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

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(B) આપેલ સદિશો $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}, \overrightarrow{b}=\hat{i}-\hat{j}+\hat{k}, \overrightarrow{c}=\hat{i}+\hat{j}-\hat{k}$ અન

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) આપેલ સદિશો $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}, \overrightarrow{b}=\hat{i}-\hat{j}+\hat{k}, \overrightarrow{c}=\hat{i}+\hat{j}-\hat{k}$ અને $\overrightarrow{d}=\hat{i}-\hat{j}-\hat{k}$ છે.$(i)$ $\overrightarrow{a} \cdot \overrightarrow{b} = (1)(1) + (1)(-1) + (1)(1) = 1 - 1 + 1 = 1$.$\overrightarrow{b} \cdot \overrightarrow{d} = (1)(1) + (-1)(-1) + (1)(-1) = 1 + 1 - 1 = 1$.આમ, $\overrightarrow{a} \cdot \overrightarrow{b} = \overrightarrow{b} \cdot \overrightarrow{d}$, જે $(C)$ ને અનુરૂપ છે.(ii) $\overrightarrow{b} \cdot \overrightarrow{c} = (1)(1) + (-1)(1) + (1)(-1) = 1 - 1 - 1 = -1$.$\overrightarrow{a} \cdot \overrightarrow{d} = (1)(1) + (1)(-1) + (1)(-1) = 1 - 1 - 1 = -1$.આમ, $\overrightarrow{b} \cdot \overrightarrow{c} = \overrightarrow{a} \cdot \overrightarrow{d}$, જે $(A)$ ને અનુરૂપ છે.(iii) $[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}] = \begin{vmatrix} 1 & 1 & 1 \ 1 & -1 & 1 \ 1 & 1 & -1 \end{vmatrix} = 1(1-1) - 1(-1-1) + 1(1+1) = 0 + 2 + 2 = 4$, જે $(F)$ ને અનુરૂપ છે.(iv) $\overrightarrow{b} 	imes \overrightarrow{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 1 \ 1 & 1 & -1 \end{vmatrix} = \hat{i}(1-1) - \hat{j}(-1-1) + \hat{k}(1+1) = 2\hat{j} + 2\hat{k}$, જે $(E)$ ને અનુરૂપ છે.તેથી, સાચી જોડ $i-C, ii-A, iii-F, iv-E$ છે.

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) $\overrightarrow{b} \times \overrightarrow{c}$	$(D)$ $2\hat{i}-2\hat{k}$
	$(E)$ $2\hat{j}+2\hat{k}$
	$(F)$ $4$

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

$A$. $a-b$ ની વિરુદ્ધ દિશામાં એકમ સદિશ	$(i) \ 5 \hat{i} + 3 \hat{j} - 3 \hat{k}$
$B$. જો $\vec{AB} = a, \vec{BC} = b$ હોય, તો $\vec{CA} =$	$(ii) \ 2 \hat{i} - \frac{8}{3} \hat{k}$
$C$. જો $a, b, c$ ત્રિકોણના શિરોબિંદુઓના સ્થાન સદિશ હોય, તો તેનું મધ્યકેન્દ્ર	$(iii) \ -3 \hat{i} + 4 \hat{k}$
$D$. જો $d$ એ $2 \sqrt{14}$ માન ધરાવતો અને $a$ ને સમાંતર સદિશ હોય, તો $b + d =$	$(iv) \ -\frac{\hat{i}}{\sqrt{73}} - \frac{6 \hat{j}}{\sqrt{73}} - \frac{6 \hat{k}}{\sqrt{73}}$
	$(v) \ 3 \hat{i} + 5 \hat{j} - 3 \hat{k}$

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