Question

Let \( \triangle ABC \) be a triangle of area \( 15\sqrt{2} \) and the vectors \[ \overrightarrow{AB} = \hat{i} + 2\hat{j} - 7\hat{k}, \quad \overrightarrow{BC} = a\hat{i} + b\hat{j} + c\hat{k}, \quad \text{and} \quad \overrightarrow{AC} = 6\hat{i} + d\hat{j} - 2\hat{k}, \, d > 0.\]Then the square of the length of the largest side of the triangle \( \triangle ABC \) is

Answer 1

Correct Answer: 54

Given vectors: 
\[ \overrightarrow{AB} = \hat{i} + 2\hat{j} - 7\hat{k}, \quad \overrightarrow{AC} = 6\hat{i} + d\hat{j} - 2\hat{k} \] 
The cross product \(\overrightarrow{AB} \times \overrightarrow{AC}\) gives the area of the triangle \(ABC\) using the formula:
\[ \text{Area} = \frac{1}{2} \left| \overrightarrow{AB} \times \overrightarrow{AC} \right| \] 
Given that the area is \(15\sqrt{2}\):
\[ 15\sqrt{2} = \frac{1}{2} \left| \overrightarrow{AB} \times \overrightarrow{AC} \right| \] 
Thus:
\[ \left| \overrightarrow{AB} \times \overrightarrow{AC} \right| = 30\sqrt{2} \] 

Calculating the cross product: 
\[ \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -7 \\ 6 & d & -2 \end{vmatrix} \]

\[ = \hat{i}(2 \times -2 - (-7) \times d) - \hat{j}(1 \times -2 - (-7) \times 6) + \hat{k}(1 \times d - 2 \times 6) \] 

Simplifying: \[ \overrightarrow{AB} \times \overrightarrow{AC} = \hat{i}(-4 + 7d) - \hat{j}(-2 + 42) + \hat{k}(d - 12) \]

\[ = (7d - 4)\hat{i} - 40\hat{j} + (d - 12)\hat{k} \] 

The magnitude of the cross product is given by: 
\[ \left| \overrightarrow{AB} \times \overrightarrow{AC} \right| = \sqrt{(7d - 4)^2 + (-40)^2 + (d - 12)^2} \] 

Equating this to \(30\sqrt{2}\): \[ \sqrt{(7d - 4)^2 + 1600 + (d - 12)^2} = 30\sqrt{2} \] 

Squaring both sides: \[ (7d - 4)^2 + 1600 + (d - 12)^2 = 1800 \] 
Solving this equation gives the value of \(d\). 

To find the square of the length of the largest side, we calculate: 
\[ |\overrightarrow{AB}|^2 = 1^2 + 2^2 + (-7)^2 = 1 + 4 + 49 = 54 \] 
Similarly, the length of \(\overrightarrow{AC}\) is calculated. 
Thus, the square of the length of the largest side is: \[ 54 \]