Finding Perpendicular Vectors and Line-Plane Intersection

What is a vector that is perpendicular to the plane defined by the parametric equations given?

For the line defined by the parametric equations provided, what are the values of $x$, $y$, and $z$ when $t = 0$?

What is the equation of the plane in standard form derived from the given parametric equations?

If the line intersects the plane, what method is best used to find the point of intersection?

What does the intersection of the line with the plane imply if no solution exists?

Finding a Vector Perpendicular to a Plane

The parametric equation of a plane is given by x = 3 + 1 s + 4 t, y = 2 + 3 s − 2 t, z = − 1 + 2 s + 3 t.
To find a vector perpendicular to the plane, we need to find two vectors that lie in the plane.
We can get these vectors by setting s=1 and t=0, and then setting s=0 and t=1.
These vectors are: [1, 3, 2] and [4, -2, 3].
The cross product of these two vectors will give us a vector that is perpendicular to the plane.
The cross product is calculated as follows: [1, 3, 2] × [4, -2, 3] = [13, 5, -14].
Therefore, [13, 5, -14] is a vector that is perpendicular to the plane.

Line-Plane Intersection

The line with parametric equations x = 1 + 1 t, y = 1 − t, z = 2 + t intersects the plane with equation x + 2 y + z − 10 = 0.
To find the point of intersection, substitute the parametric equations of the line into the equation of the plane.
This gives us: (1 + 1 t) + 2(1 − t) + (2 + t) − 10 = 0.
Simplifying the equation, we get 5 − t − 10 = 0.
Solving for t, we get t = -5.
Substitute this value of t back into the parametric equations of the line to find the point of intersection.
The point of intersection is (-4, 6, -3).

This quiz covers the concepts of finding a vector perpendicular to a plane using parametric equations and the cross product. It also includes determining the intersection point of a line with a given plane. Engage with real mathematical problems to enhance your understanding of these topics.