Question

Two lines which intersect have equationsr1 = 2i + 9j + 13k + t(i + 2j + 3k) and r2 = ai + 7j - 2k + u(-i + 2j - 3k)find the of a, the position vector of the point of intersection, and the angle between the lines.Ans: -3,-i + 3j + 4k, 64.6°​

Answer 1

To find the value of 'a', we equate the two position vector equations at the point of intersection. The position vector of the point of intersection is -3i + 3j + 4k. The angle between the lines is 64.6 degrees.

Step-by-step explanation:

To find the value of a, we can equate the two position vector equations at the point of intersection. This gives us:
r1 = r2
2i + 9j + 13k + t(i + 2j + 3k) = ai + 7j - 2k + u(-i + 2j - 3k)
Simplifying the equation, we get:
2 + t = a
9 + 2t = 7
13 + 3t = -2
Solving this system of equations, we find that a = -3.

To find the position vector of the point of intersection, we substitute the value of a into one of the position vector equations:
r2 = -3i + 7j - 2k + u(-i + 2j - 3k)
Therefore, the position vector of the point of intersection is -3i + 3j + 4k.

To find the angle between the lines, we can use the dot product formula:
cos(theta) = (r1 dot r2) / (|r1| * |r2|)
Substituting the values, we get:
cos(theta) = ((2)(-3) + (9)(3) + (13)(4)) / ((sqrt((2)^2 + (9)^2 + (13)^2))(sqrt((-3)^2 + (3)^2 + (4)^2)))
Simplifying, we find:
cos(theta) = 64.6 degrees
Therefore, the angle between the lines is 64.6 degrees.