Vectors help Watch

word2yamother
Badges: 1
Rep:
?
#1
Report Thread starter 7 years ago
#1
Can anyone please help me on this question.

The point P,Q and R are defined by the position vectors (2,0,1), (3,-1,2) and (2,3, a) respectively.

Find the value of a such that the line through P and R is perpendicular to the line through P and Q.

I have no ideas where to start, I know the dot product = 0 for normal but i have no clue as how to apply it in this question. Pls help
0
quote
reply
vc94
Badges: 10
Rep:
?
#2
Report 7 years ago
#2
For the two lines, their direction vectors must have a dot product of zero.
Can you work out the direction vectors?
0
quote
reply
Kareir
Badges: 2
Rep:
?
#3
Report 7 years ago
#3
(Original post by word2yamother)
Can anyone please help me on this question.

The point P,Q and R are defined by the position vectors (2,0,1), (3,-1,2) and (2,3, a) respectively.

Find the value of a such that the line through P and R is perpendicular to the line through P and Q.

I have no ideas where to start, I know the dot product = 0 for normal but i have no clue as how to apply it in this question. Pls help

PR = 0i + 3j + (a-1)k
PQ = 1i - 1j + 1k

So, dot product has to equal 0..

E.g. (1x0) + (-1)(3) + (1)(a-1) = 0

_Kar.
0
quote
reply
word2yamother
Badges: 1
Rep:
?
#4
Report Thread starter 7 years ago
#4
(Original post by Kareir)
PR = 0i + 3j + (a-1)k
PQ = 1i - 1j + 1k

So, dot product has to equal 0..

E.g. (1x0) + (-1)(3) + (1)(a-1)k = 0

_Kar.
Arent you meant to multiply the magnitude of the two vectors (directional vector) when doing the dot product? Im confused
0
quote
reply
dknt
Badges: 12
Rep:
?
#5
Report 7 years ago
#5
(Original post by word2yamother)
Arent you meant to multiply the magnitude of the two vectors (directional vector) when doing the dot product? Im confused
The magnitude of vectors  \not = The direction vector

The equation is \mathrm{cos} \theta = \dfrac{a \cdot b}{|a||b|}, so

a \cdot b = |a||b| \mathrm{cos} \theta

The modulus parts are your magnitudes.
0
quote
reply
babz07
Badges: 0
Rep:
?
#6
Report 7 years ago
#6
Not 100% sure, im studying vectors at the moment too, but i think you do this:

* find PQ (which is just q-p)
* find PR (which is r-p)
* then find dot product: PQ.PR = 0
* rearrange to find a
0
quote
reply
Kareir
Badges: 2
Rep:
?
#7
Report 7 years ago
#7
(Original post by word2yamother)
Arent you meant to multiply the magnitude of the two vectors (directional vector) when doing the dot product?
From my data booklet:

v.w = /v/ /w/ Cos (Theta).

Where cos (theta) is the angle between the two.

So yes, if they're perpendicular, you do multiply the two magnitudes.

But also, if you dont know the magnitudes/angle, for a vector ai + bj + ck and xi + yj + zk,

The dot product is
ax + by + cz.

_Kar.
0
quote
reply
word2yamother
Badges: 1
Rep:
?
#8
Report Thread starter 7 years ago
#8
(Original post by dknt)
The magnitude of vectors  \not = The direction vector

The equation is \mathrm{cos} \theta = \dfrac{a \cdot b}{|a||b|}
Sorry i didnt make myself clear. What i meant is that the vectors being multiplied in this case are the directional vectors hence we should find the magnitude of these vectors then multiply. Im just asking if thats the way to do it as im a little confused
0
quote
reply
dknt
Badges: 12
Rep:
?
#9
Report 7 years ago
#9
(Original post by word2yamother)
Sorry i didnt make myself clear. What i meant is that the vectors being multiplied in this case are the directional vectors hence we should find the magnitude of these vectors then multiply. Im just asking if thats the way to do it as im a little confused
Well with vectors, if  a \cdot b = 0 \Leftrightarrow they are perpendicular. This is because cos 90 is 0, so when multiplying the magnitudes together you'll multiply it by 0 anyway, as  a \cdot b = |a||b| \mathrm{cos} \thetaso you don't need to worry about it when dealing with perpendicular vectors and the dot product, so you only need to work out the dot product by multiplying the co-efficients of i, j and k.
0
quote
reply
word2yamother
Badges: 1
Rep:
?
#10
Report Thread starter 7 years ago
#10
(Original post by dknt)
Well with vectors, if  a \cdot b = 0 \Leftrightarrow they are perpendicular. This is because cos 90 is 0, so when multiplying the magnitudes together you'll multiply it by 0 anyway, as  a \cdot b = |a||b| \mathrm{cos} \thetaso you don't need to worry about it when dealing with perpendicular vectors and the dot product, so you only need to work out the dot product by multiplying the co-efficients of i, j and k.
I get it now. Thank you and everyone else who has contributed.
0
quote
reply
X

Quick Reply

Attached files
Write a reply...
Reply
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
    UEA Mini Open Day Undergraduate
    Fri, 4 Jan '19
  • University of Lincoln
    Mini Open Day at the Brayford Campus Undergraduate
    Mon, 7 Jan '19
  • Bournemouth University
    Undergraduate Mini Open Day Undergraduate
    Wed, 9 Jan '19

Did you get less than your required grades and still get into university?

Yes (76)
29.69%
No - I got the required grades (147)
57.42%
No - I missed the required grades and didn't get in (33)
12.89%

Watched Threads

View All