(adsbygoogle = window.adsbygoogle || []).push({}); 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). [itex]|N|=\sqrt{14}[/itex].

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

[tex]n= \frac{3i-2j+k}{\sqrt{14}}[/tex]

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 [itex]PQcos\theta[/itex]=|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)

[tex]=\frac{2}{\sqrt{14}}[/tex]

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

The normal vector,n_{2}for the plane 2x+y-z-5=0 is (2,1,-1)

Thus [itex]n_1.n_2= (1,-2,3).(2,1,-1)=-3[/itex]

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

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

thus

[tex]cos \theta = \frac{-3}{\sqrt{14} \sqrt{6}}[/tex]

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# Homework Help: Vector proofs for triangles and some vector plane questions

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