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

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In summary, the homework statement asks for the distance from a point (1, 0, 0) to a line. The equation for the line is found given the vector equation and the unit vector in direction of the line. The distance is found to be 5/13 of the length of the line.
  • #1
mgnymph
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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!
 
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  • #2
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).
 
  • #3
mgnymph said:
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, [itex]\vec{r}(0)= 0\vec{i}+ 0\vec{j}+ 0\vec{z}[/itex].

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?
 
  • #4
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...
 

1. How do I find the magnitude of a vector?

The magnitude of a vector is the length of the vector. To find the magnitude, you can use the Pythagorean theorem, which states that the magnitude is equal to the square root of the sum of the squares of the vector's components. In other words, if a vector has components (a, b), its magnitude is equal to √(a² + b²).

2. What is the difference between a scalar and a vector?

A scalar is a quantity that has magnitude (size) only, while a vector is a quantity that has both magnitude and direction. Scalars are represented by a single number, while vectors are represented by magnitude and direction, often using an arrow to indicate the direction.

3. How do I add or subtract vectors?

To add or subtract vectors, you first need to make sure they are the same type of vector (e.g. both 2D or both 3D). To add or subtract, simply add or subtract the corresponding components of the vectors. For example, to add (a, b) and (c, d), the result would be (a + c, b + d). To subtract, it would be (a - c, b - d).

4. How do I find the dot product of two vectors?

The dot product of two vectors is a scalar quantity that is equal to the product of the magnitudes of the vectors and the cosine of the angle between them. In other words, the dot product of vectors (a, b) and (c, d) is ac + bd. This can be used to find the angle between two vectors as well.

5. What are unit vectors?

Unit vectors are vectors with a magnitude of 1 that are used to represent direction. They are often denoted by a hat symbol (î, ĵ, k̂) and are used to describe the direction of a vector without changing its magnitude. Unit vectors are important in vector operations and can be used to represent any vector by multiplying its magnitude by the unit vector in the desired direction.

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