The lines frac{x-1}{3}=frac{y+1}{2}=frac{z-1} | vedclass

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(B) To check if the lines intersect,we calculate the shortest distance ($S$.$D$.) between them. The lines are given by $\vec{r_1} = (1, -1, 1) + \lamb

Vedclass · Accepted Answer

do not intersect.

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(B) To check if the lines intersect,we calculate the shortest distance ($S$.$D$.) between them. The lines are given by $\vec{r_1} = (1, -1, 1) + \lambda(3, 2, 5)$ and $\vec{r_2} = (-2, 1, -1) + \mu(4, 3, -2)$.The shortest distance is given by the formula $S.D. = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} 	imes \vec{b_2})}{|\vec{b_1} 	imes \vec{b_2}|} ight|$.Here,$\vec{a_2} - \vec{a_1} = (-2-1, 1-(-1), -1-1) = (-3, 2, -2)$.The cross product $\vec{b_1} 	imes \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 2 & 5 \ 4 & 3 & -2 \end{vmatrix} = \hat{i}(-4-15) - \hat{j}(-6-20) + \hat{k}(9-8) = -19\hat{i} + 26\hat{j} + 1\hat{k}$.Now,$(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} 	imes \vec{b_2}) = (-3)(-19) + (2)(26) + (-2)(1) = 57 + 52 - 2 = 107$.Since the shortest distance is $\frac{107}{\sqrt{(-19)^2 + 26^2 + 1^2}} = \frac{107}{\sqrt{361 + 676 + 1}} = \frac{107}{\sqrt{1038}} 
eq 0$,the lines do not intersect.

The lines $\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5}$ and $\frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{-2}$:

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