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: \[ 54 \]

Answer 2

Step 1: Given information.
We are given that the area of \( \triangle ABC = 15\sqrt{2} \).
Vectors are:
\[ \overrightarrow{AB} = \hat{i} + 2\hat{j} - 7\hat{k}, \quad \overrightarrow{BC} = a\hat{i} + b\hat{j} + c\hat{k}, \quad \overrightarrow{AC} = 6\hat{i} + d\hat{j} - 2\hat{k}, \; d > 0. \]

Step 2: Express all vectors from the same origin.
We know that:
\[ \overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}. \] Substitute the given vectors:
\[ 6\hat{i} + d\hat{j} - 2\hat{k} = (\hat{i} + 2\hat{j} - 7\hat{k}) + (a\hat{i} + b\hat{j} + c\hat{k}). \] Comparing coefficients, we get:
\[ a = 5 - 1 = 5 - 1 = 5 - 1 = \text{actually from } 6 - 1 = 5. \] \[ b = d - 2, \quad c = 5. \] Hence: \[ \overrightarrow{BC} = 5\hat{i} + (d - 2)\hat{j} + 12\hat{k}. \] (Small correction — since \(-2 = -7 + c \Rightarrow c = 5\), yes confirmed.)

Step 3: Area of the triangle using vector formula.
The area of a triangle formed by vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) is given by: \[ \text{Area} = \frac{1}{2} |\overrightarrow{AB} \times \overrightarrow{AC}|. \] Given: \[ \frac{1}{2} |\overrightarrow{AB} \times \overrightarrow{AC}| = 15\sqrt{2}. \] \[ |\overrightarrow{AB} \times \overrightarrow{AC}| = 30\sqrt{2}. \]

Step 4: Compute cross product.
\[ \overrightarrow{AB} = (1, 2, -7), \quad \overrightarrow{AC} = (6, d, -2). \] \[ \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -7 \\ 6 & d & -2 \end{vmatrix} = \hat{i}(2(-2) - (-7)d) - \hat{j}(1(-2) - (-7)(6)) + \hat{k}(1d - 2(6)). \] \[ = \hat{i}(-4 + 7d) - \hat{j}(-2 + 42) + \hat{k}(d - 12). \] \[ = \hat{i}(7d - 4) - \hat{j}(40) + \hat{k}(d - 12). \] Hence: \[ |\overrightarrow{AB} \times \overrightarrow{AC}|^2 = (7d - 4)^2 + (-40)^2 + (d - 12)^2. \] Given that \( |\overrightarrow{AB} \times \overrightarrow{AC}| = 30\sqrt{2} \): \[ (7d - 4)^2 + 1600 + (d - 12)^2 = 1800. \] \[ 49d^2 - 56d + 16 + 1600 + d^2 - 24d + 144 = 1800. \] \[ 50d^2 - 80d + 1760 = 1800. \] \[ 50d^2 - 80d - 40 = 0. \] \[ 5d^2 - 8d - 4 = 0. \] \[ d = \frac{8 \pm \sqrt{64 + 80}}{10} = \frac{8 \pm \sqrt{144}}{10} = \frac{8 \pm 12}{10}. \] \[ d_1 = 2, \quad d_2 = -\frac{2}{5}. \] Given \( d > 0 \Rightarrow d = 2. \)

Step 5: Find the lengths of sides.
\[ \overrightarrow{AB} = (1, 2, -7), \quad \overrightarrow{BC} = (5, 0, 5), \quad \overrightarrow{AC} = (6, 2, -2). \] Compute squares of their lengths:
\[ AB^2 = 1^2 + 2^2 + (-7)^2 = 1 + 4 + 49 = 54. \] \[ AC^2 = 6^2 + 2^2 + (-2)^2 = 36 + 4 + 4 = 44. \] \[ BC^2 = 5^2 + 0^2 + 5^2 = 25 + 25 = 50. \] Hence, the largest side is \( AB \) with square of its length = 54.

Final Answer:
\[ \boxed{54} \]