-5)+(k+9)=5
\displaystyle \Rightarrow k+1-k+5+k+9=5
\displaystyle \Rightarrow k+15=5
\displaystyle \Rightarrow k=-10
\displaystyle \text{So, the point of intersection of line (1) and the given plane is }(-10+1,\ -10-5,\ -10+9)
\displaystyle =(-9,\ -15,\ -1)
\displaystyle \therefore\ \text{Required distance }=\text{Distance between points }(1,-5,9)\text{ and }(-9,-15,-1)
\displaystyle =\sqrt{(1+9)^2+(-5+15)^2+(9+1)^2}
\displaystyle =\sqrt{100+100+100}
\displaystyle =10\sqrt{3}\ \text{units}

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