text{જો } vec{a} = hat{i} + hat{j} + hat{k}, | vedclass

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(B) આપેલ છે: $\vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = 2\hat{i} - \hat{j} + 3\hat{k}, \vec{c} = \hat{i} - \hat{j}$.પ્રથમ,સદિશ ગુણાકાર શોધો:$\v

Vedclass · Accepted Answer

$(\frac{4}{5}, \frac{11}{5}, \frac{19}{5})$

Vedclass · Answer

(B) આપેલ છે: $\vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = 2\hat{i} - \hat{j} + 3\hat{k}, \vec{c} = \hat{i} - \hat{j}$.પ્રથમ,સદિશ ગુણાકાર શોધો:$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ 2 & -1 & 3 \end{vmatrix} = 4\hat{i} - \hat{j} - 3\hat{k}$.$\vec{b} 	imes \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -1 & 3 \ 1 & -1 & 0 \end{vmatrix} = 3\hat{i} + 3\hat{j} - \hat{k}$.$\vec{c} 	imes \vec{a} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 0 \ 1 & 1 & 1 \end{vmatrix} = -\hat{i} - \hat{j} + 2\hat{k}$.હવે,$6\hat{i} + 2\hat{j} + 3\hat{k} = \lambda_1(4\hat{i} - \hat{j} - 3\hat{k}) + \lambda_2(3\hat{i} + 3\hat{j} - \hat{k}) + \lambda_3(-\hat{i} - \hat{j} + 2\hat{k})$ ને સરખાવતા:$4\lambda_1 + 3\lambda_2 - \lambda_3 = 6$ $(i)$$-\lambda_1 + 3\lambda_2 - \lambda_3 = 2$ (ii)$-3\lambda_1 - \lambda_2 + 2\lambda_3 = 3$ (iii)$(i)$ માંથી (ii) બાદ કરતા: $5\lambda_1 = 4 \Rightarrow \lambda_1 = \frac{4}{5}$.$\lambda_1$ ની કિંમત (ii) અને (iii) માં મૂકતા:$3\lambda_2 - \lambda_3 = \frac{14}{5}$ (iv)$-\lambda_2 + 2\lambda_3 = \frac{27}{5}$ $(v)$(iv) ને $2$ વડે ગુણી $(v)$ માં ઉમેરતા: $5\lambda_2 = 11 \Rightarrow \lambda_2 = \frac{11}{5}$.(iv) પરથી: $\lambda_3 = \frac{19}{5}$.આમ,$(\lambda_1, \lambda_2, \lambda_3) = (\frac{4}{5}, \frac{11}{5}, \frac{19}{5})$.

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બિંદુઓ $(1, -1, 2), (2, 0, -1)$ અને $(0, 2, 1)$ માંથી પસાર થતા સમતલને લંબ એકમ સદિશ શોધો.

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