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Question

Find the intersection of the line and plane: What are the points?

2z-(3x+4y)=7, r(t)=<-3,2,-3> +t<2,-3,-1>

Answer 1

Let's start by simplifying the plane equation a little.

2z-(3x+4y)=7 -> 2z - 3x - 4y = 7

Now, we have a line with x, y, and z relative to one variable t. Let's rewrite this line equation into x, y, and z components:

r_x = -3 + 2t

r_y = 2 - 3t

r_z = -3 - t

These are our x, y, and z components of the line, written as functions of t.

Now we can plug in r_x, r_y, and r_z in for x, y, and z to find where the functions intersect:

2(r_z) - 3(r_x) -4(r_y) = 7

Using our above component equations, this becomesL

2(-3 - t) - 3(-3 + 2t) -4(2 - 3t) = 7

Now, we find what value of t satisfies this equation. Let's rearrange and solve:

-6 - 2t + 9 - 6t - 8 + 12t = 7

-5 + 4t = 7

4t = 12

t = 3

Now, all we have to do is plug t = 3 into our component functions to find the x,y, and z components of intersection:

r_x = -3 + 2(3) = -3 + 6 = 3

r_y = 2 - 3(3) = 2 - 9 = -7

r_z = -3 - (3) = -6

So, our point of intersection is <3, -7, -6> at t=3.

If you have any questions, feel free to respond!

Answer 2

To find the intersection points between the plane given by the equation 2z - (3x + 4y) = 7 and the line parameterized by r(t) = <-3, 2, -3> + t<2, -3, -1>, you'll want to substitute the values of r(t) into the equation of the plane and solve for t.

Let's first substitute the values from r(t) into the equation of the plane:

2z - (3x + 4y) = 7

2(-3 - 2t) - [3(-3) + 4(2 - 3t)] = 7

Now, simplify and solve for t:

-6 - 4t - [(-9) + (8 - 12t)] = 7

-6 - 4t - (-9 + 8 - 12t) = 7

-6 - 4t + 9 - 8 + 12t = 7

Now, combine like terms:

3 + 8t = 7

Subtract 3 from both sides:

8t = 4

Now, divide by 8:

t = 4 / 8

t = 1/2

So, the value of t is 1/2. Now, plug this value back into the parameterization of the line to find the point of intersection:

r(1/2) = <-3, 2, -3> + (1/2)<2, -3, -1>

r(1/2) = <-3, 2, -3> + <1, -3/2, -1/2>

Now, add the corresponding components:

x = -3 + 1 = -2

y = 2 - (3/2) = 1/2

z = -3 - (1/2) = -7/2

So, the intersection point of the line and the plane is (-2, 1/2, -7/2).