Question

A line passing through the points \((9,7,5)\) and \((2,10,0)\) is perpendicular to a plane \(\pi\) passing through the point \((200,30,116)\). If the plane \(\pi\) cuts the \(X\)-, \(Y\)-, \(Z\)-axes at the points \(A\), \(B\), \(C\) respectively, then the centroid of \(\triangle ABC\) is

• A. \((70,-220,127)\)
• B. \((80,-200,125)\)
• C. \((90,-210,126)\)
• D. \((75,-205,128)\)

Hint:

Whenever a line is perpendicular to a plane, immediately use the direction ratios of the line as the normal vector of the plane. Once the plane equation is obtained, axis intercepts can be found by setting the remaining variables equal to zero.

Answer 1

The Correct Option is C

Concept: If a line is perpendicular to a plane, then the direction vector of the line acts as a normal vector to that plane. Therefore, the equation of the plane can be obtained using the point-normal form. If a plane cuts the coordinate axes at \[ A(a,0,0), \quad B(0,b,0), \quad C(0,0,c), \] then the centroid of \(\triangle ABC\) is given by \[ \left(\frac{a}{3},\frac{b}{3},\frac{c}{3}\right). \] Hence, we first determine the plane equation, then its intercepts, and finally the centroid.

Step 1: Determine the normal vector to the plane. The given line passes through the points \[ (9,7,5) \quad \text{and} \quad (2,10,0). \] Therefore, its direction vector is \[ \vec{n} = (2-9,\;10-7,\;0-5) = (-7,3,-5). \] Since the line is perpendicular to the plane, this direction vector is a normal vector to the plane. Thus, \[ \vec{n}=(-7,3,-5). \]

Step 2: Form the equation of the plane. The plane passes through the point \[ (200,30,116). \] Using the point-normal form, \[ -7(x-200)+3(y-30)-5(z-116)=0. \] Expanding, \[ -7x+1400+3y-90-5z+580=0. \] Combining constants, \[ -7x+3y-5z+1890=0. \] Multiplying by \(-1\), \[ 7x-3y+5z=1890. \] Hence the required plane equation is \[ 7x-3y+5z=1890. \]

Step 3: Find the intercept on the \(X\)-axis. For the \(X\)-axis intercept, put \[ y=0, \quad z=0. \] Then \[ 7x=1890. \] Hence, \[ x=270. \] Therefore, \[ A=(270,0,0). \]

Step 4: Find the intercept on the \(Y\)-axis. For the \(Y\)-axis intercept, put \[ x=0, \quad z=0. \] Then \[ -3y=1890. \] Hence, \[ y=-630. \] Therefore, \[ B=(0,-630,0). \]

Step 5: Find the intercept on the \(Z\)-axis. For the \(Z\)-axis intercept, put \[ x=0, \quad y=0. \] Then \[ 5z=1890. \] Hence, \[ z=378. \] Therefore, \[ C=(0,0,378). \]

Step 6: Compute the centroid of \(\triangle ABC\). The centroid of a triangle is obtained by taking the average of the coordinates of its vertices. Therefore, \[ G = \left( \frac{270+0+0}{3}, \frac{0-630+0}{3}, \frac{0+0+378}{3} \right). \] Thus, \[ G=(90,-210,126). \] Hence, the centroid of \(\triangle ABC\) is \[ \boxed{(90,-210,126)}. \]