# P3 vectorsWatch

This discussion is closed.
#1
I would appreciate any help with this question:

Find a vector which is perpendicular to both a and b where
a = i +3j - k
b = 3i - j - k

I'm not sure about the method to use for the question.

Thank you!
0
14 years ago
#2
(Original post by Mussycat)
I would appreciate any help with this question:

Find a vector which is perpendicular to both a and b where
a = i +3j - k
b = 3i - j - k

I'm not sure about the method to use for the question.

Thank you!
The perpendicular vector (c) is the cross product of the two vectors a and b.

Hence, c = axb

c = (-3-1)i - (-1+3)j + (-1-9)k
=> c = -4i - 2j - 10k

Euclid

Edit: Sorry I think vector cross products are in P6. Another way could be to let c = xi + yj + zk and solve for c.a = 0 and c.b = 0. You will get simultaneous equations for x y and z.
0
14 years ago
#3
(Original post by Euclid)
The perpendicular vector (c) is the cross product of the two vectors a and b.

Hence, c = axb

c = (-3-1)i - (-1+3)j + (-1-9)k
=> c = -4i - 2j - 10k

Euclid

Edit: Sorry I think vector cross products are in P6. Another way could be to let c = xi + yj + zk and solve for c.a = 0 and c.b = 0. You will get simultaneous equations for x y and z.
yup...but it's useful to learn vector product. my teacher taught us that for P3 vectors as well. and then later he showed us the dot product way just in case we forget how to do vector product haha
0
14 years ago
#4
(Original post by Euclid)
The perpendicular vector (c) is the cross product of the two vectors a and b.

Hence, c = axb

c = (-3-1)i - (-1+3)j + (-1-9)k
=> c = -4i - 2j - 10k

Euclid

Edit: Sorry I think vector cross products are in P6. Another way could be to let c = xi + yj + zk and solve for c.a = 0 and c.b = 0. You will get simultaneous equations for x y and z.
o btw the answer can be reduced to 2i + j + 5k
0
#5
ok thanks! i will ask my teacher about the cross product rule tomorrow then (back to college, aaaghh!)
i did try simultaneous equations but things got a bit complicated...
0
14 years ago
#6
(Original post by eurasianfeline)
o btw the answer can be reduced to 2i + j + 5k
Glad to hear you can divide by two!
0
14 years ago
#7
(Original post by Euclid)
Glad to hear you can divide by two!
it's - 2

and thank you btw haha
0
13 years ago
#8
(Original post by Galois)
The perpendicular vector (c) is the cross product of the two vectors a and b.

Hence, c = axb

c = (-3-1)i - (-1+3)j + (-1-9)k
=> c = -4i - 2j - 10k

Euclid

Edit: Sorry I think vector cross products are in P6. Another way could be to let c = xi + yj + zk and solve for c.a = 0 and c.b = 0. You will get simultaneous equations for x y and z.

is there any other rules from p6 we can use in p3?! like easier to do than the stuff in p3!
0
13 years ago
#9
(Original post by ossoss87)
is there any other rules from p6 we can use in p3?! like easier to do than the stuff in p3!
I don't think there is.
The dot product way (in this example) is just longer, but I wouldn't say it's harder.
0
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