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

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(B) Let the two lines be $L_1: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $L_2: \frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$.The lines intersect if th

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$\frac{9}{2}$

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(B) Let the two lines be $L_1: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $L_2: \frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$.The lines intersect if the shortest distance between them is $0$. This condition is equivalent to the scalar triple product of the vector connecting a point on each line and the direction vectors of the lines being $0$.Let $P_1 = (1, -1, 1)$ be a point on $L_1$ and $P_2 = (3, k, 0)$ be a point on $L_2$.The direction vectors are $\vec{v_1} = (2, 3, 4)$ and $\vec{v_2} = (1, 2, 1)$.The condition for intersection is given by the determinant:$\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \ a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \end{vmatrix} = 0$Substituting the values:$\begin{vmatrix} 3-1 & k-(-1) & 0-1 \ 2 & 3 & 4 \ 1 & 2 & 1 \end{vmatrix} = 0$$\begin{vmatrix} 2 & k+1 & -1 \ 2 & 3 & 4 \ 1 & 2 & 1 \end{vmatrix} = 0$Expanding the determinant along the first row:$2(3(1) - 4(2)) - (k+1)(2(1) - 4(1)) - 1(2(2) - 3(1)) = 0$$2(3-8) - (k+1)(2-4) - 1(4-3) = 0$$2(-5) - (k+1)(-2) - 1(1) = 0$$-10 + 2k + 2 - 1 = 0$$2k - 9 = 0$$k = \frac{9}{2}$

If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ intersect,then the value of $k$ is

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