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

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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)

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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(Original post by

Dot product

**Idg a damn**)Dot product

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#4

(Original post by

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)

**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)

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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#5

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(Original post by

This is wrong. Should be .. but it doesn't really matter, I suppose, if you're finding the eq. of a perpendicular line.

**RDKGames**)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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#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

though to my understanding you should be asked to find an equation of a plane rather than a line

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#9

the vector I said just now is:

(x-3)i + (y-1)j + ( z-2)k

(x-3)i + (y-1)j + ( z-2)k

Last edited by Idg a damn; 1 year ago

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(Original post by

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

**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

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#12

(Original post by

Yes it's the equation of the plane. So is it 2a-b-2b ?

**E--**)Yes it's the equation of the plane. So is it 2a-b-2b ?

What two pieces of information do you need to construct the equation of a plane in vector form?

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#13

(Original post by

Yes it's the equation of the plane. So is it 2a-b-2b ?

**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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(Original post by

Hey OP is your question solved?

**Idg a damn**)Hey OP is your question solved?

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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#18

Now you just need to simplify the equation and write it in the form of ax+by+cz=d

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#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.

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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#20

(Original post by

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.

**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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