Vector-Valued Function Question

[faslickit]

(a) Find the y-intercept of the line in 2-space that is represented by the vector equation r = (3+2t)i+5tj
(b) Find the coordinates of the points where the line r = ti + (1+2t)j - 3tk intersects the plane 3x-y-z=2



For a, I'm pretty sure it's (0,-7.5), just took (3,0) and went in the direction v=2i+5j to the y-intercept.
For b, I listed the point (0,1,0) from the coordinates of r and the vector v=i+2j-3k. After combining them with guess and check, I came up with the answer of (3/4, 10/4, -9/4).

I'm pretty sure I have the right answers, but I technically ended both with guess and check. What is the proper way to finish both problems?

 

[NBAJam100]

For part (a) think about it this way- you are looking to find the y-intercept, so looking in terms of a 2-D plane, if you are looking at a y-intercept what is your x-value? So take that value and set it equal to X and then solve for the time(t) when you are at that point. Then plug that t into your equation.

for (b)- try kind of thinking along the same lines of the other question in terms of what you can imply from the question.

 

[Mark44]

a) To y-intercept, so set 3 + 2t = 0 and solve for t. Then use that value to get the y value (5t) at the y-intercept.
b) This one is a bit more involved. The vector equation of the line is r(t) = (t, 1 + 2t, -3t). The equation of the plane is 3x - y - z = 2, and this can be put in the form of a vector equation this way:
3x = y + z + 2
or
x = 1/3 * y + 1/3 * z + 2/3
y = y
z = z

The last two equations are trivially true, and are thrown into give us a vector equation, namely
(x, y, z) = y*(1/3, 1, 0) + z*(1/3, 0, 1) + (2/3, 0, 0).

At any point of intersection, the point (x, y, z) on the line must also be on the plane, so the x-coordinates have to be equal, as do the y-coordinates and the z-coordinates.

This means that
(t, 1 + 2t, -3t) = y*(1/3, 1, 0) + z*(1/3, 0, 1) + (2/3, 0, 0), or
t = 1/3 * y + 1/3 * z + 2/3
1 + 2t = y
-3t = z

Solving the last two equation for y and z gives us y = 2/3*t + 1/3 and z = -t.
Substituting these values into the first equation just above yields t = 3/4 at the point of intersection, which gives us the point (3/4, 5/2, -9/4), which is what you got.

 

[HallsofIvy]

(a) Find the y-intercept of the line in 2-space that is represented by the vector equation r = (3+2t)i+5tj
(b) Find the coordinates of the points where the line r = ti + (1+2t)j - 3tk intersects the plane 3x-y-z=2



For a, I'm pretty sure it's (0,-7.5), just took (3,0) and went in the direction v=2i+5j to the y-intercept.

For both of these you need to know that the position vector, r, of the point (x, y, z) is always xi+ yj+ zk. I hope you knew that but didn't think to use it.

The x intercept is when x= 3+ 2t= 0. Solve for t and put that into y= 5t.

For b, I listed the point (0,1,0) from the coordinates of r and the vector v=i+2j-3k. After combining them with guess and check, I came up with the answer of (3/4, 10/4, -9/4).

I'm pretty sure I have the right answers, but I technically ended both with guess and check. What is the proper way to finish both problems?

x= t, y= 1+ 2t, z= -3t. Put those into the equation 3x- y- z= 2 to get an equation for t.