If overrightarrow{a}=hat{i}+hat{j}+hat{k}, ov | vedclass

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(B) Given vectors are $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}, \overrightarrow{b}=\hat{i}-\hat{j}+\hat{k}, \overrightarrow{c}=\hat{i}+\hat{j}-\hat

Vedclass · Accepted Answer

$i-C, ii-A, iii-F, iv-E$

Vedclass · Answer

(B) Given vectors are $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}, \overrightarrow{b}=\hat{i}-\hat{j}+\hat{k}, \overrightarrow{c}=\hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{d}=\hat{i}-\hat{j}-\hat{k}$.$(i)$ $\overrightarrow{a} \cdot \overrightarrow{b} = (1)(1) + (1)(-1) + (1)(1) = 1 - 1 + 1 = 1$.$\overrightarrow{b} \cdot \overrightarrow{d} = (1)(1) + (-1)(-1) + (1)(-1) = 1 + 1 - 1 = 1$.Thus, $\overrightarrow{a} \cdot \overrightarrow{b} = \overrightarrow{b} \cdot \overrightarrow{d}$, which corresponds to $(C)$.(ii) $\overrightarrow{b} \cdot \overrightarrow{c} = (1)(1) + (-1)(1) + (1)(-1) = 1 - 1 - 1 = -1$.$\overrightarrow{a} \cdot \overrightarrow{d} = (1)(1) + (1)(-1) + (1)(-1) = 1 - 1 - 1 = -1$.Thus, $\overrightarrow{b} \cdot \overrightarrow{c} = \overrightarrow{a} \cdot \overrightarrow{d}$, which corresponds to $(A)$.(iii) $[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}] = \begin{vmatrix} 1 & 1 & 1 \ 1 & -1 & 1 \ 1 & 1 & -1 \end{vmatrix} = 1(1-1) - 1(-1-1) + 1(1+1) = 0 + 2 + 2 = 4$, which corresponds to $(F)$.(iv) $\overrightarrow{b} 	imes \overrightarrow{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 1 \ 1 & 1 & -1 \end{vmatrix} = \hat{i}(1-1) - \hat{j}(-1-1) + \hat{k}(1+1) = 2\hat{j} + 2\hat{k}$, which corresponds to $(E)$.Therefore, the correct match is $i-C, ii-A, iii-F, iv-E$.

List-$I$	List-$II$
$(i)$ $\overrightarrow{a} \cdot \overrightarrow{b}$	$(A)$ $\overrightarrow{a} \cdot \overrightarrow{d}$
(ii) $\overrightarrow{b} \cdot \overrightarrow{c}$	$(B)$ $3$
(iii) $[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]$	$(C)$ $\overrightarrow{b} \cdot \overrightarrow{d}$
(iv) $\overrightarrow{b} \times \overrightarrow{c}$	$(D)$ $2\hat{i}-2\hat{k}$
	$(E)$ $2\hat{j}+2\hat{k}$
	$(F)$ $4$

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