# Vector algebra: Find the equation of line perpendicular...

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#1
Find the equation of the plane passing through Q and perpendicular to the line PQ

OP = a
OQ = b

a = (1,2,4) and b = (3,1,2)

My working out:
I found the equation of vector PQ which is (-2,1,2)
Last edited by E--; 1 year ago
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1 year ago
#2
Dot product
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#3
(Original post by Idg a damn)
Dot product
dot product between a and b ?
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1 year ago
#4
(Original post by E--)
Find the equation of the line passing through Q and perpendicular to the line PQ

OP = a
OQ = b

a = (1,2,4) and b = (3,1,2)

My working out:
I found the equation of vector PQ which is (-2,1,2)
There are infinitely many of these.

But if you want one of them, just choose any such that and that will be your perpendicular vector, i.e. the direction vector of your line.

Then just construct the eq. of the line.
Last edited by RDKGames; 1 year ago
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1 year ago
#5
(Original post by E--)
My working out:
I found the equation of vector PQ which is (-2,1,2)
This is wrong. Should be .. but it doesn't really matter, I suppose, if you're finding the eq. of a perpendicular line.
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#6
(Original post by RDKGames)
This is wrong. Should be .. but it doesn't really matter, I suppose, if you're finding the eq. of a perpendicular line.
oh thank you
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1 year ago
#7
is it equation of plane or line?
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1 year ago
#8
you take the dot product between vector pq and the vector on the line spawning from q

though to my understanding you should be asked to find an equation of a plane rather than a line
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1 year ago
#9
the vector I said just now is:

(x-3)i + (y-1)j + ( z-2)k
Last edited by Idg a damn; 1 year ago
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1 year ago
#10
you take the dot product of this vector and vector pq then equate it to zero
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#11
(Original post by Idg a damn)
you take the dot product between vector pq and the vector on the line spawning from q

though to my understanding you should be asked to find an equation of a plane rather than a line
Yes it's the equation of the plane. So is it 2a-b-2b ?
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1 year ago
#12
(Original post by E--)
Yes it's the equation of the plane. So is it 2a-b-2b ?
Well that's a massive difference to the question...

What two pieces of information do you need to construct the equation of a plane in vector form?
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1 year ago
#13
(Original post by E--)
Yes it's the equation of the plane. So is it 2a-b-2b ?

but basically you just need to take the dot product between the vector I gave you and vector PQ.

set the dot product to be equal to zero and you should get equation of plane in Cartesian form
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1 year ago
#14
sorry I made a mistake I edited the vector I wrote just now
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1 year ago
#15
Hey OP is your question solved?
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#16
(Original post by Idg a damn)
Hey OP is your question solved?
I think so
Sorry, i got consfued by the question. The question was Find the plane passing through Q and perpendicular to PQ
So the dot product between Q and PQ would be 0.

(2 ,-1, 2) . [(x,y,z)-(3, 1 ,2)]

(x-3, y-1, z-2) . (2,-1,2) = 0
2(x-3) - 1(y-1) + 2(z-2) = 0 heres what ive got

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1 year ago
#17
And thats exactly the way to solve it!
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1 year ago
#18
Now you just need to simplify the equation and write it in the form of ax+by+cz=d
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1 year ago
#19
vector pq is a line perpendicular to the plane.

there is no need whatsoever to do dot product.

cartesian eqn of a plane is of form ax + by + cz = p

you should have a,b & c quite easily from the work you've done so far, then subst to find p.
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1 year ago
#20
(Original post by begbie68)
vector pq is a line perpendicular to the plane.

there is no need whatsoever to do dot product.

cartesian eqn of a plane is of form ax + by + cz = p

you should have a,b & c quite easily from the work you've done so far, then subst to find p.

The method I suggested is

Assume that the position vector from the origin to the plane is (x,y,z), label this point A.

The position vector can then be written as OA=OQ+QA

Of which vector QA will be perpendicular to vector QP, and that both of them have a common origin, which is point Q.

Since vector QA is perpendicular to QP, dot product between the two is zero.

So we need to derive QA first, which is ( x-3, y-1, z-2)

Then take the dot product of this with QP, which is (-2,1,2)
Last edited by Idg a damn; 1 year ago
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