# Vectors - Really do not understand this question about dot products.

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Click on the link for an image of the question:

https://gyazo.com/8ea6314bdd06af425be1bd4788c35e30

I really do not understand part (b). The vector b is getting dotted with position vectors, not the direction vectors so I don't understand how this statement can be true. Anyone care to help?

https://gyazo.com/8ea6314bdd06af425be1bd4788c35e30

I really do not understand part (b). The vector b is getting dotted with position vectors, not the direction vectors so I don't understand how this statement can be true. Anyone care to help?

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#2

(Original post by

Click on the link for an image of the question:

https://gyazo.com/8ea6314bdd06af425be1bd4788c35e30

I really do not understand part (b). The vector b is getting dotted with position vectors, not the direction vectors so I don't understand how this statement can be true. Anyone care to help?

**Barath0974**)Click on the link for an image of the question:

https://gyazo.com/8ea6314bdd06af425be1bd4788c35e30

I really do not understand part (b). The vector b is getting dotted with position vectors, not the direction vectors so I don't understand how this statement can be true. Anyone care to help?

You need to use the line equations and what you are told in part (a).

Last edited by SherlockHolmes; 1 month ago

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#3

**Barath0974**)

Click on the link for an image of the question:

https://gyazo.com/8ea6314bdd06af425be1bd4788c35e30

I really do not understand part (b). The vector b is getting dotted with position vectors, not the direction vectors so I don't understand how this statement can be true. Anyone care to help?

b.x = d

where x is a point on the plane and b si the normal and d is a scalar.

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Ah, I see. So does this mean that b dotted with any point on the line is equal to 0? Or that b dotted with a point on a line (which is perpendicular to b) is equal to b dotted another point on a line that it is also perpendicular to b?

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#5

b.x is a constant (usually non-zero) for any point in the plane. That is the definition of a plane. So b.x is constant for any point on either of the two lines.

b.v where v is one of the direction vectors is zero, by definition.

Note a direction vector is equal to the difference between two points on the line, so the interpretations are consistent.

b.v where v is one of the direction vectors is zero, by definition.

Note a direction vector is equal to the difference between two points on the line, so the interpretations are consistent.

Last edited by mqb2766; 1 month ago

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So, how do I use the fact that the lines are intersecting to show the statement is true?

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Also, I don’t see how there’s a plane as the two directional vectors are not parallel.

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#8

(Original post by

So, how do I use the fact that the lines are intersecting to show the statement is true?

**Barath0974**)So, how do I use the fact that the lines are intersecting to show the statement is true?

Last edited by mqb2766; 1 month ago

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#9

(Original post by

Also, I don’t see how there’s a plane as the two directional vectors are not parallel.

**Barath0974**)Also, I don’t see how there’s a plane as the two directional vectors are not parallel.

Or more simply, a flat piece of paper is a plane. Can you draw two intersecting lines on there?

Last edited by mqb2766; 1 month ago

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Would you say that the position vectors are scalar multiples of their direction vectors?

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#11

(Original post by

Would you say that the position vectors are scalar multiples of their direction vectors?

**Barath0974**)Would you say that the position vectors are scalar multiples of their direction vectors?

x0 + l*v

where x0 is a point on the line, v is a non-zero direction vector and l is a scalar multiplier.

https://tutorial.math.lamar.edu/clas...sofplanes.aspx

It would only be a scalar multple of v if x0 is the zero vector so the line passes through the origin.

It should be in your textbook?

Last edited by mqb2766; 1 month ago

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Nevermind, I understand the question, I just equate the two equations of the line, solve for a, sub it back in to show that the statement is true.

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