Vector Properties and Area Analysis involving Cross and Scalar Products

To determine which statements are true, we first analyze the given vectors and their properties.

Let the given vectors be: $\vec{OA} = \vec{a} = 2\hat{i} + 2\hat{j} + \hat{k}$ $\vec{OB} = \vec{b} = \hat{i} - 2\hat{j} + 2\hat{k}$ $\vec{OC} = \vec{c} = \frac{1}{2}(\vec{b} - \lambda \vec{a}) \quad \text{for } \lambda > 0$

First, let's calculate the magnitudes of $\vec{a}$ and $\vec{b}$: $|\vec{a}| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{9} = 3$ $|\vec{b}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{9} = 3$

Next, we evaluate the scalar product $\vec{a} \cdot \vec{b}$: $\vec{a} \cdot \vec{b} = (2)(1) + (2)(-2) + (1)(2) = 2 - 4 + 2 = 0$ Since $\vec{a} \cdot \vec{b} = 0$, the vectors $\vec{OA}$ and $\vec{OB}$ are orthogonal ($\vec{OA} \perp \vec{OB}$).

The magnitude of the cross product $\vec{a} \times \vec{b}$ is: $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin(90^\circ) = 3 \times 3 \times 1 = 9$

Now, using the expression for $\vec{OC}$, we evaluate $\vec{OB} \times \vec{OC}$: $\vec{OB} \times \vec{OC} = \vec{b} \times \left[\frac{1}{2}(\vec{b} - \lambda \vec{a})\right] = \frac{1}{2}(\vec{b} \times \vec{b}) - \frac{\lambda}{2}(\vec{b} \times \vec{a}) = \frac{\lambda}{2}(\vec{a} \times \vec{b})$

Taking the magnitude on both sides: $|\vec{OB} \times \vec{OC}| = \frac{\lambda}{2} |\vec{a} \times \vec{b}| = \frac{9\lambda}{2}$

We are given that $|\vec{OB} \times \vec{OC}| = \frac{9}{2}$. Therefore: $\frac{9\lambda}{2} = \frac{9}{2} \implies \lambda = 1$

Thus, the vector $\vec{OC}$ becomes: $\vec{OC} = \vec{c} = \frac{1}{2}(\vec{b} - \vec{a})$

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Step-by-step evaluation of the options:

Option (A): Projection of $\vec{OC}$ on $\vec{OA}$

The projection of $\vec{OC}$ on $\vec{OA}$ is given by: $\text{Projection} = \frac{\vec{OC} \cdot \vec{OA}}{|\vec{OA}|} = \frac{\frac{1}{2}(\vec{b} - \vec{a}) \cdot \vec{a}}{3} = \frac{\frac{1}{2}(\vec{b} \cdot \vec{a} - |\vec{a}|^2)}{3}$ Since $\vec{b} \cdot \vec{a} = 0$ and $|\vec{a}|^2 = 9$: $\text{Projection} = \frac{\frac{1}{2}(0 - 9)}{3} = \frac{-\frac{9}{2}}{3} = -\frac{3}{2}$ Thus, Statement (A) is TRUE.

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Option (B): Area of triangle $OAB$

The area of $\triangle OAB$ is: $\text{Area}(\triangle OAB) = \frac{1}{2} |\vec{OA} \times \vec{OB}| = \frac{1}{2} |\vec{a} \times \vec{b}| = \frac{1}{2} \times 9 = \frac{9}{2}$ Thus, Statement (B) is TRUE.

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Option (C): Area of triangle $ABC$

The position vectors of the vertices are $\vec{a}$, $\vec{b}$, and $\vec{c} = \frac{1}{2}(\vec{b} - \vec{a})$.

The vector sides from vertex $A$ are: $\vec{AB} = \vec{b} - \vec{a}$ $\vec{AC} = \vec{c} - \vec{a} = \frac{1}{2}(\vec{b} - \vec{a}) - \vec{a} = \frac{1}{2}\vec{b} - \frac{3}{2}\vec{a}$

The cross product $\vec{AB} \times \vec{AC}$ is: $\vec{AB} \times \vec{AC} = (\vec{b} - \vec{a}) \times \left(\frac{1}{2}\vec{b} - \frac{3}{2}\vec{a}\right)$ $= -\frac{3}{2}(\vec{b} \times \vec{a}) - \frac{1}{2}(\vec{a} \times \vec{b}) = \frac{3}{2}(\vec{a} \times \vec{b}) - \frac{1}{2}(\vec{a} \times \vec{b}) = \vec{a} \times \vec{b}$

Therefore, the area of $\triangle ABC$ is: $\text{Area}(\triangle ABC) = \frac{1}{2} |\vec{AB} \times \vec{AC}| = \frac{1}{2} |\vec{a} \times \vec{b}| = \frac{9}{2}$ Thus, Statement (C) is TRUE.

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Option (D): Acute angle between the diagonals of the parallelogram with adjacent sides $\vec{OA}$ and $\vec{OC}$

The adjacent sides are $\vec{u} = \vec{a}$ and $\vec{v} = \frac{1}{2}(\vec{b} - \vec{a})$.

The diagonals are given by: $\vec{D}_1 = \vec{u} + \vec{v} = \vec{a} + \frac{1}{2}(\vec{b} - \vec{a}) = \frac{1}{2}(\vec{a} + \vec{b})$ $\vec{D}_2 = \vec{u} - \vec{v} = \vec{a} - \frac{1}{2}(\vec{b} - \vec{a}) = \frac{1}{2}(3\vec{a} - \vec{b})$

Calculating their magnitudes: $|\vec{D}_1| = \frac{1}{2} \sqrt{|\vec{a}|^2 + |\vec{b}|^2} = \frac{1}{2} \sqrt{9 + 9} = \frac{3}{\sqrt{2}}$ $|\vec{D}_2| = \frac{1}{2} \sqrt{9|\vec{a}|^2 + |\vec{b}|^2} = \frac{1}{2} \sqrt{81 + 9} = \frac{3\sqrt{10}}{2}$

Calculating their dot product: $\vec{D}_1 \cdot \vec{D}_2 = \frac{1}{4} (\vec{a} + \vec{b}) \cdot (3\vec{a} - \vec{b}) = \frac{1}{4} (3|\vec{a}|^2 - |\vec{b}|^2) = \frac{1}{4}(27 - 9) = \frac{9}{2}$

Let $\theta$ be the angle between the diagonals: $\cos \theta = \frac{|\vec{D}_1 \cdot \vec{D}_2|}{|\vec{D}_1| |\vec{D}_2|} = \frac{\frac{9}{2}}{\left(\frac{3}{\sqrt{2}}\right)\left(\frac{3\sqrt{10}}{2}\right)} = \frac{\frac{9}{2}}{\frac{9\sqrt{5}}{2}} = \frac{1}{\sqrt{5}}$

Since $\cos \theta = \frac{1}{\sqrt{5}} \neq \frac{1}{2}$, we have $\theta \neq \frac{\pi}{3}$. Thus, Statement (D) is FALSE.

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Conclusion:

The correct statements are (A), (B), and (C).