\displaystyle \textbf{Question 1: }~\text{Write the equation of the plane whose intercepts on the coordinate} \\ \text{axes are }2,\ -3\text{ and }4.
\displaystyle \text{Answer:}
\displaystyle \text{We know that the equation of the plane with }a,b\text{ and }c\text{ intercepts on the coordinate axes is given by}
\displaystyle \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1
\displaystyle \text{Given that}
\displaystyle a=2;b=-3;c=4
\displaystyle \text{So, the equation of the required plane is}
\displaystyle \frac{x}{2}+\frac{y}{-3}+\frac{z}{4}=1
\displaystyle \Rightarrow 6x-4y+3z=12

\displaystyle \textbf{Question 2: }~\text{Reduce the equations of the following planes in intercept form} \\ \text{and find its intercepts on the coordinate axes:}
\displaystyle \text{(i) }\ 4x+3y-6z-12=0
\displaystyle \text{(ii) }\ 2x+3y-z=6
\displaystyle \text{(iii) }\ 2x-y+z=5
\displaystyle \text{Answer:}
\displaystyle \text{(i)  }
\displaystyle \text{Equation of the given plane is}
\displaystyle 4x+3y-6z-12=0
\displaystyle \Rightarrow 4x+3y-6z=12
\displaystyle \text{Dividing both sides by }12,\text{ we get}
\displaystyle \frac{4x}{12}+\frac{3y}{12}+\frac{-6z}{12}=\frac{12}{12}
\displaystyle \Rightarrow \frac{4x}{12}+\frac{3y}{12}-\frac{6z}{12}=\frac{12}{12}
\displaystyle \Rightarrow \frac{x}{3}+\frac{y}{4}+\frac{z}{-2}=1\quad ...(1)
\displaystyle \text{We know that the equation of the plane whose intercepts on the coordinate axes are }a,b\text{ and }c\text{ is}
\displaystyle \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\quad ...(2)
\displaystyle \text{Comparing }(1)\text{ and }(2),\text{ we get}
\displaystyle a=3;b=4;c=-2

\displaystyle \text{(ii)  }
\displaystyle \text{The equation of the given plane is}
\displaystyle 2x+3y-z=6
\displaystyle \text{Dividing both sides by }6,\text{ we get}
\displaystyle \frac{2x}{6}+\frac{3y}{6}-\frac{z}{6}=\frac{6}{6}
\displaystyle \Rightarrow \frac{x}{3}+\frac{y}{2}+\frac{z}{-6}=1\quad ...(1)
\displaystyle \text{We know that the equation of the plane whose intercepts on the coordinate axes are }a,b\text{ and }c\text{ is}
\displaystyle \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\quad ...(2)
\displaystyle \text{Comparing }(1)\text{ and }(2),\text{ we get}
\displaystyle a=3;b=2;c=-6

\displaystyle \text{(iii)  }
\displaystyle \text{Equation of the given plane is}
\displaystyle 2x-y+z=5
\displaystyle \text{Dividing both sides by }5,\text{ we get}
\displaystyle \frac{2x}{5}-\frac{y}{5}+\frac{z}{5}=\frac{5}{5}
\displaystyle \Rightarrow \frac{x}{\frac{5}{2}}+\frac{y}{-5}+\frac{z}{5}=1\quad ...(1)
\displaystyle \text{We know that the equation of the plane whose intercepts on the coordinate axes are }a,b\text{ and }c\text{ is}
\displaystyle \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\quad ...(2)
\displaystyle \text{Comparing }(1)\text{ and }(2),\text{ we get}
\displaystyle a=\frac{5}{2};b=-5;c=5

\displaystyle \textbf{Question 3: }~\text{Find the equation of a plane which meets the axes in }A,B\text{ and } \\ C,\ \text{given that the centroid of the triangle }ABC\text{ is the point }(\alpha,\beta,\gamma).
\displaystyle \text{Answer:}
\displaystyle \text{Let }a,b\text{ and }c\text{ be the intercepts of the given plane on the coordinate axes.}
\displaystyle \text{Then the plane meets the coordinate axes at}
\displaystyle A(a,0,0),B(0,b,0)\text{ and }C(0,0,c)
\displaystyle \text{Given that the centroid of the triangle is }(\alpha,\beta,\gamma)
\displaystyle \Rightarrow \left(\frac{a+0+0}{3},\frac{0+b+0}{3},\frac{0+0+c}{3}\right)=(\alpha,\beta,\gamma)
\displaystyle \Rightarrow \left(\frac{a}{3},\frac{b}{3},\frac{c}{3}\right)=(\alpha,\beta,\gamma)
\displaystyle \Rightarrow \frac{a}{3}=\alpha,\frac{b}{3}=\beta,\frac{c}{3}=\gamma
\displaystyle \Rightarrow a=3\alpha,b=3\beta,c=3\gamma\quad ...(1)
\displaystyle \text{The equation of the plane whose intercepts on the coordinate axes are }a,b\text{ and }c\text{ are}
\displaystyle \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1
\displaystyle \Rightarrow \frac{x}{3\alpha}+\frac{y}{3\beta}+\frac{z}{3\gamma}=1\ \text{[From (1)]}
\displaystyle \Rightarrow \frac{x}{\alpha}+\frac{y}{\beta}+\frac{z}{\gamma}=3

\displaystyle \textbf{Question 4: }~\text{Find the equation of the plane passing through the point }(2,4,6) \\ \text{ and making equal intercepts on the coordinate axes.}
\displaystyle \text{Answer:}
\displaystyle \text{We know that the equation of the plane whose intercepts on the coordinate axes are }a,b\text{ and }c\text{ is}
\displaystyle \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1
\displaystyle \text{Given that the plane makes equal intercepts on the coordinate axes}
\displaystyle \text{So, }a=b=c
\displaystyle \text{So, the equation of the plane is}
\displaystyle \frac{x}{a}+\frac{y}{a}+\frac{z}{a}=1
\displaystyle \Rightarrow x+y+z=a\quad ...(1)
\displaystyle \text{This plane passes through the point }(2,4,6).
\displaystyle \text{Substituting this point in }(1),\text{ we get}
\displaystyle 2+4+6=a
\displaystyle \Rightarrow a=12
\displaystyle \text{Substituting this in }(1),\text{ we get}
\displaystyle x+y+z=12

\displaystyle \textbf{Question 5: }~\text{A plane meets the coordinate axes at }A,\ B\text{ and }C \text{ respectively such} \\ \text{that the centroid of triangle }ABC\text{ is }(1,-2,3).\text{ Find the equation of the plane.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }a,b\text{ and }c\text{ be the intercepts of the given plane on the coordinate axes.}
\displaystyle \text{Then the plane meets the coordinate axes at}
\displaystyle A(a,0,0),B(0,b,0)\text{ and }C(0,0,c)
\displaystyle \text{Given that the centroid of the triangle is }(1,-2,3)
\displaystyle \Rightarrow \left(\frac{a+0+0}{3},\frac{0+b+0}{3},\frac{0+0+c}{3}\right)=(1,-2,3)
\displaystyle \Rightarrow \left(\frac{a}{3},\frac{b}{3},\frac{c}{3}\right)=(1,-2,3)
\displaystyle \Rightarrow \frac{a}{3}=1,\frac{b}{3}=-2,\frac{c}{3}=3
\displaystyle \Rightarrow a=3,b=-6,c=9\quad ...(1)
\displaystyle \text{Equation of the plane whose intercepts on the coordinate axes are }a,b\text{ and }c\text{ is}
\displaystyle \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1
\displaystyle \Rightarrow \frac{x}{3}+\frac{y}{-6}+\frac{z}{9}=1\ \text{[From (1)]}
\displaystyle \Rightarrow 6x-3y+2z=18

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.