Which plane goes through the origin and is perpendicular to the line r=(2,-2,1) + s(2,3,-4), seR?

a) 2x-2y+z=0
b) 2x+3y-4z=0
c) 2x+3y+z-4=0

d) none of the above.

I got D, none of the above. I have substituted and couldn't find the answer equal to zero.

Well, I hate to burst your bubble, but it looks like you've got the wrong answer.

The correct answer is actually (a) 2x - 2y + z = 0. Let me clown-explain why!

The equation of the line is r = (2, -2, 1) + s(2, 3, -4). To find a plane perpendicular to this line, we need a normal vector of the plane that is perpendicular to the direction vector of the line.

The direction vector of the line is (2, 3, -4). To find a normal vector to the line, we take the cross product of this vector with any other vector not parallel to the line. So, let's take the cross product with (1, 1, 1), just for fun.

(2, 3, -4) cross (1, 1, 1) = (7, -6, -1)

So, the normal vector of the plane is (7, -6, -1). Now, we just need to find the equation of the plane.

A plane is determined by a point on the plane (the origin in this case) and a normal vector to the plane. The equation of the plane is therefore:

7x - 6y - z = 0

But wait, this isn't one of the choices! Don't panic, we can rearrange this equation to match one of the given choices.

Rearranging, we get:

2x - 2y + z = 0

Ta-da! Option (a) is the correct answer. So, hold on to your clown nose next time and choose wisely!

To find the plane that goes through the origin and is perpendicular to the given line, we need to find the normal vector of the plane.

The line r = (2, -2, 1) + s(2, 3, -4) consists of direction vector (2, 3, -4). The normal vector of the plane perpendicular to this line would be a vector that is perpendicular to the direction vector.

We can find the normal vector by taking the cross product of the direction vector with any vector not parallel to it. We'll choose the vector (1, 0 , 0) as it is not parallel to the direction vector. Taking the cross product:

(2, 3, -4) x (1, 0, 0) = (-3, -4, -3).

So, the normal vector of the plane is (-3, -4, -3).

Now, to find the equation of the plane passing through the origin and with this normal vector, we can substitute the values into the general equation of a plane:

Ax + By + Cz = 0.

Substituting the values, we have:

-3x - 4y - 3z = 0.

Therefore, the correct equation of the plane that goes through the origin and is perpendicular to the given line is:

-3x - 4y - 3z = 0.

So, the answer is d) none of the above.

To determine which plane goes through the origin and is perpendicular to the given line, we can make use of the concept that a plane is perpendicular to a line if and only if the direction vector of the line is orthogonal (perpendicular) to the normal vector of the plane.

So, let's first find the direction vector of the given line r = (2, -2, 1) + s(2, 3, -4). The vector (2, 3, -4) is the direction vector of the line.

Now, we need to find the normal vector of the plane. Since the plane goes through the origin, any vector lying on the plane can serve as a normal vector. So, we can choose any vector that is perpendicular to the direction vector of the line.

One way to obtain a normal vector is by taking the cross product of the direction vector of the line and any other vector. Let's choose the vector (1, 0, 0) as the second vector. So, the normal vector of the plane can be computed as follows:

(2, 3, -4) x (1, 0, 0) = (3, 4, 3)

Now, if the normal vector is perpendicular to the direction vector of the line, their dot product should be zero. Let's check if it is indeed zero:

(2, 3, -4) · (3, 4, 3) = 6 + 12 - 12 = 6

Since the dot product is not zero, it means that the plane is not perpendicular to the line. Therefore, the correct answer is option D, none of the above.

1.

The line has direction numbers (2,3,-4)

The direction numbers of the normal of plane b)
are (2,3, -4)

Since the normal is perpendicular to the plane
the given line is perpendicular to the plane

2.
Does (0,0,0 satisfy the equation of plane b) ?
LS = 2(0) + 3(0) - 4(0) = 0
RS = 0 , so YES, it does

your correct answer is b)