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.

Answer 1

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.

Answer 2

how do you find the direction vectors?

Answer 3

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).