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

To determine the area of triangle \( \triangle ABC \), we start by analyzing the given information:

• Point \( B \) is at coordinates \((1, 2, 3)\). 
• The line equation on which points \( A \) and \( C \) lie is: \(\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{-2}\).
• The length of side \( AC \) is 6 units.

The line can be parameterized as follows: let the parameter be \( t \).

• For point \( A \):
  • \( x = 3t + 6 \)
  • \( y = 2t + 7 \)
  • \( z = -2t + 7 \)
• For point \( C \):
  • \( x = 3s + 6 \)
  • \( y = 2s + 7 \)
  • \( z = -2s + 7 \)

The distance between points \( A(t) \) and \( C(s) \) is given by:

\[\sqrt{(3(s-t))^2 + (2(s-t))^2 + (-2(s-t))^2} = 6\]

This simplifies to:

\[\sqrt{9(s-t)^2 + 4(s-t)^2 + 4(s-t)^2} = 6 \rightarrow \sqrt{17(s-t)^2} = 6\]

Squaring both sides, we get:

\[17(s-t)^2 = 36 \rightarrow (s-t)^2 = \frac{36}{17}\]

Thus, \( s-t = \pm \sqrt{\frac{36}{17}} \).

The area of triangle \( \triangle ABC \) is calculated using the cross product of vectors \( \overrightarrow{AB} \) and \( \overrightarrow{BC} \):

• Vector \( \overrightarrow{AB} = (3t + 6 - 1, 2t + 7 - 2, -2t + 7 - 3)\) = \( (3t + 5, 2t + 5, -2t + 4)\)
• Vector \( \overrightarrow{BC} = (3s + 6 - 1, 2s + 7 - 2, -2s + 7 - 3)\) = \( (3s + 5, 2s + 5, -2s + 4)\)

The cross product \(\overrightarrow{AB} \times \overrightarrow{BC}\) is computed and its magnitude used to find the area:

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

Simplifying, we find the magnitude of this cross product. Solving gives:

\[\text{Area} = \frac{1}{2} | \overrightarrow{AB} \times \overrightarrow{BC} |\]

Finally, substituting geometric and trigonometric components, this culminates in an area of 21 square units.

Thus, the area of \( \triangle ABC \) is 21 square units, making option 21 the correct choice.

Answer 2

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.