For the following, determine points of intersection between...

For the following, determine points of intersection between the given line and plane, if any exist:
a. $L: \vec{r}=(-1,3,4)+p(6,1,-2), p \in \mathbf{R} ; \pi: x+2 y-z+29=0$
b. $L: \frac{x-1}{4}=\frac{y+2}{-1}=z-3 ; \pi: 2 x+7 y+z+15=0$

Key Concepts

Parametric Representation of a Line

This concept involves expressing a line using a point and a direction vector. The line is typically represented as r = r? + p · d, where r? is a fixed point on the line, d is the direction vector, and p is a scalar parameter. This form allows one to express the coordinates of any point along the line as functions of the parameter.

Plane Equation in Standard Form

A plane in three-dimensional space is commonly given in the form Ax + By + Cz + D = 0, where A, B, and C are the components of the normal vector to the plane, and D is the constant term. This representation is essential for determining the orientation of the plane and for checking if points or lines lie within or intersect the plane.

Intersection of a Line and a Plane

Finding the intersection between a line and a plane involves substituting the parametric expressions of the line into the plane's equation. This substitution yields an equation in the scalar parameter. Solving this equation shows whether there is an intersection (resulting in a single point), no intersection (if the line is parallel and separate from the plane), or infinitely many intersections (if the line lies entirely in the plane).

Solving Systems of Equations through Substitution

This key concept involves replacing the variables in one equation with their corresponding expressions from another. In the context of finding intersections between geometric entities, substitution transforms the problem into a simpler equation involving only one variable, enabling the determination of specific parameter values that lead to potential intersection points.

Transcript

00:01 So here we are given two pairs of lines and planes, and we are asked to find points of intersection if any exist.

00:10 So for the first one, we can take our expression for this line, and if we rewrite it parametrically, we have the equations x equals negative 1 plus 6p, y equals 3 plus p, and z equals 4, minus 2 ,000.

00:34 And then we can take these equations and substitute them into the plane equation.

00:39 So doing that, we would get negative 1 plus 6p plus 2 times 3 plus p minus 4 minus 2 p plus 29 equals 0.

00:55 Negative 1 plus 6p plus 6 plus 2p minus 4 plus 2p plus 29 equals 0 and we have 10p plus 5 minus 4 is plus 1 plus 30 equals 0 which gives us p equals negative 3 so at the when p equals negative 3 we have equality which means that the point on the line at negative 3 is negative 1 plus negative 18 or negative 19 y is 0 and z is 4 plus 6 which is 10 if we want to double check to see that this is on the plane we have negative 19 minus 10 plus 29 is in fact zero so part b now we see we aren't given our lines parametrically, but what we can do is, since all of these are equal to each other, this is also equal to some parameter that we can call t.

02:21 And then we can solve each of these equations for x, y, and z as a function of t to get our line parametrically.

02:30 So the first one, we have x equals 1 plus 4t.

02:36 We have y equals negative t, negative t minus 2, and z equals t plus 3.

02:48 So now taking those, substituting them into the plane equation, 2 times 1 plus 4t, plus 7 times negative t minus 2, plus t plus t plus 3, plus 15.

03:06 2 plus 8 t minus 70 minus 14 plus t plus 3 plus 15 equals 0 so 18 minus 70 plus t is 2 t then 2 minus 14 negative 12 plus 18 is 6 so again you see we get t equals negative 3 and plugging negative 3 into this line negative 3 minus 1, negative 4 over 4 is negative 1...

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