Question

Let \( \vec a = 2\hat i + 3\hat j + 3\hat k \) and \( \vec b = 6\hat i + 3\hat j + 3\hat k \). Then the square of the area of the triangle with adjacent sides determined by the vectors \( (2\vec a + 3\vec b) \) and \( (\vec a - \vec b) \) is :

• A. \(450\)
• B. \(900\)
• C. \(1800\)
• D. \(2400\)

Answer 1

The Correct Option is A

Concept:
If two adjacent sides of a triangle are represented by vectors \( \vec{u} \) and \( \vec{v} \), then: \[ \text{Area of triangle} = \frac{1}{2} |\vec{u} \times \vec{v}| \] Thus, \[ (\text{Area})^2 = \frac{1}{4} |\vec{u} \times \vec{v}|^2 \]
Step 1: Find the vectors. 
\[ \vec{u} = 2\vec{a} + 3\vec{b} \] \[ = 2(2,3,3) + 3(6,3,3) \] \[ = (4,6,6) + (18,9,9) \] \[ = (22,15,15) \] \[ \vec{v} = \vec{a} - \vec{b} \] \[ = (2,3,3) - (6,3,3) \] \[ = (-4,0,0) \] 
Step 2: Compute cross product.
\[ \vec{u} \times \vec{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 22 & 15 & 15 \\ -4 & 0 & 0 \end{vmatrix} \] \[ = (0)\mathbf{i} - (-60)\mathbf{j} + (60)\mathbf{k} \] \[ = (0, 60, -60) \] 
Step 3: Find magnitude squared.
\[ |\vec{u} \times \vec{v}|^2 = 0^2 + 60^2 + (-60)^2 = 7200 \] 
Step 4: Compute square of area.
\[ (\text{Area})^2 = \frac{1}{4} \times 7200 = 1800 \] Hence, the square of the area of the triangle is \(1800\).