Vector proofs for triangles and some vector plane questions

1. Oct 6, 2008

rock.freak667

1. The problem statement, all variables and given/known data

1) Show by the use of vectors that the three altitudes of a triangle pass through the same point.

2) Show using vectos that the bisectors of the angles of a triangle pass through thr same point.

3)Find the distance from the point (1,0,-2) to the plane 3x-2y+z+1=0

4) Find the distance between the lines r=(-1,-2,0)+t(1,0,-1) and r=(1,0,1)+t(1,-1,2)

5) Find the cosine of the angle between the two planes x-2y+3z-4=0 and 2x+y-z-5=0
2. Relevant equations

3. The attempt at a solution
For the first two I just need some hints on how to prove them.

3) The equation of the plane is 3x-2y+z+1=0, meaning that the normal vector,N, to the plane is (3,-2,1). $|N|=\sqrt{14}$.

So the unit vector in the direction of the normal vector:

$$n= \frac{3i-2j+k}{\sqrt{14}}$$

Let P=(1,0,-2).

I need to choose any point on the given plane.
Let Q=(1,2,0)

The distance I need is $PQcos\theta$=|PQ.n|

So the vector PQ=Q-P=(1,2,0)-(1,0,-2)=(0,2,2)

therefore |PQ.n|=[(0,2,2).(3,-2,1)]/sqrt(14)

$$=\frac{2}{\sqrt{14}}$$

5) The normal vector,n1 for the plane x-2y+3z-4=0 is (1,-2,3).
The normal vector,n2 for the plane 2x+y-z-5=0 is (2,1,-1)

Thus $n_1.n_2= (1,-2,3).(2,1,-1)=-3$

$|n_1|=\sqrt{14}]/itex] [itex]|n_2|=\sqrt{6}$

thus

$$cos \theta = \frac{-3}{\sqrt{14} \sqrt{6}}$$

2. Oct 6, 2008

rock.freak667

4)Direction of first line,u1=(1,0,-1)
2nd line,u2=(1,-1,2)

$$\vec{OA}=(1,-2,0)$$

$$\vec{OB}=(1,0,1)$$

$$\vec{AB}=(0,2,1)$$

$$u_1 \times u_2=-i-3j-k \Rightarrow |u_1 \times u_2|=\sqrt{11}$$

Thus the required distance is

$$|\frac{(-1,-3,-1).(0,2,1)}{\sqrt{14}}|=\frac{7}{\sqrt{11}}$$

6) Find the equation of a plane such that any point on this plane is equidistant from points (1,1,1) and (3,1,5).

Solution:
Midpt. AB i.e. a point on the req'd plane= ( (1+3)/2, (1+1)/2, (1+5)/2 )=(2,1,3)

The normal vector is the vector AB=-(1,1,1)+(3,1,5)=(2,0,4). Thus (1,0,2) is also a normal vector.

Using the defintion that (r-a).n=0

(x-2,y-1,z-3).(1,0,2)=0
=> x-2+2z-6=0
so x+2z=8.

Correct?

Last edited: Oct 6, 2008
3. Oct 11, 2008

rock.freak667

Can someone tell me if I am going correct for the 2nd one?

I created a vector triangle, ABC such that a+b+c=0

Drew the bisectors,AP;BR;CQ such that AP=b/2 ; BR=c/2 and CQ=a/2.

Finding the equation of the line passing through $\vec{AP}$

$$l_1: r=a+ t(a+ \frac{b}{2})$$

the line through $\vec{BR}$

$$l_2: r= c+ u(c+ \frac{a}{2})$$

Of they meet,then

$$a+ t(a+ \frac{b}{2})= c+ u(c+ \frac{a}{2})$$

If they are equal, then a=c.

and a(1+t)=c(1+u) as well. i.e. t=u. So they intersect. Good so far?

4. Oct 11, 2008

rock.freak667

Here is what I tried for the 1st question. I drew a triangle ABC, with H as the point of intersection of the three altitudes.

AH.BC=0

(AO+OH).(BO+OC)=0
(OA.OB-OB.OH-OA.OC+OC.OH)=0 .......1

BH.AC=0
(BO+OH).(AO+OC)=0
(OA.OB-OA.OH-OC.OB+OC.OH)=0.........2

eq'n 2-1 gives

-OB.OH-OA.OC-OA.OH-OC.OB=0

-OB.(OH-OC)+OA.(OH-OC)=0

-OB.CH+OA.CH=0

CH.(OA-OB)=0

CH.BA=0

Meaning that CH is perpendicular to AB. Thus all three altitudes intersect at the same point,H.

Correct??