Question

The equation of the plane containing two lines $\frac{x+1}{3}=\frac{y+3}{3}=\frac{z+5}{7}$ and $\frac{x-1}{1}=\frac{y+3}{3}=\frac{z+7}{\lambda}$ is (where $\lambda \in R-\{0\}$)

• A. 3x - 10y + 3z - 12 = 0
• B. 3x + 10y + 3z + 12 = 0
• C. 3x - 10y - 3z - 12 = 0
• D. 3x + 10y - 3z + 12 = 0

Answer 1

Answer: (A)

3x - 10y + 3z - 12 = 0

Answer 2

To find the equation of the plane containing two lines, we first identify a point on each line and their direction vectors.

For the first line, $L_1: \frac{x+1}{3}=\frac{y+3}{3}=\frac{z+5}{7}$, a point $P_1$ is $(-1, -3, -5)$ and its direction vector $\vec{v_1}$ is $\langle 3, 3, 7 \rangle$.

For the second line, $L_2: \frac{x-1}{1}=\frac{y+3}{3}=\frac{z+7}{\lambda}$, a point $P_2$ is $(1, -3, -7)$ and its direction vector $\vec{v_2}$ is $\langle 1, 3, \lambda \rangle$.

For the plane to contain both lines, the lines must be coplanar. Since the direction vectors are not proportional, the lines must intersect. The condition for coplanarity is that the scalar triple product of the vector joining a point on each line and their direction vectors is zero: $[\vec{P_1P_2}, \vec{v_1}, \vec{v_2}] = 0$.

The vector $\vec{P_1P_2} = P_2 - P_1 = \langle 1 - (-1), -3 - (-3), -7 - (-5) \rangle = \langle 2, 0, -2 \rangle$.

Now, we compute the scalar triple product using a determinant: $\begin{vmatrix} 2 & 0 & -2 \\ 3 & 3 & 7 \\ 1 & 3 & \lambda \end{vmatrix} = 0$ Expanding the determinant: $2(3\lambda - 7 \times 3) - 0(3\lambda - 7 \times 1) + (-2)(3 \times 3 - 3 \times 1) = 0$ $2(3\lambda - 21) - 0 - 2(9 - 3) = 0$ $6\lambda - 42 - 2(6) = 0$ $6\lambda - 42 - 12 = 0$ $6\lambda - 54 = 0$ $6\lambda = 54 \implies \lambda = 9$. This value $\lambda = 9$ satisfies the condition $\lambda \in R-\{0\}$.

Now, with $\lambda=9$, the direction vector for $L_2$ is $\vec{v_2} = \langle 1, 3, 9 \rangle$. The equation of the plane can be found using the determinant form with a point on the plane (e.g., $P_1$) and the two direction vectors: $\begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$ Using $P_1(-1, -3, -5)$, $\vec{v_1} = \langle 3, 3, 7 \rangle$, and $\vec{v_2} = \langle 1, 3, 9 \rangle$: $\begin{vmatrix} x-(-1) & y-(-3) & z-(-5) \\ 3 & 3 & 7 \\ 1 & 3 & 9 \end{vmatrix} = 0$ $\begin{vmatrix} x+1 & y+3 & z+5 \\ 3 & 3 & 7 \\ 1 & 3 & 9 \end{vmatrix} = 0$ Expanding this determinant: $(x+1)(3 \times 9 - 7 \times 3) - (y+3)(3 \times 9 - 7 \times 1) + (z+5)(3 \times 3 - 3 \times 1) = 0$ $(x+1)(27 - 21) - (y+3)(27 - 7) + (z+5)(9 - 3) = 0$ $(x+1)(6) - (y+3)(20) + (z+5)(6) = 0$ Divide by 2: $3(x+1) - 10(y+3) + 3(z+5) = 0$ $3x + 3 - 10y - 30 + 3z + 15 = 0$ $3x - 10y + 3z - 12 = 0$.