# Let y = [4,-5,-7] and u = [-4,-6,-7] Compute the distance d from y to the line through u and the origin?

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Linear Algebra question.

• 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

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