Vector Question?

Question

Points P (2,1,0) and Q (1,3,1) and R (1,-1,2).
1. Find the area of point PQR.
2. Find the vector equation of line L through P and Q.
3. If S is the point (1,1,1), show that S does not lie on the line L and fined the vector equation of the line joining S to L which is perpendicular to L.

Answer 1

Problem #1
Method #1
You can use Heron’s formula to find the area of a triangle.
A=sqrt(s(s-a)(s-b)(s-c)) where a, b, c are the sides of the triangle and s equal to the half of its perimeter s=(a + b + c)/2.
For P (2,1,0) and Q (1,3,1) and R (1,-1,2) a=||P-Q||, b=||Q-R||, c=||P-R||
a= SQRT( (2-1)^2+( (1-3)^2( (0-1)^2)=SQRT(1+4+1)=SQRT(6)=2...
Similarly you can compute b, c and s.
Substitute into A= SQRT (s(s-a)(s-b)(s-c)) to get your answer.
Note: a=||P-Q||=|PQ| - absolute value of vector PQ

Method #2
Using vector math length of the cross product can be interpreted as the unsigned area of the parallelogram having PQ and RQ as sides:
A= ||PQ X RQ|| for the triangle will be A= ||PQ X RQ|| /2 (see also ref 4 for the vector product)
Where:
PQ=(2-1)i+(1-3)j+(0-1)k = i - 2j – k
RQ=(1-1)i+(-1-3)j+(2-1)k=-4j+k
The vector product PQ x RQ=-6i – j -4k
The magnitude of this vector = ||PQ X RQ||=
=SQRT(36+1+16)=SQRT(53)
A= SQRT(53)/2

Method # 3
Refer to ref #2 Height of triangle h = ||PQ X RQ||/||RQ||
Then A= ||RQ|| h/2 = ||RQ||(||PQ X RQ||/(2||RQ||))= ||PQ X RQ|| /2 (which is the same as in method# 2)

Problem #2
The line through P (2,1,0) and Q (1,3,1)
x=x1+x2(t) y=y1+y2(t) z=z1+z2(t)
and so x=2+ t y=1+3t z=t
Or
(2+t, 1+3t, t)

Problem #3
We know that a vector normal to the plane on which the line in question lies can be found by using a vector product. See ref (4) using results in method 2 of problem #1 we have
PQ x RQ=-6i – j -4k
The family of planes is -6x - y- 4k =C where C for any point on the plane say P(2,1,0) is equal C=–6(2)-(1) – 4(0)= -13

So the equation for the plane -6x - y- 4z =-13 or 6x + y + 4z =13 is also fine.

Iff S (1,1,1) is on the plane the equation must be satisfied. Substitute S we have 6+1+4 = 11 not 13 so the point is not on the plane (6x + y + 4z =13).

To determine the vector equation of the line joining S to L which is perpendicular to L we find a projection of any vector trough point S onto unit normal to the plane upon which L lies.
Unit normal =n= PQ x RQ/| PQ x RQ| =-6i – j -4k/ SQRT(53)

Let B=i + j + k (using S) then V the vector in question is

V= (B dot n ) n= (-6 – 1 - 4) (-6i – j -4k)/ 53 =(-11/53) (-6i – j -4k)
Note: ‘dot’ is the dot product.

Well this is it.

I hope it helps.

Answer 2

1. Compute the vectors PQ=<-1,2,1> and PR=<-1,-2,2>. The area of the triangle is 1/2 of the length of the cross product PQxPR. The cross product is <6,1,4>, so the length is sqrt(53). THe area is sqrt(53)/2.

2. The vector equation of the line will be =P+t*PQ=<2-t, 1+2t, t>

3. If <1,1,1>=<2-t, 1+2t,t>, we would need t=1, which doesn't work. The vector from <1,1,1> to <2-t,1+2t, t> is <1-t, 2t, -1+t>. The value of t which makes this vector perpendicular to PQ is t=1/3 (use dot products to find this). This gives a vector from <1,1,1> to the point on the line as <2/3, 2/3, -2/3>, so the euqation of the perpendicular line is =<1+2t/3, 1+2t/3, 1=2t/3>

Answer 3

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Answer 4

1. find the area of a point? hmm.. zero?... or did you mean the triangle PQR?

ok... area of a triangle in 3d space is still the same as in 2d space.. namely... (alt * length)/2

you will need to calculate the length of the sides... and then find the altitude and the length...

L = sqrt[(x1-x0)^2 + (y1-y0)^2 + (z1-y0)^2] ... or something like that...

I'll let you do the rest..

Answer 5

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Answer 6

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