Find the distance of the point (4,6,−4)(4,6,−4) from the line 𝐫(𝑡)=⟨−1+4𝑡,−1+2𝑡,3−2𝑡⟩

To find the distance from a point to a line in 3D space, we can use the formula for the distance from a point to a line, which is given by:

d = |(P - A) x (P - B)| / |B - A|

where:

• P is the point in space,
• A is a point on the line,
• B is another point on the line,
• "x" denotes the cross product, and
• "|" denotes the magnitude of a vector.

First, we need to find two points on the line. We can do this by plugging in two different values of t into the equation of the line. Let's use t = 0 and t = 1.

When t = 0, r(t) = ⟨-1, -1, 3⟩. So, A = (-1, -1, 3). When t = 1, r(t) = ⟨3, 1, 1⟩. So, B = (3, 1, 1).

Next, we need to calculate the vectors P - A and P - B. Given that P = (4, 6, -4), we find:

P - A = (4 - (-1), 6 - (-1), -4 - 3) = (5, 7, -7) P - B = (4 - 3, 6 - 1, -4 - 1) = (1, 5, -5)

Now, we calculate the cross product of these two vectors:

(P - A) x (P - B) = (7*(-5) - (-7)5, -71 - 5*(-7), 55 - 17) = (-35 + 35, -7 + 35, 25 - 7) = (0, 28, 18)

Next, we find the magnitude of this cross product:

|(P - A) x (P - B)| = sqrt(0^2 + 28^2 + 18^2) = sqrt(784 + 324) = sqrt(1108) = 33.27 (approx)

Finally, we calculate the vector B - A and find its magnitude:

B - A = (3 - (-1), 1 - (-1), 1 - 3) = (4, 2, -2) |B - A| = sqrt(4^2 + 2^2 + (-2)^2) = sqrt(16 + 4 + 4) = sqrt(24) = 4.90 (approx)

So, the distance from the point to the line is:

d = |(P - A) x (P - B)| / |B - A| = 33.27 /