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

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(D) The given lines are $L_1: \frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5}$ and $L_2: \frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{2}$.Two lines are coplanar a

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do not intersect.

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(D) The given lines are $L_1: \frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5}$ and $L_2: \frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{2}$.Two lines are coplanar and intersect if the shortest distance between them is $0$.The condition for intersection is given by the determinant of the vector connecting points on the lines and the direction vectors of the lines being $0$.Let $(x_1, y_1, z_1) = (1, -1, 1)$ and $(x_2, y_2, z_2) = (-2, 1, -1)$.The direction vectors are $(a_1, b_1, c_1) = (3, 2, 5)$ and $(a_2, b_2, c_2) = (4, 3, 2)$.We calculate the determinant:$\Delta = \begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \ a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \end{vmatrix} = \begin{vmatrix} -3 & 2 & -2 \ 3 & 2 & 5 \ 4 & 3 & 2 \end{vmatrix}$$\Delta = -3(2 	imes 2 - 5 	imes 3) - 2(3 	imes 2 - 5 	imes 4) - 2(3 	imes 3 - 2 	imes 4)$$\Delta = -3(4 - 15) - 2(6 - 20) - 2(9 - 8)$$\Delta = -3(-11) - 2(-14) - 2(1)$$\Delta = 33 + 28 - 2 = 59$.Since $\Delta 
eq 0$,the lines are skew and 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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