Let O be the origin and overline{OA} = 2hat{i | vedclass

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(C) Given $\overline{OA} = 2\hat{i} + 2\hat{j} + \hat{k}$ and $\overline{OB} = \hat{i} - 2\hat{j} + 2\hat{k}$.$|\overline{OA}| = \sqrt{2^2 + 2^2 + 1^2

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$A, B, C$

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(C) Given $\overline{OA} = 2\hat{i} + 2\hat{j} + \hat{k}$ and $\overline{OB} = \hat{i} - 2\hat{j} + 2\hat{k}$.$|\overline{OA}| = \sqrt{2^2 + 2^2 + 1^2} = 3$ and $|\overline{OB}| = \sqrt{1^2 + (-2)^2 + 2^2} = 3$.$\overline{OA} \cdot \overline{OB} = (2)(1) + (2)(-2) + (1)(2) = 2 - 4 + 2 = 0$,so $\overline{OA} \perp \overline{OB}$.$\overline{OB} 	imes \overline{OC} = \overline{OB} 	imes \frac{1}{2}(\overline{OB} - \lambda\overline{OA}) = \frac{1}{2}(\overline{OB} 	imes \overline{OB} - \lambda(\overline{OB} 	imes \overline{OA})) = \frac{\lambda}{2}(\overline{OA} 	imes \overline{OB})$.$|\overline{OB} 	imes \overline{OC}| = \frac{\lambda}{2} |\overline{OA}| |\overline{OB}| = \frac{\lambda}{2} (3)(3) = \frac{9\lambda}{2}$.Given $|\overline{OB} 	imes \overline{OC}| = \frac{9}{2}$,so $\frac{9\lambda}{2} = \frac{9}{2} \implies \lambda = 1$.Thus,$\overline{OC} = \frac{1}{2}(\overline{OB} - \overline{OA})$.$(A)$ Projection of $\overline{OC}$ on $\overline{OA} = \frac{\overline{OC} \cdot \overline{OA}}{|\overline{OA}|} = \frac{\frac{1}{2}(\overline{OB} - \overline{OA}) \cdot \overline{OA}}{3} = \frac{1}{6}(\overline{OB} \cdot \overline{OA} - |\overline{OA}|^2) = \frac{1}{6}(0 - 9) = -\frac{3}{2}$. Statement $(A)$ is $TRUE$.$(B)$ Area of $	riangle OAB = \frac{1}{2} |\overline{OA} 	imes \overline{OB}| = \frac{1}{2} |\overline{OA}| |\overline{OB}| = \frac{1}{2} (3)(3) = \frac{9}{2}$. Statement $(B)$ is $TRUE$.$(C)$ Area of $	riangle ABC = \frac{1}{2} |\overline{AB} 	imes \overline{AC}|$. Since $\overline{OC} = \frac{1}{2}(\overline{OB} - \overline{OA}) = \frac{1}{2}\overline{AB}$,then $\overline{AB} = 2\overline{OC}$. $\overline{AC} = \overline{OC} - \overline{OA}$.Area $= \frac{1}{2} |2\overline{OC} 	imes (\overline{OC} - \overline{OA})| = |\overline{OC} 	imes \overline{OC} - \overline{OC} 	imes \overline{OA}| = |\overline{OA} 	imes \overline{OC}| = |\overline{OA} 	imes \frac{1}{2}(\overline{OB} - \overline{OA})| = \frac{1}{2} |\overline{OA} 	imes \overline{OB}| = \frac{9}{2}$. Statement $(C)$ is $TRUE$.

$A$. Unit vector in the direction opposite to that $a-b$ is	$(i) \ 5 \hat{i} + 3 \hat{j} - 3 \hat{k}$
$B$. If $\vec{AB} = a, \vec{BC} = b$,then $\vec{CA} =$	$(ii) \ 2 \hat{i} - \frac{8}{3} \hat{k}$
$C$. If $a, b, c$ are the position vectors of the vertices of a triangle then,its centroid is	$(iii) \ -3 \hat{i} + 4 \hat{k}$
$D$. If $d$ is a vector of magnitude $2 \sqrt{14}$ and parallel to the vector $a$,then $b + d =$	$(iv) \ -\frac{\hat{i}}{\sqrt{73}} - \frac{6 \hat{j}}{\sqrt{73}} - \frac{6 \hat{k}}{\sqrt{73}}$
	$(v) \ 3 \hat{i} + 5 \hat{j} - 3 \hat{k}$

List-$I$	List-$II$
$(A)$ $[\mathbf{a} \mathbf{b} \mathbf{c}]$	$1. \|\mathbf{a}\|\|\mathbf{b}\|\cos(\mathbf{a}, \mathbf{b})$
$(B)$ $(\mathbf{c} \times \mathbf{a}) \times \mathbf{b}$	$2. (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}$
$(C)$ $\mathbf{a} \times (\mathbf{b} \times \mathbf{c})$	$3. \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$
$(D)$ $\mathbf{a} \cdot \mathbf{b}$	$4. \|\mathbf{a}\|\|\mathbf{b}\|$
	$5. (\mathbf{b} \cdot \mathbf{c})\mathbf{a} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}$

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