The line defined by (x, y, z) = (2 + t, 10 − 2t, 6 − t) intersects the line defined by (x, y, z) = (9 + 2u, 2 − u, 5 + u). Find coordinates for the point of intersection. To the nearest degree, find the size of the acute angle formed by these lines.

3 answers

To find the point of intersection, we can set the x, y, and z coordinates of the two lines equal to each other and solve for t and u.

From the first line, we have:
x = 2 + t
y = 10 - 2t
z = 6 - t

From the second line, we have:
x = 9 + 2u
y = 2 - u
z = 5 + u

Setting the x coordinates equal to each other, we have:
2 + t = 9 + 2u
t = 7 + 2u

Setting the y coordinates equal to each other, we have:
10 - 2t = 2 - u
2t = u + 8
t = (u + 8)/2

Since t = 7 + 2u and t = (u + 8)/2, we can set them equal to each other:
7 + 2u = (u + 8)/2

Multiplying both sides of the equation by 2, we get rid of the fraction:
14 + 4u = u + 8

Subtracting u from both sides of the equation, we have:
4u - u = 8 - 14
3u = -6

Dividing both sides of the equation by 3, we find:
u = -2

Now that we have the value for u, we can substitute it back into one of the equations to solve for t. Let's use the first equation:
t = 7 + 2u
t = 7 + 2(-2)
t = 7 - 4
t = 3

Now we have the values for t and u. Substituting them back into any of the original equations, we find the point of intersection:
x = 2 + t
x = 2 + 3
x = 5

y = 10 - 2t
y = 10 - 2(3)
y = 10 - 6
y = 4

z = 6 - t
z = 6 - 3
z = 3

So the point of intersection is (5, 4, 3).

To find the angle between the two lines, we can find the dot product of their direction vectors and use the formula:

cosθ = dot product / (magnitude of vector 1 * magnitude of vector 2)

The direction vectors of the two lines are given by the coefficients of t and u in the parametric equations. For the first line, the direction vector is (1, -2, -1) and for the second line, the direction vector is (2, -1, 1).

The dot product of the direction vectors is:

(1)(2) + (-2)(-1) + (-1)(1) = 2 + 2 - 1 = 3

The magnitude of the first direction vector is:

sqrt(1^2 + (-2)^2 + (-1)^2) = sqrt(6)

The magnitude of the second direction vector is:

sqrt(2^2 + (-1)^2 + 1^2) = sqrt(6)

Plugging these values into the formula, we have:

cosθ = 3 / (sqrt(6) * sqrt(6)) = 3 / 6 = 0.5

To find the angle θ, we take the inverse cosine of 0.5:

θ = arccos(0.5) ≈ 60 degrees

So the size of the acute angle formed by these lines is approximately 60 degrees.
how do you find the direction vectors?
To find the direction vectors of the lines, we can look at the coefficients of t and u in the parametric equations of the lines.

For the first line, the parametric equation is:
(x, y, z) = (2 + t, 10 − 2t, 6 − t)

The coefficient of t for the x-coordinate is 1, for the y-coordinate is -2, and for the z-coordinate is -1. Therefore, the direction vector for the first line is (1, -2, -1).

For the second line, the parametric equation is:
(x, y, z) = (9 + 2u, 2 − u, 5 + u)

The coefficient of u for the x-coordinate is 2, for the y-coordinate is -1, and for the z-coordinate is 1. Therefore, the direction vector for the second line is (2, -1, 1).