Question

If the line passing through the points \( (5,1,a) \) and \( (3,b,1) \) crosses the YZ plane at the point \( \left( 0, \frac{17}{2}, \frac{-13}{2} \right) \), then \( a + b = \):

• A. \( 12 \)
• B. \( 10 \)
• C. \( 8 \)
• D. \( 4 \)

Hint:

To find the intersection with the YZ plane, set \( x = 0 \) in the line equation and solve for the corresponding values of \( y \) and \( z \).

Answer 1

The Correct Option is B

We are given two points \( (5, 1, a) \) and \( (3, b, 1) \) and the YZ plane equation \( x = 0 \), which is the condition for the line to intersect the YZ plane.
Step 1: Parametrize the line equation. The general form of the equation of a line passing through two points \( P_1(x_1, y_1, z_1) \) and \( P_2(x_2, y_2, z_2) \) is given by: \[ \left( x, y, z \right) = \left( x_1, y_1, z_1 \right) + t \left( x_2 - x_1, y_2 - y_1, z_2 - z_1 \right) \] Substitute the values of the given points \( (5, 1, a) \) and \( (3, b, 1) \): \[ x = 5 + t(3 - 5), \quad y = 1 + t(b - 1), \quad z = a + t(1 - a) \] Step 2: Apply the YZ plane condition. At the point where the line intersects the YZ plane, \( x = 0 \). So, substitute \( x = 0 \): \[ 0 = 5 - 2t \quad \Rightarrow \quad t = \frac{5}{2} \] Step 3: Find \( a \) and \( b \). Now, substitute \( t = \frac{5}{2} \) into the equations for \( y \) and \( z \): \[ y = 1 + \frac{5}{2}(b - 1) = \frac{17}{2}, \quad z = a + \frac{5}{2}(1 - a) = \frac{-13}{2} \] From the first equation: \[ 1 + \frac{5}{2}(b - 1) = \frac{17}{2} \quad \Rightarrow \quad \frac{5}{2}(b - 1) = \frac{15}{2} \quad \Rightarrow \quad b - 1 = 3 \quad \Rightarrow \quad b = 4 \] From the second equation: \[ a + \frac{5}{2}(1 - a) = \frac{-13}{2} \quad \Rightarrow \quad a + \frac{5}{2} - \frac{5a}{2} = \frac{-13}{2} \] Multiply through by 2 to eliminate the fractions: \[ 2a + 5 - 5a = -13 \quad \Rightarrow \quad -3a = -18 \quad \Rightarrow \quad a = 6 \] Step 4: Find \( a + b \). Now that we have \( a = 6 \) and \( b = 4 \), we can calculate \( a + b \): \[ a + b = 6 + 4 = 10 \]