Let vec{w}=hat{i}+hat{j}-2 hat{k},and vec{u} | vedclass

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(A) Given $\vec{w} = \hat{i} + \hat{j} - 2\hat{k}$. Since $\vec{u} 	imes \vec{v} = \vec{w}$,we have $\vec{u} \perp \vec{w}$ and $\vec{v} \perp \vec{w

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$(P) \rightarrow (2), (Q) \rightarrow (1), (R) \rightarrow (4), (S) \rightarrow (5)$

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(A) Given $\vec{w} = \hat{i} + \hat{j} - 2\hat{k}$. Since $\vec{u} 	imes \vec{v} = \vec{w}$,we have $\vec{u} \perp \vec{w}$ and $\vec{v} \perp \vec{w}$.Also,$\vec{v} 	imes \vec{w} = \vec{u}$,which implies $\vec{u} \perp \vec{w}$ and $\vec{v} \perp \vec{w}$.The system of equations $-t\alpha + \beta + \gamma = 0$,$\alpha - t\beta + \gamma = 0$,$\alpha + \beta - t\gamma = 0$ has a non-trivial solution if the determinant is zero:$\begin{vmatrix} -t & 1 & 1 \ 1 & -t & 1 \ 1 & 1 & -t \end{vmatrix} = 0 \Rightarrow -(t^3 - 1) - 1(-t - 1) + 1(1 + t) = 0 \Rightarrow -t^3 + 1 + t + 1 + 1 + t = 0 \Rightarrow t^3 - 2t - 3 = 0$.Factoring gives $(t+1)(t^2 - t - 3) = 0$. For the given conditions,$t = -1$ or $t = 2$.If $t = 2$,$\alpha = \beta = \gamma$. Since $\vec{u} \cdot \vec{w} = 0$,$\alpha + \beta - 2\gamma = 0 \Rightarrow 2\alpha - 2\alpha = 0$,which is consistent.Using $|\vec{u}| = |\vec{w}| = \sqrt{6}$,$3\alpha^2 = 6 \Rightarrow \alpha^2 = 2$. Thus $\alpha = \pm \sqrt{2}$. For $t=2$,$t+3 = 5$.If $t = -1$,$\alpha + \beta + \gamma = 0$. Also $\vec{u} \cdot \vec{w} = 0 \Rightarrow \alpha + \beta - 2\gamma = 0$.Subtracting gives $3\gamma = 0 \Rightarrow \gamma = 0$. Then $\beta = -\alpha$.For $\alpha = \sqrt{3}$,$\gamma^2 = 0$ and $(\beta + \gamma)^2 = (-\sqrt{3} + 0)^2 = 3$.Thus,$(P) ightarrow (2), (Q) ightarrow (1), (R) ightarrow (4), (S) ightarrow (5)$.

List-$I$	List-$II$
$(P)$ $\|\vec{v}\|^2$ is equal to	$(1)$ $0$
$(Q)$ If $\alpha=\sqrt{3}$,then $\gamma^2$ is equal to	$(2)$ $1$
$(R)$ If $\alpha=\sqrt{3}$,then $(\beta+\gamma)^2$ is equal to	$(3)$ $2$
$(S)$ If $\alpha=\sqrt{2}$,then $t+3$ is equal to	$(4)$ $3$
	$(5)$ $5$

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

$A$. 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 $b + d =$	$(iv) \ -\frac{\hat{i}}{\sqrt{73}} - \frac{6 \hat{j}}{\sqrt{73}} - \frac{6 \hat{k}}{\sqrt{73}}$
	$(v) \ 3 \hat{i} + 5 \hat{j} - 3 \hat{k}$

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