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

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E--
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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)
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Idg a damn
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Dot product
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E--
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(Original post by Idg a damn)
Dot product
dot product between a and b ?
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RDKGames
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(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 (\alpha,\beta,\gamma) such that -2\alpha + \beta + 2\gamma = 0 and that will be your perpendicular vector, i.e. the direction vector of your line.

Then just construct the eq. of the line.
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RDKGames
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(Original post by E--)
My working out:
I found the equation of vector PQ which is (-2,1,2)
This is wrong. Should be PQ  = OQ - OP = (2, -1, -2).. but it doesn't really matter, I suppose, if you're finding the eq. of a perpendicular line.
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E--
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(Original post by RDKGames)
This is wrong. Should be PQ  = OQ - OP = (2, -1, -2).. 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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Idg a damn
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is it equation of plane or line?
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Idg a damn
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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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Idg a damn
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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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Idg a damn
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you take the dot product of this vector and vector pq then equate it to zero
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E--
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(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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RDKGames
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(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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Idg a damn
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(Original post by E--)
Yes it's the equation of the plane. So is it 2a-b-2b ?
I don't understand your question...

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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Idg a damn
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sorry I made a mistake I edited the vector I wrote just now
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Idg a damn
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Hey OP is your question solved?
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E--
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(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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Idg a damn
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And thats exactly the way to solve it!
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Idg a damn
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Now you just need to simplify the equation and write it in the form of ax+by+cz=d
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begbie68
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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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(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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