Help with Pre-Calculus Vector Questions from a 16 Year Old in NZ

[mgnymph]

I don't know what pre-calculus is, I live in New Zealand and I'm 16 if that helps... I apologize in advance if this is the wrong area!

Homework Statement

Find the distance of A (1,0,0) from the line r = t(12i - 3j -4k).

Homework Equations

a.b = lal.lbl.cos(theta)

The Attempt at a Solution

Well, since the vector equation only has direction vector.. I'm assuming it passes through origin?

So I found OA (= A - O) = (1, 0, 0) and line OA squared = 1 (the length)

then I found the unit vector in direction Ot, which came out to be (12/13, -3/13, -4/13).

So, Ot = OAcos(theta)
= OA x 1 x cos(theta)
= (a-o).u
= 12/13

At quared = OA squared - Ot squared...
= 1 - (12/13)^2
= 25/169
At = 5/13

But the answer says (root69) / 13 !

What am I doing wrong!

 

[CompuChip]

Hint: you are given a vector a = (1, 0, 0) and a set of vectors r(t) = (12t, -3t, -4t). For any given t, the distance of a to r(t) is the length of the difference vector. What is it? For which t is this minimal?

(As for the topic, I think it fits best in linear algebra).

 

[HallsofIvy]

I don't know what pre-calculus is, I live in New Zealand and I'm 16 if that helps... I apologize in advance if this is the wrong area!

Homework Statement

Find the distance of A (1,0,0) from the line r = t(12i - 3j -4k).

Homework Equations

a.b = lal.lbl.cos(theta)

The Attempt at a Solution

Well, since the vector equation only has direction vector.. I'm assuming it passes through origin?

You don't need to assume it. If t= 0, \vec{r}(0)= 0\vec{i}+ 0\vec{j}+ 0\vec{z}.

So I found OA (= A - O) = (1, 0, 0) and line OA squared = 1 (the length)

then I found the unit vector in direction Ot, which came out to be (12/13, -3/13, -4/13).

So, Ot = OAcos(theta)
= OA x 1 x cos(theta)
= (a-o).u
= 12/13

At quared = OA squared - Ot squared...
= 1 - (12/13)^2
= 25/169
At = 5/13

But the answer says (root69) / 13 !

What am I doing wrong!

There are many different distances from A to different points on the line. The distance from A to the line is defined as the shortest of all these distances.
The line from A to the point on the line nearest A must be perpendicular to the line
(Do you see why?), and so A must lie in a plane perpendicular to A.

Any plane perpendicular to the given vector must be of the form 12x- 3y- 4z= C for some number C. In order that A= (1, 0, 0) be in that plane we must have 12(1)- 3(1)- 4(1)= 12- 3- 4= 5= C. That is, the plane perpendicular to the given line, containing A has equaiton 12x- 3y- 4z= 5. Now where does the given line cross that plane? What is the distance from A to that point?

 

[mgnymph]

Yea that's why I did the a^2 = b^2 + c^2 rule thing because the lines were at right angles to each other, so I got a right angled triangle...