Question

Let \( A(1, 6, 3) \) and point \( B \) and \( C \) lie on the line \[ \frac{x - 1}{1} = \frac{y - 1}{2} = \frac{z - 2}{3} \] where \( B(4, 9, \alpha) \) and point \( C \) is 10 units from \( B \). Find the area of \( \triangle ABC \):

• A. \( 6\sqrt{13} \)
• B. \( 5\sqrt{13} \)
• C. \( 7\sqrt{13} \)
• D. \( 8\sqrt{13} \)

Hint:

To calculate the area of a triangle formed by three points, use the cross product of two vectors from the origin and then calculate the magnitude of the resulting vector.

Answer 1

The Correct Option is C

Step 1: Parametrize the Line.
The equation of the line passing through points \( A \) and \( B \) is given as: \[ \frac{x - 1}{1} = \frac{y - 1}{2} = \frac{z - 2}{3} = t \] So, the parametric coordinates for any point on the line are: \[ x = 1 + t, \quad y = 1 + 2t, \quad z = 2 + 3t \] Thus, point \( B(4, 9, \alpha) \) satisfies: \[ 4 = 1 + t \quad \text{and} \quad 9 = 1 + 2t \] Solving for \( t \), we get \( t = 3 \), and substituting this into \( z \), we get: \[ \alpha = 2 + 3(3) = 11 \] Thus, point \( B(4, 9, 11) \).
Step 2: Find the Coordinates of Point C.
Since point \( C \) lies on the line, its parametric coordinates are: \[ C(x, y, z) = (1 + t, 1 + 2t, 2 + 3t) \] We are given that the distance between \( B(4, 9, 11) \) and \( C \) is 10 units, so we use the distance formula: \[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \] Substituting the coordinates for \( B \) and \( C \), we find that the value of \( t \) for which the distance is 10 is \( t = 4 \). Thus, the coordinates of point \( C \) are: \[ C(5, 9, 14) \]
Step 3: Find the Area of Triangle ABC.
The area of triangle \( ABC \) can be found using the formula for the area of a triangle given by its vertices: \[ \text{Area} = \frac{1}{2} \left| \overrightarrow{AB} \times \overrightarrow{AC} \right| \] We first find the vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \): \[ \overrightarrow{AB} = (4 - 1, 9 - 6, 11 - 3) = (3, 3, 8) \] \[ \overrightarrow{AC} = (5 - 1, 9 - 6, 14 - 3) = (4, 3, 11) \] Now, compute the cross product \( \overrightarrow{AB} \times \overrightarrow{AC} \): \[ \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
3 & 3 & 8
4 & 3 & 11 \end{vmatrix} \] Expanding this determinant, we get: \[ \overrightarrow{AB} \times \overrightarrow{AC} = \hat{i}(33 - 24) - \hat{j}(33 - 32) + \hat{k}(9 - 12) \] \[ = 9\hat{i} - \hat{j} - 3\hat{k} \] Now, find the magnitude of this vector: \[ \left| \overrightarrow{AB} \times \overrightarrow{AC} \right| = \sqrt{9^2 + (-1)^2 + (-3)^2} = \sqrt{81 + 1 + 9} = \sqrt{91} \] Thus, the area is: \[ \text{Area} = \frac{1}{2} \sqrt{91} = 7\sqrt{13} \] Final Answer: \[ \boxed{7\sqrt{13}} \]