The lines frac{x-3}{1}=frac{y-2}{1}=frac{z-5} | 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 $\frac{x-x_2}{a_2} = \frac{y-y_2}{b_2} = \frac{z-z_2}{c_2}$ are coplanar

Vedclass · Accepted Answer

$1, 2$

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

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

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