- #1
brochesspro
- 155
- 22
- Homework Statement
- Given below.
- Relevant Equations
- Given below.
Do you mean dot product?ergospherical said:(can you solve it faster another way? what are the properties of the cross product?)
brochesspro said:Do you mean dot product?
I think they meant the cross product!brochesspro said:Do you mean dot product?
You gave me the answer too soon. It would have been better if you would have slowly led me to that conclusion.PhDeezNutz said:Yeah I’m going to have to echo the comment about using cross products.
Take the cross product of the two vectors and find its unit vector. Then negate it and you have your two answers.
I was only focusing on the perpendicular part and the fact that perpendicular vectors have dot product 0. And so, I completely forgot about cross products.PeroK said:I think they meant the cross product!
brochesspro said:You gave me the answer too soon. It would have been better if you would have slowly led me to that conclusion.
That's a good point. You could look generate simultanenous equations by looking for a vector whose dot product with both those vectors is zero. Using the cross product could be quicker.brochesspro said:I was only focusing on the perpendicular part and the fact that perpendicular vectors have dot product 0. And so, I completely forgot about cross products.
No need for it now, I already got the answer. But, thank you.PhDeezNutz said:Fair enough. Let’s back track. Can you prove that. Let’s justify this approach.
##(\vec{a} \times \vec{b}) \cdot \vec{b} = (\vec{a} \times \vec{b}) \cdot \vec{a} = 0##View attachment 295004
I tried to do that itself. Could you once go through the attachments and guide me to what I have done wrong?PeroK said:That's a good point. You could look generate simultanenous equations by looking for a vector whose dot product with both those vectors is zero. Using the cross product could be quicker.
Never mind, I got the answer. Again, thank you for all your responses.brochesspro said:I tried to do that itself. Could you once go through the attachments and guide me to what I have done wrong?
You got as far as $$2a -3b = 0$$$$-a + 4b - 5c =0$$And then things seemed to go a bit wrong. I just multiplied the second equation by ##2## and then added them.brochesspro said:I tried to do that itself. Could you once go through the attachments and guide me to what I have done wrong?
I am not sure about it. But, many people write it that way, I think it is used for points rather than vectors, but I may be wrong.PeroK said:PS I like to write vectors as ##(2, -3, 0), (-1, 4, -5), (a, b, c)##. Has this gone out of fashion? Anyway, it seems very simple to write down the dot products in this way.
Yes, I noticed that I forgot to multiply the 4 by 2 and things went downhill from there. But, I can now do this both ways, the dot product way, and the cross product way. Goodbye for now.PeroK said:You got as far as $$2a -3b = 0$$$$-a + 4b - 5c =0$$And then things seemed to go a bit wrong. I just multiplied the second equation by ##2## and then added them.
Note that you are always going to have a free parameter here. In this case ##c## can be anything.
PeroK said:You got as far as $$2a -3b = 0$$$$-a + 4b - 5c =0$$And then things seemed to go a bit wrong. I just multiplied the second equation by ##2## and then added them.
Note that you are always going to have a free parameter here. In this case ##c## can be anything.
That is what I did, but with ##c##.vcsharp2003 said:Wouldn't the following be another equation in the variables a, b and c due to the fact that the unknown vector is a unit vector? If yes, then we end up with 3 equations for the three variables, which now can be solved for a, b and c.
$$a^2 + b^2 + c^2= 1$$We could find ##a## in terms of ##b## from first equation. Then from second equation, substituting ##a## in terms of ##b##, we will get ##c## in terms of ##b##. Now, we substitute in above equation the variables ##a## and ##c##, and solve for ##b##. Knowing ##b## means we can determine ##a## and ##c##.
That should give you two different unit vectors.brochesspro said:That is what I did, but with ##c##.
A unit vector is a vector with a magnitude of 1. It represents the direction of a given vector without any influence from its magnitude.
To find the unit vector perpendicular to a given vector, you can use the cross product or dot product method. Both methods involve mathematical calculations using the given vector's components.
Unit vectors perpendicular to a given vector help in determining the direction of that vector in a three-dimensional space. They are also useful in physics and engineering applications, such as calculating forces and determining the direction of motion.
Yes, there can be infinitely many unit vectors perpendicular to a given vector. This is because there are infinite directions in a three-dimensional space that can be perpendicular to a given vector.
A unit vector has a magnitude of 1 and represents the direction of a given vector, while a perpendicular vector is a vector that forms a 90-degree angle with the given vector. A unit vector can also be perpendicular to a given vector, but not all perpendicular vectors are unit vectors.