Question

Let $P_1$ and $P_2$ be two planes given by $ P_1: 10 x+15 y+12 z-60=0,$ $ P_2:-2 x+5 y+4 z-20=0 $ Which of the following straight lines can be an edge of some tetrahedron whose two faces lie on $P_1$ and $P_2$ ?

• A. $\frac{x-1}{0}=\frac{y-1}{0}=\frac{z-1}{5}$
• B. $\frac{x-6}{-5}=\frac{y}{2}=\frac{z}{3}$
• C. $\frac{x}{-2}=\frac{y-4}{5}=\frac{z}{4}$
• D. $\frac{x}{1}=\frac{y-4}{-2}=\frac{z}{3}$

Answer 1

The Correct Option is A, B, D

Step 1: Understanding the Planes

The equation of a plane is given by:

\( ax + by + cz + d = 0 \), where \( a, b, c \) are the coefficients of the normal vector to the plane.

For plane \( P_1 \):

We have the equation \( P_1: 10x + 15y + 12z - 60 = 0 \). The normal vector to this plane is:

\( \mathbf{n_1} = (10, 15, 12) \).

For plane \( P_2 \):

We have the equation \( P_2: -2x + 5y + 4z - 20 = 0 \). The normal vector to this plane is:

\( \mathbf{n_2} = (-2, 5, 4) \).

Step 2: Finding the Line of Intersection

The line of intersection of two planes is given by the cross product of their normal vectors.

Let’s find the cross product of \( \mathbf{n_1} \) and \( \mathbf{n_2} \):

\( \mathbf{n_1} = (10, 15, 12) \), \( \mathbf{n_2} = (-2, 5, 4) \)

We compute the cross product \( \mathbf{n_1} \times \mathbf{n_2} \) as follows:

\[ \mathbf{n_1} \times \mathbf{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 10 & 15 & 12 \\ -2 & 5 & 4 \end{vmatrix} \]

Expanding the determinant:

\[ \mathbf{n_1} \times \mathbf{n_2} = \hat{i}(15 \times 4 - 12 \times 5) - \hat{j}(10 \times 4 - 12 \times (-2)) + \hat{k}(10 \times 5 - 15 \times (-2)) \]

\(= \hat{i}(60 - 60) - \hat{j}(40 + 24) + \hat{k}(50 + 30)\)

\(= \hat{i}(0) - \hat{j}(64) + \hat{k}(80)\)

Thus, the direction vector of the line of intersection is:

\( \mathbf{d} = (0, -64, 80) \).

Step 3: Equation of the Line

The line of intersection can be written in parametric form as: \[ \frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c} \] where \( (x_0, y_0, z_0) \) is a point on the line, and \( (a, b, c) \) is the direction vector. Here, the direction vector is \( (0, -64, 80) \). The parametric equations of the line become: \[ \frac{x - x_0}{0} = \frac{y - y_0}{-64} = \frac{z - z_0}{80} \] This gives the line equations as: \[ x = x_0, \quad y = -64t + y_0, \quad z = 80t + z_0 \] where \( t \) is a parameter. This corresponds to the straight line that could be an edge of the tetrahedron formed by the intersection of the two planes.

Step 4: Matching with Given Options

From the given options, we check the direction vectors and equations:

Option A:

\[ \frac{x - 1}{0} = \frac{y - 1}{-1} = \frac{z - 1}{1} \] This matches the form we derived, and so it is correct.

Option B:

\[ \frac{x + 6}{5} = \frac{y}{3} \] This doesn't match the derived equations, so it’s not correct.

Option C:

\[ \frac{x - 2}{4} = \frac{y - 4}{5} = \frac{z}{3} \] This matches the form, and so it is correct.

Option D:

\[ \frac{x + 4}{2} = \frac{z - 3}{3} \] This matches the form, and so it is correct.

Final Answer:

The correct options are: A, C, D.