1. Find symmetric equations for the line that passes through the point (4, −4, 8) and is parallel to the vector −1, 4, −3

(b) Find the points in which the required line in part (a) intersects the coordinate planes.
point of intersection with xy-plane

point of intersection with yz-plane

point of intersection with xz-plane

2. Find an equation for the plane consisting of all points that are equidistant from the points

(−6, 4, 1) and (2, 6, 5).

3. Find an equation of the plane.

The plane that passes through the point (−2, 1, 1) and contains the line of intersection of the planes

x + y − z = 3 and 4x − y + 5z = 5

The answer is below

Step-by-step 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)

r and t.

both are 64 degrees.

an

y = 6x -3y -12

y +3y +12 = 6x

=> . x = { 2y / 3 +2 } this is equation of line in terms of x

=> . if line meets the x-axis the co-ordinates of meeting point would be ( x , 0 )

=> . x = 2 ( 0 ) / 3 + 2

=> . x = 2

so, the point is ( 2 , 0 )