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: Parametrize the Line Equation

The given equation of the line is:

\( \frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{-2}. \)

Let the common ratio be \( t \). Then the coordinates of points \( A \) and \( C \) on the line can be written as:

\( x = 3t + 6, \quad y = 2t + 7, \quad z = -2t + 7. \)

Thus, the coordinates of \( A \) and \( C \) are parametrized as:

\( A(3t + 6, 2t + 7, -2t + 7) \quad \text{and} \quad C(3s + 6, 2s + 7, -2s + 7). \)

Step 2: Find the Vector \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \)

Let \( B(1, 2, 3) \) be the coordinates of point B. Now, the vector \( \overrightarrow{AB} \) is:

\[ \overrightarrow{AB} = (1 - (3t + 6), 2 - (2t + 7), 3 - (-2t + 7)) = (-3t - 5, -2t - 5, 2t - 4). \]

The vector \( \overrightarrow{AC} \) is:

\[ \overrightarrow{AC} = (3s + 6 - (3t + 6), 2s + 7 - (2t + 7), -2s + 7 - (-2t + 7)) = (3(s - t), 2(s - t), -2(s - t)). \]

Step 3: Use the Formula for the Area of a Triangle

The area of triangle \( ABC \) is given by:

\[ \text{Area} = \frac{1}{2} \left| \overrightarrow{AB} \times \overrightarrow{AC} \right|. \]

Now, compute the cross product \( \overrightarrow{AB} \times \overrightarrow{AC} \):

\[ \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3t - 5 & -2t - 5 & 2t - 4 \\ 3(s - t) & 2(s - t) & -2(s - t) \end{vmatrix}. \]

Step 4: Calculate the Determinant and Simplify

After simplifying the determinant, we find the magnitude of the cross product and compute the area. \[ \text{Area} = 21 \, \text{square units}. \]

Answer:

The area of triangle \( ABC \) is 21 square units.