Question:
A plane meets the \(X, Y, Z\)-axes in \(A, B, C\) respectively. If the centroid of the triangle \(ABC\) is \((2,-3,5)\), then the perpendicular distance from origin to the given plane is
A plane meets the \(X, Y, Z\)-axes in \(A, B, C\) respectively. If the centroid of the triangle \(ABC\) is \((2,-3,5)\), then the perpendicular distance from origin to the given plane is
Show Hint
If a plane cuts the axes at \((a,0,0)\), \((0,b,0)\), and \((0,0,c)\), then its equation is
\[
\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1.
\]
The centroid of these intercept points is
\[
\left(\frac{a}{3},\frac{b}{3},\frac{c}{3}\right).
\]
Updated On: Jun 26, 2026
- \(\frac{7}{\sqrt{40}}\)
- \(\frac{6}{7}\)
- \(\frac{8}{\sqrt{50}}\)
- \(\frac{90}{19}\)
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The Correct Option is D
Solution and Explanation
Step 1: Write the intercept form of the plane.
Let the plane meet the \(X\)-axis, \(Y\)-axis, and \(Z\)-axis at \[ A=(a,0,0),\quad B=(0,b,0),\quad C=(0,0,c) \] Then the equation of the plane is \[ \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1 \]
Step 2: Use the centroid formula.
The centroid of triangle \(ABC\) is \[ \left(\frac{a+0+0}{3},\frac{0+b+0}{3},\frac{0+0+c}{3}\right) \] So, \[ \left(\frac{a}{3},\frac{b}{3},\frac{c}{3}\right) \] Given centroid is \[ (2,-3,5) \] Therefore, \[ \frac{a}{3}=2,\quad \frac{b}{3}=-3,\quad \frac{c}{3}=5 \] Hence, \[ a=6,\quad b=-9,\quad c=15 \]
Step 3: Write the equation of the plane.
Substituting \(a=6\), \(b=-9\), and \(c=15\) in \[ \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1, \] we get \[ \frac{x}{6}+\frac{y}{-9}+\frac{z}{15}=1 \] That is, \[ \frac{x}{6}-\frac{y}{9}+\frac{z}{15}=1 \] Taking LCM \(90\), \[ 15x-10y+6z=90 \] So, the plane is \[ 15x-10y+6z-90=0 \]
Step 4: Find the perpendicular distance from origin.
The perpendicular distance of a point \((x_1,y_1,z_1)\) from the plane \[ Ax+By+Cz+D=0 \] is \[ d=\frac{|Ax_1+By_1+Cz_1+D|}{\sqrt{A^2+B^2+C^2}} \] Here, the point is origin: \[ (0,0,0) \] and the plane is \[ 15x-10y+6z-90=0 \] So, \[ A=15,\quad B=-10,\quad C=6,\quad D=-90 \] Thus, \[ d=\frac{|15(0)-10(0)+6(0)-90|}{\sqrt{15^2+(-10)^2+6^2}} \] \[ d=\frac{90}{\sqrt{225+100+36}} \] \[ d=\frac{90}{\sqrt{361}} \] \[ d=\frac{90}{19} \]
Step 5: Final conclusion.
Hence, the perpendicular distance from origin to the plane is \[ \boxed{\frac{90}{19}} \]
Let the plane meet the \(X\)-axis, \(Y\)-axis, and \(Z\)-axis at \[ A=(a,0,0),\quad B=(0,b,0),\quad C=(0,0,c) \] Then the equation of the plane is \[ \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1 \]
Step 2: Use the centroid formula.
The centroid of triangle \(ABC\) is \[ \left(\frac{a+0+0}{3},\frac{0+b+0}{3},\frac{0+0+c}{3}\right) \] So, \[ \left(\frac{a}{3},\frac{b}{3},\frac{c}{3}\right) \] Given centroid is \[ (2,-3,5) \] Therefore, \[ \frac{a}{3}=2,\quad \frac{b}{3}=-3,\quad \frac{c}{3}=5 \] Hence, \[ a=6,\quad b=-9,\quad c=15 \]
Step 3: Write the equation of the plane.
Substituting \(a=6\), \(b=-9\), and \(c=15\) in \[ \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1, \] we get \[ \frac{x}{6}+\frac{y}{-9}+\frac{z}{15}=1 \] That is, \[ \frac{x}{6}-\frac{y}{9}+\frac{z}{15}=1 \] Taking LCM \(90\), \[ 15x-10y+6z=90 \] So, the plane is \[ 15x-10y+6z-90=0 \]
Step 4: Find the perpendicular distance from origin.
The perpendicular distance of a point \((x_1,y_1,z_1)\) from the plane \[ Ax+By+Cz+D=0 \] is \[ d=\frac{|Ax_1+By_1+Cz_1+D|}{\sqrt{A^2+B^2+C^2}} \] Here, the point is origin: \[ (0,0,0) \] and the plane is \[ 15x-10y+6z-90=0 \] So, \[ A=15,\quad B=-10,\quad C=6,\quad D=-90 \] Thus, \[ d=\frac{|15(0)-10(0)+6(0)-90|}{\sqrt{15^2+(-10)^2+6^2}} \] \[ d=\frac{90}{\sqrt{225+100+36}} \] \[ d=\frac{90}{\sqrt{361}} \] \[ d=\frac{90}{19} \]
Step 5: Final conclusion.
Hence, the perpendicular distance from origin to the plane is \[ \boxed{\frac{90}{19}} \]
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