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. \( \frac{7}{\sqrt{40}} \)
• B. \( \frac{6}{7} \)
• C. \( \frac{8}{\sqrt{50}} \)
• D. \( \frac{90}{19} \)

Hint:

Equation of a plane in intercept form: \(x/a + y/b + z/c = 1\).
Centroid of a triangle with vertices \((x_1,y_1,z_1), (x_2,y_2,z_2), (x_3,y_3,z_3)\) is \((\frac{\sum x_i}{3}, \frac{\sum y_i}{3}, \frac{\sum z_i}{3})\).
Perpendicular distance from origin \((0,0,0)\) to plane \(Ax+By+Cz+D=0\) is \(|D|/\sqrt{A^2+B^2+C^2}\).

Answer 1

The Correct Option is D

Let the plane intersect the X, Y, Z axes at points A, B, C respectively. Let A = \((a,0,0)\), B = \((0,b,0)\), C = \((0,0,c)\). These are the x, y, z intercepts. The equation of the plane in intercept form is \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \). The centroid of \(\triangle ABC\) has coordinates: \(G = \left( \frac{a+0+0}{3}, \frac{0+b+0}{3}, \frac{0+0+c}{3} \right) = \left( \frac{a}{3}, \frac{b}{3}, \frac{c}{3} \right)\). Given that the centroid is \((2, -3, 5)\). So, \(\frac{a}{3} = 2 \Rightarrow a = 6\). \(\frac{b}{3} = -3 \Rightarrow b = -9\). \(\frac{c}{3} = 5 \Rightarrow c = 15\). The equation of the plane is \( \frac{x}{6} + \frac{y}{-9} + \frac{z}{15} = 1 \). To find the perpendicular distance from the origin (0,0,0) to this plane, we can convert the equation to the form \(Ax+By+Cz+D=0\). Multiply by the LCM of 6, 9, 15. LCM(6,9,15): \(6=2 \cdot 3\), \(9=3^2\), \(15=3 \cdot 5\). LCM = \(2 \cdot 3^2 \cdot 5 = 2 \cdot 9 \cdot 5 = 90\). Multiply by 90: \( \frac{90x}{6} + \frac{90y}{-9} + \frac{90z}{15} = 90 \) \( 15x - 10y + 6z = 90 \) \( 15x - 10y + 6z - 90 = 0 \). The perpendicular distance \(d\) from a point \((x_0, y_0, z_0)\) to the plane \(Ax+By+Cz+D=0\) is given by: \[ d = \frac{|Ax_0+By_0+Cz_0+D|}{\sqrt{A^2+B^2+C^2}} \] Here, the point is origin \((0,0,0)\), and the plane is \(15x - 10y + 6z - 90 = 0\). So, \(A=15, B=-10, C=6, D=-90\). \[ d = \frac{|15(0) - 10(0) + 6(0) - 90|}{\sqrt{15^2+(-10)^2+6^2}} = \frac{|-90|}{\sqrt{225+100+36}} \] \[ d = \frac{90}{\sqrt{361}} \] We need to find \(\sqrt{361}\). \(10^2=100, 20^2=400\). So it's between 10 and 20. Numbers ending in 1 when squared end in 1 (e.g., 1, 9). Try \(19^2 = (20-1)^2 = 400 - 40 + 1 = 361\). So \(\sqrt{361}=19\). \[ d = \frac{90}{19} \] This matches option (d). \[ \boxed{\frac{90}{19}} \]