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(A) We analyze each expression:$(A)$ The scalar triple product $[\mathbf{a} \mathbf{b} \mathbf{c}]$ is defined as $\mathbf{a} \cdot (\mathbf{b} 	imes

Vedclass · Accepted Answer

$A-3, B-5, C-2, D-1$

Vedclass · Answer

(A) We analyze each expression:$(A)$ The scalar triple product $[\mathbf{a} \mathbf{b} \mathbf{c}]$ is defined as $\mathbf{a} \cdot (\mathbf{b} 	imes \mathbf{c})$. Thus,$(A)$ matches with $3$.$(B)$ Using the vector triple product formula $(\mathbf{x} 	imes \mathbf{y}) 	imes \mathbf{z} = (\mathbf{x} \cdot \mathbf{z})\mathbf{y} - (\mathbf{y} \cdot \mathbf{z})\mathbf{x}$,we have $(\mathbf{c} 	imes \mathbf{a}) 	imes \mathbf{b} = (\mathbf{c} \cdot \mathbf{b})\mathbf{a} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c} = (\mathbf{b} \cdot \mathbf{c})\mathbf{a} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}$. Thus,$(B)$ matches with $5$.$(C)$ Using the vector triple product formula $\mathbf{a} 	imes (\mathbf{b} 	imes \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}$. Thus,$(C)$ matches with $2$.$(D)$ The dot product $\mathbf{a} \cdot \mathbf{b}$ is defined as $|\mathbf{a}||\mathbf{b}|\cos(	heta)$,where $	heta$ is the angle between $\mathbf{a}$ and $\mathbf{b}$. Thus,$(D)$ matches with $1$.Therefore,the correct matching is $A-3, B-5, C-2, D-1$.

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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