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Homework Statement
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
Homework Equations
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,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 [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]