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

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(A) Two lines $\frac{x - x_1}{a_1} = \frac{y - y_1}{b_1} = \frac{z - z_1}{c_1}$ and $\frac{x - x_2}{a_2} = \frac{y - y_2}{b_2} = \frac{z - z_2}{c_2}$

Vedclass · Accepted Answer

$k = 0, -3$

Vedclass · Answer

(A) Two lines $\frac{x - x_1}{a_1} = \frac{y - y_1}{b_1} = \frac{z - z_1}{c_1}$ and $\frac{x - x_2}{a_2} = \frac{y - y_2}{b_2} = \frac{z - z_2}{c_2}$ are coplanar if the determinant $\begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \ a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \end{vmatrix} = 0$.Here, $(x_1, y_1, z_1) = (2, 3, 4)$, $(a_1, b_1, c_1) = (1, 1, -k)$, $(x_2, y_2, z_2) = (1, 4, 5)$, and $(a_2, b_2, c_2) = (k, 2, 1)$.The condition becomes $\begin{vmatrix} 1-2 & 4-3 & 5-4 \ 1 & 1 & -k \ k & 2 & 1 \end{vmatrix} = 0$.$\begin{vmatrix} -1 & 1 & 1 \ 1 & 1 & -k \ k & 2 & 1 \end{vmatrix} = 0$.Expanding along the first row: $-1(1 - (-2k)) - 1(1 - (-k^2)) + 1(2 - k) = 0$.$-1(1 + 2k) - 1(1 + k^2) + 2 - k = 0$.$-1 - 2k - 1 - k^2 + 2 - k = 0$.$-k^2 - 3k = 0$.$k(k + 3) = 0$.Thus, $k = 0$ or $k = -3$.

The lines $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{-k}$ and $\frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1}$ are coplanar if

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