Question

\( A(1,2,3) \), \( B(3,4,k) \), \( C(2,1,4) \) form an isosceles triangle. If \( AB=BC \), then the area of \( \triangle ABC \) is

• A. \( \frac{\sqrt{165}}{4} \)
• B. \( \frac{15}{4} \)
• C. \( \frac{7}{2} \)
• D. \( \frac{\sqrt{114}}{4} \)

Hint:

For an isosceles triangle with \( AB=BC \), the midpoint of side \( AC \) forms a right angle with vertex \( B \). Calculating the base length and height using this geometric property is an elegant alternative to computing the vector cross product.

Answer 1

The Correct Option is D

Concept: We use the 3D distance formula to set up the equation \( AB^2 = BC^2 \) to solve for the unknown parameter \( k \). Once all vertex coordinates are determined, the area of the triangle can be evaluated using the cross product of two adjacent side vectors: \[ \text{Area} = \frac{1}{2} |\vec{AB} \times \vec{AC}| \]

Step 1: Solving for parameter \( k \) using the side length condition.
\[ AB^2 = (3 - 1)^2 + (4 - 2)^2 + (k - 3)^2 = 2^2 + 2^2 + (k - 3)^2 = 8 + (k - 3)^2 \] \[ BC^2 = (2 - 3)^2 + (1 - 4)^2 + (4 - k)^2 = (-1)^2 + (-3)^2 + (4 - k)^2 = 10 + (4 - k)^2 \] Equating the two lengths: \[ 8 + k^2 - 6k + 9 = 10 + 16 - 8k + k^2 \] \[ -6k + 17 = 26 - 8k \implies 2k = 9 \implies k = \frac{9}{2} = 4.5 \]

Step 2: Forming side vectors.
Using \( k = 4.5 \): \[ \vec{AB} = (2, 2, 1.5) \implies \vec{AB} = 2\hat{i} + 2\hat{j} + \frac{3}{2}\hat{k} \] \[ \vec{AC} = (2-1)\hat{i} + (1-2)\hat{j} + (4-3)\hat{k} = 1\hat{i} - 1\hat{j} + 1\hat{k} \]

Step 3: Computing the cross product and area.
\[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & 3/2 \\ 1 & -1 & 1 \end{vmatrix} = \hat{i}\left(2 + \frac{3}{2}\right) - \hat{j}\left(2 - \frac{3}{2}\right) + \hat{k}(-2 - 2) \] \[ = \frac{7}{2}\hat{i} - \frac{1}{2}\hat{j} - 4\hat{k} \] Now find its magnitude: \[ |\vec{AB} \times \vec{AC}| = \sqrt{\left(\frac{7}{2}\right)^2 + \left(-\frac{1}{2}\right)^2 + (-4)^2} = \sqrt{\frac{49}{4} + \frac{1}{4} + 16} = \sqrt{\frac{50 + 64}{4}} = \frac{\sqrt{114}}{2} \]