# Vectors helpWatch

#1

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
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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?
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7 years ago
#3
(Original post by word2yamother)

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.
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#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
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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 The direction vector

The equation is , so

The modulus parts are your magnitudes.
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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
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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.
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#8
(Original post by dknt)
The magnitude of vectors The direction vector

The equation is
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
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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 they are perpendicular. This is because cos 90 is 0, so when multiplying the magnitudes together you'll multiply it by 0 anyway, as so 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.
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#10
(Original post by dknt)
Well with vectors, if they are perpendicular. This is because cos 90 is 0, so when multiplying the magnitudes together you'll multiply it by 0 anyway, as so 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.
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