Vector dot product and parallel vectors

[late347]

Homework Statement

show that points P, Q and R are in a straight line
P (1, -3, 4)
Q ( 2, 2, 1)
R (3, 7, -2)

and find the vectors $\vec{PQ}$ and $\vec{QR}$

Homework Equations

The Attempt at a Solution

In proving that the points are in a straight line, we might be able to use dot product.
In a straight line, means that the two vectors are parallel, (either in the same direction or opposite direction)

According to my math textbook it would appear that in the case of angle between vectors being zero degress, from there it follows that the vectors are parallel.

I was wondering about the second case when the vectors are in opposite direction but also parallel, wouldn't the angle between the vectors be 180 degrees?

From my understanding of parallel ( or in a straight line) in geometry, both cases may be possible for these vectors.

My textbook mentiosn only the 0 degree angle case

In any case it would seem to be prudent to compute the dot product of PQ and QR, and then find out the lengths of the vectors, and then finally find out what the angle actually is.

my textbook was Engineering Mathematics: Croft, Davison and Hargreaves.

Here is a quote page 219

If vector a and vector b are parallel vectors, show that a⋅b = |a| |b| . If a and b are orthogonal show that their scalar product is zero.

solution:
If a and b are parallel then the angle between them is zero. Therefore a ⋅b = |a| |b| cos(0deg)

 

[DrClaude]

I was wondering about the second case when the vectors are in opposite direction but also parallel, wouldn't the angle between the vectors be 180 degrees?

Yes, but point $Q$ is between points $P$ and $R$, therefore $\vec{PQ}$ and $\vec{QR}$ point in the same direction.

In any case it would seem to be prudent to compute the dot product of PQ and QR, and then find out the lengths of the vectors, and then finally find out what the angle actually is.

my textbook was Engineering Mathematics: Croft, Davison and Hargreaves.

Here is a quote page 219

Indeed, since $\mathbf{a} \cot \mathbf{b} = | \mathbf{a} | | \mathbf{b} | \cos \theta$, if $\mathbf{a} \cot \mathbf{b} = | \mathbf{a} | | \mathbf{b} |$, then $\cos \theta = 1$, meaning $\theta = 0, 180^\circ$, and the vectors are parallel.

 

[Chestermiller]

Do you know how to determine the equation for the straight line passing through two arbitrary points in 3D space?

Also, are you familiar with the cross product of two vectors? If so, do you recall that the cross product of two parallel vectors is equal to zero?

 

[late347]

Do you know how to determine the equation for the straight line passing through two arbitrary points in 3D space?

Also, are you familiar with the cross product of two vectors? If so, do you recall that the cross product of two parallel vectors is equal to zero?

not really. I was taught about this dot product thing on Wednesday. I was just a little bit confused by the quote in the textbook which discussed a general casse of vectors a and b.

There was the statement that if a and b are parallel, then it follows that the angle is zero... But that doesn't seem to be the case that it can only be one angle, the zero degree angle?

For opposite sided vectors, those are still parallel, and it would seem that the angle between them (if you put the vectors tail to tail, wouldn't the angle be the 180 degrees?)

Or is there a definition of the angle between vectors that I missed somewhere?

 

[DrClaude]

It's probably a question of nomenclature. In the case where the angle is 180°, some would call that anti-parallel instead of parallel.

 

[late347]

It's probably a question of nomenclature. In the case where the angle is 180°, some would call that anti-parallel instead of parallel.

I see, thanks for your input though. Studyin math always seems to bring out new tidbits like this. Especially studying math in English (not my native language)

 

[Chestermiller]

not really. I was taught about this dot product thing on Wednesday. I was just a little bit confused by the quote in the textbook which discussed a general casse of vectors a and b.

There was the statement that if a and b are parallel, then it follows that the angle is zero... But that doesn't seem to be the case that it can only be one angle, the zero degree angle?

For opposite sided vectors, those are still parallel, and it would seem that the angle between them (if you put the vectors tail to tail, wouldn't the angle be the 180 degrees?)

Or is there a definition of the angle between vectors that I missed somewhere?

If the dot product of a and b is equal to the product of the magnitudes of a and b, then the angle is zero.

The algebraic equation for a straight line in 3D is $$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$$