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

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#1
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?
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1 month ago
#2
(Original post by 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?
Regardless of what those two vectors represent, the dot product can still be used. i.e. the dot product of two vectors simply gives a number.

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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1 month ago
#3
(Original post by 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 is the normal to the plane containing the two lines. The two position vectors are both points on the plane, of which there are many. So the dot product of each with the normal (b) must be constant. Thats how a plane is defined.
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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#4
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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1 month ago
#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.
Last edited by mqb2766; 1 month ago
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#6
So, how do I use the fact that the lines are intersecting to show the statement is true?
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#7
Also, I don’t see how there’s a plane as the two directional vectors are not parallel.
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1 month ago
#8
(Original post by Barath0974)
So, how do I use the fact that the lines are intersecting to show the statement is true?
Simply write down the fact that the lines are intersecting using the vectors, then take the dot product with b. Its just a few lines of work which uses the property that b is perpendicular to the two direction vectors.
Last edited by mqb2766; 1 month ago
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1 month ago
#9
(Original post by Barath0974)
Also, I don’t see how there’s a plane as the two directional vectors are not parallel.
Its not necessarily relevant for the question, but just have a look in your textbook about a plane definition.
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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#10
Would you say that the position vectors are scalar multiples of their direction vectors?
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1 month ago
#11
(Original post by Barath0974)
Would you say that the position vectors are scalar multiples of their direction vectors?
No. The definition should be clear in the question. A point on the line (in parametric form) is given by
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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#12
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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