Linear Algebra question.
2 Answers

Let’s use the name “L” for the line through u and the origin. Any plane perpendicular to L has an equation of the form 4x – 6y – 7z = C, where the C will vary depending on WHICH perpendicular plane is chosen. I want the particular plane that contains the point y, so I need C to be 4*4 + 6*5 + 7*7 = 63. I’m using the name “P” for the plane 4x – 6y – 7z = 63. This plane is perpendicular to L and contains the point y.
Now I’m going to find out exactly where P intersects L. I need a point that simultaneously satisfies the equation of P and the equation(s) of L. The equations of L may be written x = 2y/3 = 4z/7. So now I write:
4x – 6(3x/2) – 7(7x/4) = 63, implying that
4x – 9x – 49x/4 = 63 or 101x/4 = 63, and
x = 252/101, so that y = 378/101 and z = 441/101.
Please check my arithmetic up to here, as my answer is ugly (and therefore suspect). If my answer is correct, I am sure you know how to find the distance from (4,5,7) to (252/101, 378/101, 441/101).

Let a = vector Oy.
Let b = vector Ou.
Let the projection of a onto b be kb = k<4, 6, 7> = <4k, 6k, 7k>.
(kb – a) ⊥ b
(kb – a) · a = 0
(<4k, 6k, 7k> – <4, 5, 7>) · <4, 6, 7> = 0
<4k – 4, 6k + 5, 7k + 7> · <4, 6, 7> = 0
16k + 16 + 36k – 30 + 49k – 49 = 0
101k = 63
k = 63/101
The distance you want is kb – a
kb – a
= <4k – 4, 6k + 5, 7k + 7>
= <656/101, 127/101, 266/101>
<656/101, 127/101, 266/101>
= √(517221)/101
≈ 7.12