The lines frac{x - a + d}{alpha - delta} = fr | vedclass

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(C) The condition for two lines to be coplanar is given by the determinant of the differences of their points and their direction vectors being zero:

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$x - 2y + z = 0$

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(C) The condition for two lines to be coplanar is given by the determinant of the differences of their points and their direction vectors being zero: $\begin{vmatrix} a-d-b+c & a-b & a+d-b-c \ \alpha-\delta & \alpha & \alpha+\delta \ \beta-\gamma & \beta & \beta+\gamma \end{vmatrix} = 0$ Adding the $3^{rd}$ column to the $1^{st}$ column,we get the first column as $2(a-b), 2\alpha, 2\beta$,which is twice the $2^{nd}$ column. Thus,the determinant is zero,confirming they are coplanar. The equation of the plane containing these lines is: $\begin{vmatrix} x-a+d & y-a & z-a-d \ \alpha-\delta & \alpha & \alpha+\delta \ \beta-\gamma & \beta & \beta+\gamma \end{vmatrix} = 0$ Adding the $1^{st}$ and $3^{rd}$ columns and subtracting twice the $2^{nd}$ column,we get: $\begin{vmatrix} x+z-2y & y-a & z-a-d \ 0 & \alpha & \alpha+\delta \ 0 & \beta & \beta+\gamma \end{vmatrix} = 0$ Expanding along the $1^{st}$ column: $(x+z-2y) \cdot [\alpha(\beta+\gamma) - \beta(\alpha+\delta)] = 0$ $(x+z-2y) \cdot [\alpha\beta + \alpha\gamma - \beta\alpha - \beta\delta] = 0$ Assuming the term in the bracket is non-zero,we get $x - 2y + z = 0$.

The lines $\frac{x - a + d}{\alpha - \delta} = \frac{y - a}{\alpha} = \frac{z - a - d}{\alpha + \delta}$ and $\frac{x - b + c}{\beta - \gamma} = \frac{y - b}{\beta} = \frac{z - b - c}{\beta + \gamma}$ are coplanar. Find the equation of the plane in which they lie.

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