For Part 1: To find the equation of the plane containing the points, we need the normal vector to the plane. We can find it by finding two vectors that lie on the plane and then computing their cross product. How this will work is as follows:
1) If we take the coordinates of Point 2, (4,1,10), and Point 1, (0,0,0) and subtract them, this will make one of the vectors on the plane. It works as follows: 4-0=4, 1-0=1, and 10-0=10 (I'm just subtracting corresponding coordinates). The vector I get is <4,1,10>.
2) I can make a second vector if I take the coordinates of Point 3, (0,6,24), and Point 1 and subtract like with Point 2 and Point 1 earlier. It works as follows: 0-0=0, 6-0=6, and 24-0=24, giving me the vector <0,6,24>.
3) Now I compute the cross product of these two vectors I just obtained.
<4,1,10>X<0,6,24>= (i j k)
det(4 1 10)
(0 6 24)
Using the method of expansion by minors, this determinant is:
i*det(1 10)-j*det(4 10)+k*det(4 1)
(6 24) (0 24) (0 6)
=i*(1*24-6*10)-j*(4*24-0*10)+k*(4*6-0*1)
=i*(24-60)-j*(96-0)+k(24-0)
=-36i-96j+24k
=<-36, -96, 24>
The normal vector to the plane is <-36, -96, 24>.
The general form of a plane equation is a(x-x0)+b(y-y0)+c(z-z0)=0, where a,b, and c are the i, j, and k components of the normal vector and (x0,y0,z0) is a point on the plane. It is easiest to take the point to be Point 1, (0,0,0), so that we can ultimately drop the -x0, the -y0, and the -z0. Plugging in the normal vector components gives me the plane equation -36x-96y+24z=0.
For Part 2: To find the point where the rocket is perpendicular to our plane and 245,500 feet above the plane, it is easiest to take our normal vector and make a unit vector out of it, and then multiply this unit vector by 245,500. Then it will spell out the coordinates of the point with the desired distance.
To find a unit vector (denoted u) for our normal vector, we need to find the magnitude of the normal vector.
||<-36, -96, 24>||=square_root((-36)^2+(-96)^2+(24)^2)
=square_root(1,296+9,216+576)
=square_root(11,088)
11,088 does not have a nice square root. The best thing to do here would be to try to reduce the square root. Since all of the numbers in the original problem, square_root((-36)^2+(-96)^2+(24)^2), were squared, we could reduce the square root by finding the greatest common factor of -36, -96, and 24, which is 12. Since we are squaring -36, -96, and 24, the perfect square factor that we should take out of 11,088 should be 12^2, or 144. In fact, 144*77=11,088, so it works and square_root(11,088)=square_root(144*77)=12*square_root(77).
The unit vector u is then obtained by taking the normal vector and dividing each component by the magnitude we just found:
u=<-36, -96, 24>/(12*square_root(77))=<-3, -8, 2>/square_root(77)
I left the square_root(77) outside rather than writing it three times; it means the same thing in a simpler format.
Now multiply this vector by 245,500 and we will have our point coordinates for where the rocket must start its descent:
245,500*<-3, -8, 2>/square_root(77)=<-736,500, -1,964,000, 491,000>/square_root(77)
By carrying out the division by square_root(77), this is approximately the point (-83,931, -223,819, 55,955). The numbers are very big simply because they are measured in feet. If they were in miles like the next question about the displacement vector wants, then the numbers would be a lot smaller.
On to the next question about the displacement vector: first convert the 245,500 feet and the point we just found with its current coordinates of feet into miles by dividing by 5,280:
245,500/5,280=46.5 (approximately)
-83,931/5,280=-15.9 (approximately)
-223,819/5,280=-42.4 (approximately)
55,955/5,280=10.6 (approximately)
Now, our point is (-15.9, -42.4, 10.6).
To get the displacement vector in miles per hour, we need to find how many hours it takes to travel 46.5 miles at 30 miles per hour. This is obtained simply by dividing 46.5 by 30, which gives 1.55 hours. Divide the coordinates of our point by -1.55 (negative because we need to flip the direction to get the rocket starting at the point and heading to our plane):
-15.9/-1.55=10.3 (approximately)
-42.4/-1.55=27.4 (approximately)
10.6/-1.55=-6.8 (approximately)
The displacement vector is then <10.3, 27.4, -6.8>. By the way, I can use point notation and vector notation interchangeably; they mean virtually the same thing. I hope this helps!
