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

$\frac{11}{2}$

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,$(x_1, y_1, z_1) = (-1, 1, -1)$ and $(x_2, y_2, z_2) = (-2, k, 0)$. The direction ratios are $(a_1, b_1, c_1) = (2, 1, 3)$ and $(a_2, b_2, c_2) = (2, 3, 4)$. Substituting these values into the determinant: $\begin{vmatrix} -2 - (-1) & k - 1 & 0 - (-1) \ 2 & 1 & 3 \ 2 & 3 & 4 \end{vmatrix} = 0$ $\Rightarrow \begin{vmatrix} -1 & k - 1 & 1 \ 2 & 1 & 3 \ 2 & 3 & 4 \end{vmatrix} = 0$ Expanding along the first row: $-1(4 - 9) - (k - 1)(8 - 6) + 1(6 - 2) = 0$ $-1(-5) - (k - 1)(2) + 4 = 0$ $5 - 2k + 2 + 4 = 0$ $11 - 2k = 0$ $2k = 11$ $k = \frac{11}{2}$.

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

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