Let y =
9
5
1
and u =
-7
2
3
Compute the distance d from y to the line through u and the origin. just answer

The distance \( d \) from \( y \) to the line through \( u \) and the origin is:

\[
d = \frac{|y \cdot u|}{\|u\|}
\]

Let's compute it.

First, find the dot product \( y \cdot u \):

\[
y \cdot u = (9)(-7) + (5)(2) + (1)(3) = -63 + 10 + 3 = -50
\]

Now, find the magnitude of \( u \):

\[
\|u\| = \sqrt{(-7)^2 + 2^2 + 3^2} = \sqrt{49 + 4 + 9} = \sqrt{62}
\]

Thus, the distance is:

\[
d = \frac{| -50 |}{\sqrt{62}} = \frac{50}{\sqrt{62}} \approx 6.35
\]

So, \( d \approx 6.35 \).

Let y = [9, 5, 1] and u = [-7, 2, 3]. Compute the distance d from y to the line through u and the origin.

Learn how to calculate the distance from the point y = [9, 5, 1] to the line through the vector u = [-7, 2, 3] and the origin. This problem uses vector dot product, magnitude, and the projection theorem, and is suitable for students in Grades 11-12 or college-level linear algebra.

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