Let a = 2 i + 3 j + 3 k and b = 6 i + 3 j + 3 k. Then the square of the area of the triangle with adjacent sides determined by the vectors (2 a + 3…

The area of a triangle with adjacent sides $\vec{u}$u and $\vec{v}$v is given by $\Delta = \dfrac{1}{2} |\vec{u} \times \vec{v}|$Δ=21​∣u×v∣.

Here, $\vec{u} = 2\vec{a} + 3\vec{b}$u=2a+3b and $\vec{v} = \vec{a} - \vec{b}$v=a−b.

$$\vec{u} \times \vec{v} = (2\vec{a} + 3\vec{b}) \times (\vec{a} - \vec{b})$$u×v=(2a+3b)×(a−b)

$$\vec{u} \times \vec{v} = 2(\vec{a} \times \vec{a}) - 2(\vec{a} \times \vec{b}) + 3(\vec{b} \times \vec{a}) - 3(\vec{b} \times \vec{b})$$u×v=2(a×a)−2(a×b)+3(b×a)−3(b×b)

Since $\vec{a} \times \vec{a} = \vec{0}$a×a=0, $\vec{b} \times \vec{b} = \vec{0}$b×b=0, and $\vec{b} \times \vec{a} = -\vec{a} \times \vec{b}$b×a=−a×b, we get:

$$\vec{u} \times \vec{v} = -2(\vec{a} \times \vec{b}) - 3(\vec{a} \times \vec{b}) = -5(\vec{a} \times \vec{b})$$u×v=−2(a×b)−3(a×b)=−5(a×b)

The area of the triangle is $\Delta = \dfrac{1}{2} |-5(\vec{a} \times \vec{b})| = \dfrac{5}{2} |\vec{a} \times \vec{b}|$Δ=21​∣−5(a×b)∣=25​∣a×b∣.

Now, calculating $\vec{a} \times \vec{b}$a×b:

$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 3 \\ 6 & 3 & 3 \end{vmatrix}$$a×b=​i^26​j^​33​k^33​​

$$\vec{a} \times \vec{b} = \hat{i}(9 - 9) - \hat{j}(6 - 18) + \hat{k}(6 - 18) = 12\hat{j} - 12\hat{k}$$a×b=i^(9−9)−j^​(6−18)+k^(6−18)=12j^​−12k^

The square of the magnitude is:

$$|\vec{a} \times \vec{b}|^2 = 12^2 + (-12)^2 = 144 + 144 = 288$$∣a×b∣2=122+(−12)2=144+144=288

The square of the area of the triangle is:

$$\Delta^2 = \left( \dfrac{5}{2} \right)^2 |\vec{a} \times \vec{b}|^2 = \dfrac{25}{4} \times 288 = 25 \times 72 = 1800$$Δ2=(25​)2∣a×b∣2=425​×288=25×72=1800

Answer: $1800$1800