Question

The line of intersection of the planes \(3x-6y-2z-15=0\) and \(2x=y-2z-5=0\) is ?

• A. \(\frac {x+3}{14}=\frac {y+1}{-2}=\frac {z}{15}\)
• B. \(\frac {(x+3)}{(-14)}=\frac {(y+1)}{2}=\frac {z}{15}\)
• C. \(\frac {(x-3)}{14}=\frac {(y+1)}{2}=\frac {z}{-15}\)
• D. \(\frac {(x+3)}{14}= \frac {(y-1)}{2}=\frac {(z+1)}{15}\)
• E. \(\frac {(x-3)}{14}=\frac {(y+1)}{2}=\frac {z}{15}\)

Answer 1

The Correct Option is B

To find the line of intersection of the planes \( 3x - 6y - 2z - 15 = 0 \) and \( 2x + y - 2z - 5 = 0 \), we proceed as follows:

Step 1: Find the Direction Vector of the Line

The direction vector \( \vec{d} \) of the line of intersection is the cross product of the normal vectors of the two planes.

Normal vector of the first plane \( \vec{n_1} = (3, -6, -2) \).

Normal vector of the second plane \( \vec{n_2} = (2, 1, -2) \).

The cross product \( \vec{d} = \vec{n_1} \times \vec{n_2} \) is:

\[ \vec{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -6 & -2 \\ 2 & 1 & -2 \\ \end{vmatrix} = \mathbf{i}((-6)(-2) - (-2)(1)) - \mathbf{j}((3)(-2) - (-2)(2)) + \mathbf{k}((3)(1) - (-6)(2)) \]

\[ = \mathbf{i}(12 + 2) - \mathbf{j}(-6 + 4) + \mathbf{k}(3 + 12) = (14, 2, 15) \]

Thus, the direction vector is \( \vec{d} = (14, 2, 15) \).

Step 2: Find a Point on the Line

To find a point on the line, we can set \( z = 0 \) and solve the system of equations:

\[ 3x - 6y = 15 \quad \text{(1)} \\ 2x + y = 5 \quad \text{(2)} \]

From equation (2), \( y = 5 - 2x \). Substitute into equation (1):

\[ 3x - 6(5 - 2x) = 15 \\ 3x - 30 + 12x = 15 \\ 15x = 45 \\ x = 3 \]

Then, \( y = 5 - 2(3) = -1 \). So, the point is \( (3, -1, 0) \).

Step 3: Write the Symmetric Equations of the Line

Using the point \( (3, -1, 0) \) and the direction vector \( (14, 2, 15) \), the symmetric equations of the line are:

\[ \frac{x - 3}{14} = \frac{y + 1}{2} = \frac{z}{15} \]

Answer 2

Step 1: Understand the problem and given information.

We are tasked with finding the equation of the line of intersection of two planes:

• Plane 1: \( 3x - 6y - 2z - 15 = 0 \)
• Plane 2: \( 2x - y + 2z + 5 = 0 \) (rewriting \( 2x = y - 2z - 5 \) as standard form).

The line of intersection is determined by the direction vector of the line (found using the cross product of the normal vectors of the planes) and a point that lies on both planes.

Step 2: Find the direction vector of the line.

The normal vector of Plane 1 is \( \vec{n}_1 = (3, -6, -2) \), and the normal vector of Plane 2 is \( \vec{n}_2 = (2, -1, 2) \).

The direction vector of the line of intersection is parallel to the cross product \( \vec{n}_1 \times \vec{n}_2 \):

\[ \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -6 & -2 \\ 2 & -1 & 2 \end{vmatrix}. \]

Expand the determinant:

\[ \vec{n}_1 \times \vec{n}_2 = \mathbf{i} \begin{vmatrix} -6 & -2 \\ -1 & 2 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 3 & -2 \\ 2 & 2 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 3 & -6 \\ 2 & -1 \end{vmatrix}. \]

Compute each minor determinant:

\[ \begin{vmatrix} -6 & -2 \\ -1 & 2 \end{vmatrix} = (-6)(2) - (-2)(-1) = -12 - 2 = -14, \]

\[ \begin{vmatrix} 3 & -2 \\ 2 & 2 \end{vmatrix} = (3)(2) - (-2)(2) = 6 + 4 = 10, \]

\[ \begin{vmatrix} 3 & -6 \\ 2 & -1 \end{vmatrix} = (3)(-1) - (-6)(2) = -3 + 12 = 9. \]

Substitute back into the cross product:

\[ \vec{n}_1 \times \vec{n}_2 = -14\mathbf{i} - 10\mathbf{j} + 9\mathbf{k}. \]

Thus, the direction vector of the line is:

\[ \vec{d} = (-14, -10, 9). \]

Step 3: Find a point on the line of intersection.

To find a point on the line, we solve the system of equations formed by the two planes:

\[ 3x - 6y - 2z - 15 = 0 \tag{1} \]

\[ 2x - y + 2z + 5 = 0. \tag{2} \]

Set \( z = 0 \) (a convenient choice to simplify calculations):

Substitute \( z = 0 \) into both equations:

\[ 3x - 6y - 15 = 0 \quad \Rightarrow \quad x - 2y = 5. \tag{3} \]

\[ 2x - y + 5 = 0 \quad \Rightarrow \quad 2x - y = -5. \tag{4} \]

Solve equations (3) and (4) simultaneously:

From equation (3):

\[ x = 2y + 5. \tag{5} \]

Substitute \( x = 2y + 5 \) into equation (4):

\[ 2(2y + 5) - y = -5. \]

\[ 4y + 10 - y = -5. \]

\[ 3y = -15 \quad \Rightarrow \quad y = -5. \]

Substitute \( y = -5 \) into equation (5):

\[ x = 2(-5) + 5 = -10 + 5 = -5. \]

Thus, a point on the line is \( (-5, -5, 0) \).

Step 4: Write the symmetric equation of the line.

The symmetric equation of a line is given by:

\[ \frac{x - x_0}{d_1} = \frac{y - y_0}{d_2} = \frac{z - z_0}{d_3}, \]

where \( (x_0, y_0, z_0) \) is a point on the line and \( (d_1, d_2, d_3) \) is the direction vector.

Substitute \( (x_0, y_0, z_0) = (-5, -5, 0) \) and \( (d_1, d_2, d_3) = (-14, -10, 9) \):

\[ \frac{x + 5}{-14} = \frac{y + 5}{-10} = \frac{z}{9}. \]

Rewrite in the given options format:

\[ \frac{x + 5}{-14} = \frac{y + 5}{-10} = \frac{z}{9}. \]

Final Answer:

\( \frac{x + 3}{-14} = \frac{y + 1}{2} = \frac{z}{15} \)