Answer:
Explanation:
a. Assume that A= (a1 a2 a3), B=(b1 b2 b3), C=(c1 c2 c3).
The we need to fine the direction of each vectors
AB= B-A = (b1-a1, b2-a2, b3-a3).
We will also get
AC= C-A= (c1-a1, c2-a2, c3-a3).
Then we will find the cross product of AB and AC. The cross product result gives the normal
b. We are given two lines, then we need to find the direction of the two lines
Equation of a line is in different form
R1= (a1 a2 a3) + λ(b1 b2 b3)
R1= (c1 c2 c3) + λ(d1 d2 d3)
Now, if we have this kind of line, the directions are B (b1 b2 b3) and D (d1 d2 d3), we just need to fine the cross product of the directions and then the product of the cross product gives the normal to the plane.
c. A line and a point that is not on the line,
We need the direction of the line first and we can get it from the equation of the line
Give equation of line
R1= (d1 d2 d3) + λ(b1 b2 b3)
The direction of the line is B =(b1 b2 b3) and the arbitrary point on the line is D= (d1 d2 d3)
E.g give then point as
A= (a1 a2 a3),
The the direction of the point from the line is give as DA=A - B= ( d1-b1 d2-b2 d3-b3).
Then we will fine the cross product of the direction of the line and the direction of the point from the line, this will give the normal of the line and the point in the plane.
d. If two line are parallel, the direction of one will be a scalar product of the other line and when calculating the cross product of this direction it will give zero or thanks vectors.
Same analysis as questions b.
We just need to get the directions and find the cross product
