P3 vectors Watch

This discussion is closed.
Mussycat
Badges: 0
Rep:
?
#1
Report Thread starter 14 years ago
#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
Gauss
Badges: 2
Rep:
?
#2
Report 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
eurasianfeline
Badges: 13
Rep:
?
#3
Report 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
eurasianfeline
Badges: 13
Rep:
?
#4
Report 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
Mussycat
Badges: 0
Rep:
?
#5
Report Thread starter 14 years ago
#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
Gauss
Badges: 2
Rep:
?
#6
Report 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
eurasianfeline
Badges: 13
Rep:
?
#7
Report 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
ossoss87
Badges: 3
Rep:
?
#8
Report 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
m:)ckel
Badges: 2
Rep:
?
#9
Report 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
X
new posts
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • University of East Anglia
    All Departments Open 13:00-17:00. Find out more about our diverse range of subject areas and career progression in the Arts & Humanities, Social Sciences, Medicine & Health Sciences, and the Sciences. Postgraduate
    Wed, 30 Jan '19
  • Aston University
    Postgraduate Open Day Postgraduate
    Wed, 30 Jan '19
  • Solent University
    Careers in maritime Undergraduate
    Sat, 2 Feb '19

Brexit: Given the chance now, would you vote leave or remain?

Remain (867)
80.06%
Leave (216)
19.94%

Watched Threads

View All