The required plane can be found by:
1. find two distinct points A, B on the line r.
2. Form two vectors from the point P to the two points on the line. These two vectors will lie in the required plane.
3. find the normal n to the plane by the cross product of the two vectors PA and PB.
4. Determine the plane from the normal n and any point passing through the plane, i.e. P, A or B.
Here are the details:
r : (2,3,4)+μ(-1,0,3)
Set μ to 0 and 1 to get two distinct points A & B.
A(2,3,4)
B(2-1, 3-0, 4+3)=B(1,3,7)
The point P(6,0,2) is in the plane.
Now calculate vectors PA and PB:
PA(2-6,3-0,4-2)=PA(-4,3,2)
PB(-5,3,5)
The normal n to the plane is given by the cross product of PA and PB, namely
n=PA X PB
=(9,10,3)
The plane passing through the line r and the point P is therefore of the form
9x+10y+3z+k=0 ......(1)
Substitute the coordinates of P into the above equation (1) to find k=-60.
The line r=(2,3,4)+µ(-1,0,3) and the point (6,0,2) form which plane: (a) r.(-1,0,3)=10 (b) r.(9,10,3)=60 (c) r.(2,3,4)=29 (d) r=(2,3,4)+µ(-1,0,3)+ù(6,0,2)
Please help!!! Thanks!
2 answers
The plane is flying at an altitude of 8 miles and is 175 miles from the runway, as measured from the ground. What angle will the plane's path make with the runway? Round your answer to the nearest hundredth.