Questions: Find the vector equation for the line of intersection of the planes 3x-y-4z=-3 and 3x+3z=-5

Transcript text: Find the vector equation for the line of intersection of the planes $3 x-y-4 z=-3$ and $3 x+3 z=-5$

Solution

Solution Steps

To find the vector equation for the line of intersection of two planes, we need to:

Find a point that lies on both planes.
Determine the direction vector of the line, which is perpendicular to the normal vectors of both planes.

Step 1: Determine the Normal Vectors

The equations of the planes are given as:

$ 3x - y - 4z = -3 $
$ 3x + 3z = -5 $

From these equations, we can extract the normal vectors:

For the first plane, the normal vector is $ \mathbf{n_1} = \langle 3, -1, -4 \rangle $.
For the second plane, the normal vector is $ \mathbf{n_2} = \langle 3, 0, 3 \rangle $.

Step 2: Calculate the Direction Vector

The direction vector of the line of intersection can be found using the cross product of the normal vectors: \[ \mathbf{d} = \mathbf{n_1} \times \mathbf{n_2} = \langle -3, -21, 3 \rangle \]

Step 3: 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 - y = -3 $
$ 3x = -5 $

Solving these equations gives: \[ x = \frac{5}{3} - z \quad \text{and} \quad y = 2 - 7z \] Substituting $ z = 0 $ yields the point: \[ \left( \frac{5}{3}, 2, 0 \right) \]

Step 4: Write the Vector Equation

The vector equation of the line can now be expressed as: \[ \mathbf{r} = \left\langle \frac{5}{3}, 2, 0 \right\rangle + t \langle -3, -21, 3 \rangle \]

Final Answer

The vector equation for the line of intersection of the planes is: \[ \boxed{\mathbf{r} = \left\langle \frac{5}{3}, 2, 0 \right\rangle + t \langle -3, -21, 3 \rangle} \]

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