If the lines L_1: x = -1 + s, y = 3 - lambda | vedclass

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(C) The lines are coplanar if the determinant of the vector connecting points on the lines and the direction vectors is zero.Point $P_1$ on $L_1$ is $

Vedclass · Accepted Answer

$-2$

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(C) The lines are coplanar if the determinant of the vector connecting points on the lines and the direction vectors is zero.Point $P_1$ on $L_1$ is $(-1, 3, 1)$ and direction vector $\vec{v_1} = (1, -\lambda, \lambda)$.Point $P_2$ on $L_2$ is $(0, 1, 2)$ and direction vector $\vec{v_2} = (1/2, 1, -1)$.The vector $\vec{P_1P_2} = (0 - (-1), 1 - 3, 2 - 1) = (1, -2, 1)$.The condition for coplanarity is $\begin{vmatrix} 1 & -2 & 1 \ 1 & -\lambda & \lambda \ 1/2 & 1 & -1 \end{vmatrix} = 0$.Expanding the determinant: $1(\lambda - \lambda) - (-2)(-1 - \lambda/2) + 1(1 + \lambda/2) = 0$.$0 + 2(-1 - \lambda/2) + 1 + \lambda/2 = 0$.$-2 - \lambda + 1 + \lambda/2 = 0$.$-1 - \lambda/2 = 0 \implies \lambda/2 = -1 \implies \lambda = -2$.

If the lines $L_1: x = -1 + s, y = 3 - \lambda s, z = 1 + \lambda s$ and $L_2: x = \frac{t}{2}, y = 1 + t, z = 2 - t$ with parameters $s$ and $t$ are coplanar, then $\lambda =$ ?

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