In the given figure,if a vector x satisfies t | vedclass

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(B) From the triangle law of vector addition in $	riangle ADC$,we have $\vec{AD} + \vec{v} = \vec{AC}$. Given $\vec{AD} = b$ and $\vec{AC} = c$,we ge

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$-a + 2b - c$

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(B) From the triangle law of vector addition in $	riangle ADC$,we have $\vec{AD} + \vec{v} = \vec{AC}$. Given $\vec{AD} = b$ and $\vec{AC} = c$,we get $b + v = c$,so $v = c - b$.From the triangle law of vector addition in $	riangle ABD$,we have $\vec{AB} + \vec{w} = \vec{AD}$. Given $\vec{AB} = a$ and $\vec{AD} = b$,we get $a + w = b$,so $w = b - a$.The given equation is $x - w = v$,which implies $x = v + w$.Substituting the values of $v$ and $w$,we get $x = (c - b) + (b - a) = c - a$.Wait,let us re-examine the figure. In $	riangle ABD$,$\vec{AB} + \vec{BD} = \vec{AD} \implies a + w = b \implies w = b - a$.In $	riangle ADC$,$\vec{AD} + \vec{DC} = \vec{AC} \implies b + v = c \implies v = c - b$ is incorrect based on the arrow direction. The arrow for $v$ is from $C$ to $D$,so $\vec{DC} = -v$. Thus $\vec{AD} + \vec{DC} = \vec{AC} \implies b - v = c \implies v = b - c$.Then $x = v + w = (b - c) + (b - a) = 2b - a - c$.

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

In the given figure,if a vector $x$ satisfies the equation $x - w = v$,then $x = ?$

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