ધારો કે ચતુષ્ફલક ABCD ના શિરોબિંદુઓ A, B અને | vedclass

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(D) સ્થાન સદિશો $\vec{A} = \hat{i}+2\hat{j}+\hat{k}$,$\vec{B} = \hat{i}+3\hat{j}-2\hat{k}$,અને $\vec{C} = 2\hat{i}+\hat{j}-\hat{k}$ છે.$\vec{AB} = \ve

Vedclass · Accepted Answer

$\frac{1}{6}(7\hat{i}+12\hat{j}+\hat{k})$

Vedclass · Answer

(D) સ્થાન સદિશો $\vec{A} = \hat{i}+2\hat{j}+\hat{k}$,$\vec{B} = \hat{i}+3\hat{j}-2\hat{k}$,અને $\vec{C} = 2\hat{i}+\hat{j}-\hat{k}$ છે.$\vec{AB} = \vec{B} - \vec{A} = 0\hat{i}+\hat{j}-3\hat{k}$ અને $\vec{AC} = \vec{C} - \vec{A} = \hat{i}-\hat{j}-2\hat{k}$.$\vec{AB} 	imes \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & 1 & -3 \ 1 & -1 & -2 \end{vmatrix} = -5\hat{i}-3\hat{j}-\hat{k}$.$	riangle ABC$ નું ક્ષેત્રફળ = $\frac{1}{2}|\vec{AB} 	imes \vec{AC}| = \frac{\sqrt{35}}{2}$.ચતુષ્ફલકનું ઘનફળ = $\frac{1}{3} 	imes 	ext{Area}(ABC) 	imes h = \frac{\sqrt{805}}{6\sqrt{2}}$.$\frac{1}{3} 	imes \frac{\sqrt{35}}{2} 	imes h = \frac{\sqrt{805}}{6\sqrt{2}} \implies h = \sqrt{\frac{23}{2}}$.$	riangle ADE$ માં,$AD^2 = AE^2 + DE^2$. $AD = \frac{\sqrt{110}}{3} \implies AD^2 = \frac{110}{9}$.$AE^2 = \frac{110}{9} - \frac{23}{2} = \frac{13}{18}$.$F$ એ $BC$ નું મધ્યબિંદુ છે,$F = \frac{3\hat{i}+4\hat{j}-3\hat{k}}{2}$.$\vec{AF} = F-A = \frac{1}{2}(\hat{i}-5\hat{k})$.$\vec{AE} = AE \cdot \frac{\vec{AF}}{|AF|} = \sqrt{\frac{13}{18}} \cdot \frac{\hat{i}-5\hat{k}}{\sqrt{26}} = \frac{\hat{i}-5\hat{k}}{6}$.$E$ નો સ્થાન સદિશ = $\vec{A} + \vec{AE} = (\hat{i}+2\hat{j}+\hat{k}) + \frac{1}{6}(\hat{i}-5\hat{k}) = \frac{7\hat{i}+12\hat{j}+\hat{k}}{6}$.

Similar Questions

જો $a = t^2 \hat{i} + e^t \hat{j} + \hat{k}$ અને $b = 2 \hat{i} + t^2 \hat{j} + \log t \hat{k}$,અને $f(t) = a \cdot b$ હોય,તો $f^{\prime}(1)$ ની કિંમત શોધો.

જો $a \perp b$ અને $(a+b) \perp (a+mb)$ હોય,તો $m$ ની કિંમત શોધો.

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