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.
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).
To find the point of intersection between the two lines, we need to find values of t and u that satisfy the equations for both lines. Let's set the coordinates equal to each other and solve for t and u.
x: 2 + t = 9 + 2u
y: 10 − 2t = 2 − u
z: 6 − t = 5 + u
From the first equation, we can solve for t:
t = 9 + 2u - 2
Simplifying:
t = 2u + 7
Substituting this value of t into the third equation:
6 - (2u + 7) = 5 + u
Simplifying:
-2u - 1 = 5 + u
Re-arranging the equation:
3u = -6
u = -2
Now, substitute the value of u back into the first equation:
t = 2u + 7 = 2(-2) + 7 = 3
So, the values of t and u that satisfy both equations are t = 3 and u = -2.
To find the point of intersection, substitute these values of t and u back into either of the original equations:
(x, y, z) = (2 + t, 10 − 2t, 6 − t)
Plugging in t = 3, we get:
(x, y, z) = (2 + 3, 10 − 2(3), 6 − 3)
= (5, 4, 3)
Therefore, 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. The direction vectors for the two lines can be found by subtracting their initial points from their endpoint coordinates.
For the first line:
Direction vector = (9, 2, 5) - (2, 10, 6)
= (7, -8, -1)
For the second line:
Direction vector = (2, -1, 1) - (9, 2, 5)
= (-7, -3, -4)
Now, find the dot product of these two vectors:
Dot product = (7 * -7) + (-8 * -3) + (-1 * -4)
= 49 + 24 + 4
= 77
The magnitude of the first direction vector is:
sqrt((7^2 + (-8)^2 + (-1)^2) = sqrt(74)
The magnitude of the second direction vector is:
sqrt((-7^2 + (-3)^2 + (-4)^2) = sqrt(74)
The cosine of the acute angle formed by these lines can be found using the formula:
cos(angle) = dot product / (magnitude of first direction vector * magnitude of second direction vector)
cos(angle) = 77 / (sqrt(74) * sqrt(74))
cos(angle) = 77 / 74
angle = arccos(77/74)
Using a calculator, the angle is approximately 16 degrees. Therefore, the size of the acute angle formed by these lines is approximately 16 degrees (to the nearest degree).