રેખાઓ overline{r}=(2 hat{i}-hat{j})+lambda(2 | vedclass

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(A) આપેલ રેખાઓ $\overline{r}=\overline{a_1}+\lambda\overline{b_1}$ અને $\overline{r}=\overline{a_2}+\mu\overline{b_2}$ છે.અહીં,$\overline{a_1}=2\hat{i

Vedclass · Accepted Answer

$\frac{1}{\sqrt{5}}$ એકમ

Vedclass · Answer

(A) આપેલ રેખાઓ $\overline{r}=\overline{a_1}+\lambda\overline{b_1}$ અને $\overline{r}=\overline{a_2}+\mu\overline{b_2}$ છે.અહીં,$\overline{a_1}=2\hat{i}-\hat{j}$,$\overline{b_1}=2\hat{i}+\hat{j}-3\hat{k}$ અને $\overline{a_2}=\hat{i}-\hat{j}+2\hat{k}$,$\overline{b_2}=2\hat{i}+\hat{j}-5\hat{k}$ છે.સદિશ $\overline{a_2}-\overline{a_1} = (\hat{i}-\hat{j}+2\hat{k})-(2\hat{i}-\hat{j}) = -\hat{i}+2\hat{k}$ થાય.ક્રોસ પ્રોડક્ટ $\overline{b_1} 	imes \overline{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -3 \ 2 & 1 & -5 \end{vmatrix} = \hat{i}(-5+3) - \hat{j}(-10+6) + \hat{k}(2-2) = -2\hat{i}+4\hat{j}$ મળે.તેનું માન $|\overline{b_1} 	imes \overline{b_2}| = \sqrt{(-2)^2+4^2+0^2} = \sqrt{4+16} = \sqrt{20} = 2\sqrt{5}$ થાય.લઘુત્તમ અંતર $d = \left| \frac{(\overline{a_2}-\overline{a_1}) \cdot (\overline{b_1} 	imes \overline{b_2})}{|\overline{b_1} 	imes \overline{b_2}|} ight|$ છે.$d = \left| \frac{(-\hat{i}+2\hat{k}) \cdot (-2\hat{i}+4\hat{j})}{2\sqrt{5}} ight| = \left| \frac{2+0+0}{2\sqrt{5}} ight| = \frac{2}{2\sqrt{5}} = \frac{1}{\sqrt{5}}$ એકમ.

રેખાઓ $\overline{r}=(2 \hat{i}-\hat{j})+\lambda(2 \hat{i}+\hat{j}-3 \hat{k})$ અને $\overline{r}=(\hat{i}-\hat{j}+2 \hat{k})+\mu(2 \hat{i}+\hat{j}-5 \hat{k})$ વચ્ચેનું લઘુત્તમ અંતર શોધો.

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