Answer:
The answer is below
Explanation:
1
a) The symmetric equations for the line that passes through the point (a, b, c) and is parallel to the vector (e, f, g) is:

Therefore using the above equation to Find symmetric equations for the line that passes through the point (4, −4, 8) and is parallel to the vector (−1, 4, −3) we get:

b)i) The line would intersect the xy plane where z = 0. Hence:

ii) The line would intersect the yz plane where x = 0. Hence:

iii) The line would intersect the xz plane where y = 0. Hence:

2)
Let the points be A(−6, 4, 1) and B(2, 6, 5). Let O be the midpoint of the two points A and B. Therefore O is the average of the x coordinates, y coordinates and z coordinate.

The normal vector (n) in the direction of line between A and B is:
n = AB = B - A = (2, 6, 5) - (−6, 4, 1) = (8, 2, 4)
n = 8x + 2y + 4z
The equation of the plane based on the normal vector and the midpoint 0 is:
Plane = 8x + 2y + 4z = 8(-2) + 2(5) + 4(3) = 6
Therefore:
8x + 2y + 4z = 6
4x + y + 2z = 3
3) The normal vectors to the plane are:

The point of intersection O of the two planes is normal to the normal vectors, hence:

A point that lies on both plane is gotten by substituting z = 0, hence:
x + y - (0) = 3, and 4x - y + 5(0) = 5
x + y = 3 and 4x - y = 5
Solving simultaneously gives x = 8/5, y=7/5
From this two points we get:
AB = (-2-8/5, 1 - 7/5, 1-0) = (-18/5, -2/5, 1)
The vector normal to the plane (n) = (4, -9, -5) * (-18/5, -2/5, 1) = (-11, 14, -34)
