Question

The foot of the perpendicular drawn from the point \((-2, -1, 3)\) to a plane is \((1, 0, -2)\). If \(a, b, c\) are the intercepts made by the plane \(\pi\) on \(X, Y, Z\)-axes respectively, then \(3a+b+5c =\)

• A. 39
• B. 26
• C. 13
• D. 0 \vspace{0.5cm}

Hint:

- When solving for intercepts of a plane, use the method of setting variables to zero and solving for the remaining variable. The intercepts provide important information for writing the equation of the plane.

Answer 1

The Correct Option is C

Step 1: Equation of the plane with intercepts.
- The general equation of a plane with intercepts \(a\), \(b\), and \(c\) on the \(X\), \(Y\), and \(Z\)-axes, respectively, is: \[ \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \] \vspace{0.5cm} Step 2: Substitute the point \((1, 0, -2)\) into the plane equation.
- Since the point \((1, 0, -2)\) lies on the plane, substitute \(x = 1\), \(y = 0\), and \(z = -2\) into the equation: \[ \frac{1}{a} + \frac{0}{b} + \frac{-2}{c} = 1 \] \[ \frac{1}{a} - \frac{2}{c} = 1 \] \vspace{0.5cm} Step 3: Find the normal vector of the plane.
- The foot of the perpendicular from the point \((-2, -1, 3)\) to the plane is \((1, 0, -2)\), so the vector from \((-2, -1, 3)\) to \((1, 0, -2)\) is: \[ \vec{v} = (1 - (-2), 0 - (-1), -2 - 3) = (3, 1, -5) \] - This vector \(\vec{v} = (3, 1, -5)\) is parallel to the normal vector of the plane. Thus, the normal vector of the plane is \(\vec{n} = (3, 1, -5)\). \vspace{0.5cm} Step 4: Use the point-normal form of the plane equation.
- The point-normal form of the equation of the plane is: \[ 3(x - 1) + 1(y - 0) - 5(z + 2) = 0 \] Simplifying this: \[ 3(x - 1) + y - 5(z + 2) = 0 \] \[ 3x - 3 + y - 5z - 10 = 0 \] \[ 3x + y - 5z - 13 = 0 \] So, the equation of the plane is: \[ 3x + y - 5z = 13 \] \vspace{0.5cm} Step 5: Identify the intercepts of the plane.
- The intercepts of the plane are the points where the plane intersects the axes: - Set \(y = 0\) and \(z = 0\) to find the \(x\)-intercept: \[ 3x = 13 \quad \Rightarrow x = \frac{13}{3} \] - Set \(x = 0\) and \(z = 0\) to find the \(y\)-intercept: \[ y = 13 \] - Set \(x = 0\) and \(y = 0\) to find the \(z\)-intercept: \[ -5z = 13 \quad \Rightarrow z = -\frac{13}{5} \] Thus, the intercepts are: \[ a = \frac{13}{3}, \quad b = 13, \quad c = -\frac{13}{5} \] \vspace{0.5cm} Step 6: Calculate \(3a + b + 5c\).
- Finally, calculate: \[ 3a + b + 5c = 3 \times \frac{13}{3} + 13 + 5 \times \left(-\frac{13}{5}\right) \] \[ = 13 + 13 - 13 = 13 \] Thus, \(3a + b + 5c = 13\). \vspace{0.5cm}