Question

Let in a \( \triangle ABC \), the length of the side AC is 6, the vertex B is \( (1, 2, 3) \) and the vertices A, C lie on the line \[ \frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{-2}. \] Then the area (in sq. units) of \( \triangle ABC \) is:

• A. 42
• B. 21
• C. 56
• D. 17

Hint:

To find the area of a triangle in 3D, use the formula \( \frac{1}{2} \times \left| \overrightarrow{AB} \times \overrightarrow{AC} \right| \). Ensure that the coordinates are substituted correctly when finding the vectors and computing the cross product.

Answer 1

The Correct Option is B

Step 1: The coordinates of points A and C can be parameterized using the equation of the line: \[ \frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{-2} = t. \] Thus, the coordinates of points A and C are: \[ A = (6 + 3t, 7 + 2t, 7 - 2t), \quad C = (6 + 3t', 7 + 2t', 7 - 2t'). \] The distance AC is given to be 6. To find the coordinates of A and C that satisfy this, we can calculate the distance using the distance formula in 3D: \[ {Distance}(A, C) = \sqrt{(x_C - x_A)^2 + (y_C - y_A)^2 + (z_C - z_A)^2} = 6. \] Step 2: Next, we can use the coordinates of point B, \( B(1, 2, 3) \), and the area formula for the triangle in 3D: \[ {Area of} \, \triangle ABC = \frac{1}{2} \times \left| \overrightarrow{AB} \times \overrightarrow{AC} \right|. \] We calculate the vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \), then compute the cross product and magnitude to obtain the area of the triangle. Finally, after simplifying, the area of the triangle \( \triangle ABC \) is found to be 21 square units.