The number of distinct real values of lambda | vedclass

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(C) 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

$3$

Vedclass · Answer

(C) 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 and only if the determinant of the matrix formed by the difference of their points and their direction vectors is zero:$\begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \ a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \end{vmatrix} = 0$Given the lines:Line $1$: $(x_1, y_1, z_1) = (1, 2, -3)$ and $(a_1, b_1, c_1) = (1, 2, \lambda^2)$Line $2$: $(x_2, y_2, z_2) = (3, 2, 1)$ and $(a_2, b_2, c_2) = (1, \lambda^2, 2)$Substituting these into the determinant condition:$\begin{vmatrix} 3 - 1 & 2 - 2 & 1 - (-3) \ 1 & 2 & \lambda^2 \ 1 & \lambda^2 & 2 \end{vmatrix} = 0$$\begin{vmatrix} 2 & 0 & 4 \ 1 & 2 & \lambda^2 \ 1 & \lambda^2 & 2 \end{vmatrix} = 0$Expanding along the first row:$2(4 - \lambda^4) - 0 + 4(\lambda^2 - 2) = 0$$8 - 2\lambda^4 + 4\lambda^2 - 8 = 0$$-2\lambda^4 + 4\lambda^2 = 0$$-2\lambda^2(\lambda^2 - 2) = 0$This gives $\lambda^2 = 0$ or $\lambda^2 = 2$.Thus,$\lambda = 0, \sqrt{2}, -\sqrt{2}$.There are $3$ distinct real values of $\lambda$.

The number of distinct real values of $\lambda$ for which the lines $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z + 3}{\lambda^2}$ and $\frac{x - 3}{1} = \frac{y - 2}{\lambda^2} = \frac{z - 1}{2}$ are coplanar is

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