Questions: Consider the line L(t)=⟨1-3t, 2-t,-5t⟩. Then: L is neither to the plane 2y-5x-4z=2 L is to the plane 6x+2y+10z=20 L is to the plane 6z-(6x+12y)=0 L is neither to the plane 4.5x+1.5y+7.5z=-30

Consider the line L(t)=⟨1-3t, 2-t,-5t⟩. Then:
L is neither to the plane 2y-5x-4z=2
L is to the plane 6x+2y+10z=20
L is to the plane 6z-(6x+12y)=0
L is neither to the plane 4.5x+1.5y+7.5z=-30

Transcript text: Consider the line $L(t)=\langle 1-3 t, 2-t,-5 t\rangle$. Then: $L$ is neither $\square$ to the plane $2 y-5 x-4 z=2$ $L$ is $\square$ to the plane $6 x+2 y+10 z=20$ $L$ is $\square$ to the plane $6 z-(6 x+12 y)=0$ $L$ is $\square$ neither to the plane $4.5 x+1.5 y+7.5 z=-30$

Solution

Solution Steps

To determine the relationship between the line $ L(t) = \langle 1-3t, 2-t, -5t \rangle $ and the given planes, we need to check if the line is parallel or perpendicular to each plane. The direction vector of the line is $\langle -3, -1, -5 \rangle$. For a line to be parallel to a plane, its direction vector should be orthogonal to the plane's normal vector. For a line to be perpendicular to a plane, its direction vector should be parallel to the plane's normal vector.

Plane 1: $2y - 5x - 4z = 2$
- Normal vector: $\langle -5, 2, -4 \rangle$
- Check if the direction vector is orthogonal to the normal vector.
Plane 2: $6x + 2y + 10z = 20$
- Normal vector: $\langle 6, 2, 10 \rangle$
- Check if the direction vector is parallel to the normal vector.
Plane 3: $6z - (6x + 12y) = 0$
- Normal vector: $\langle -6, -12, 6 \rangle$
- Check if the direction vector is parallel to the normal vector.

Step 1: Determine the Direction Vector of the Line

The line is given by $ L(t) = \langle 1-3t, 2-t, -5t \rangle $. The direction vector of the line is obtained from the coefficients of $ t $, which is $\langle -3, -1, -5 \rangle$.

Step 2: Identify the Normal Vectors of the Planes

For each plane, the normal vector is derived from the coefficients of $ x $, $ y $, and $ z $ in the plane equation:

Plane 1: $2y - 5x - 4z = 2$ has a normal vector $\langle -5, 2, -4 \rangle$.
Plane 2: $6x + 2y + 10z = 20$ has a normal vector $\langle 6, 2, 10 \rangle$.
Plane 3: $6z - (6x + 12y) = 0$ simplifies to $6x + 12y - 6z = 0$ with a normal vector $\langle -6, -12, 6 \rangle$.

Step 3: Check Orthogonality and Parallelism

Plane 1: The dot product of the direction vector $\langle -3, -1, -5 \rangle$ and the normal vector $\langle -5, 2, -4 \rangle$ is not zero, and the cross product is not zero, indicating the line is neither parallel nor perpendicular to the plane.
Plane 2: The cross product of the direction vector $\langle -3, -1, -5 \rangle$ and the normal vector $\langle 6, 2, 10 \rangle$ is zero, indicating the line is perpendicular to the plane.
Plane 3: The cross product of the direction vector $\langle -3, -1, -5 \rangle$ and the normal vector $\langle -6, -12, 6 \rangle$ is zero, indicating the line is parallel to the plane.