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

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: $54$