Is there a pt of intersection between these two vector equations? if so how do you find it?

Question

r=(5-4t)i+(-1+2t)j+(3+4t)k
v=(2+4s)i+(3-s)j+(-5+8s)k

i gave up trying to find the answer

Answer 1

The two 3D vectors are given parametrically with i, j, k values (if you like, you can think of these as x, y, and z). What if you had two intersecting lines on a coordinate plane? You would set the two equations equal, solve for x, and then solve for y.

In this case, you can't set the equations equal exactly, but you can set each variable equal to the other, because at a common point of intersection they will be! So:

(1) 5 - 4t = 2 + 4s
(2) -1 + 2t = 3 - s
(3) 3 + 4t = -5 + 8s

You have 2 variables and 3 equations, so pick the easiest two and solve for t and s. I'll choose to add (3) to (1).

5 - 4t = 2 + 4s
3 + 4t = -5 + 8s
--------------------
8 + 0t = -3 + 12s
12s = 11
s = 11/12
t = -1/6

Substitute these values into BOTH lines (just to be sure) to find the coordinate of intersection.
in r: (17/3, -4/3, 7/3)
in v: (17/3, 25/12, 7/3)

Although i and k match (as they should, since we used those equations), the j value does not. Therefore, there is no solution and the vectors do NOT intersect.

Answer 2

Find the point of intersection between the two lines if it exists or show that the lines do not intersect.

r = (5-4t)i + (-1+2t)j + (3+4t)k
v = (2+4s)i + (3-s)j + (-5+8s)k

The coordinates of the two lines are equal at the point of intersection. So set each variable x, y, and z equal separately and solve for s and t.

x: 5 - 4t = 2 + 4s
y: -1 + 2t = 3 - s
z: 3 + 4t = -5 + 8s

Solve for s and t.

x + z: 8 = -3 + 12s
11 = 12s
s = 11/12

Plug back into the equation for z to solve for t.

z: 3 + 4t = -5 + 8s
4t = -8 + 8s = -8 + 8(11/12) = -8/12 = -2/3
t = -1/6

Plug these values into y to verify.

y: -1 + 2t = 3 - s
-1 + 2(-1/6) = 3 - 11/12
-1 - 1/3 = (36 - 11)/12
-4/3 ≠ 25/12

This is an inconsistency so there is no solution and no intersection. The lines are skew.

Answer 3

Its incredibly elementary. the 1st step, get the two equations to you are able to cancel out between the variables ie. x or y; So... 2(5x - 2y = -sixteen) 5(-2x + y = 7) ______________ 10x - 4y = -32 -10x + 5y = 35 observe: If the two coefficients of the variable are the comparable ie. 5 and six or -7 and -9 then between the the coefficients will could desire to be negated. ie. -5 and six or 7 and -9. Step 2. upload the two equations mutually. So.... 10x - 4y = -32 + -10x + 5y = 35 _______________ y= 3 observe if no longer each and every thing occurred incredibly ordinary ie. 3y = 15, in simple terms remedy as a classic equation. As in get y through itself. ie. y =5 because of the fact 15 divided through 3 equals 5. Step 3. using the respond you gained up above, placed back into on of the unique equation to remedy for the different variable. So.... -2x + y = 7 is -2x + (5) = 7 Then remedy: 7-5 = 2 -2 = 2 x= -a million