The line r= (-8,-6,-1) + s(2,2,1), sER, intersects the xz- and yz-coordinate planes at the points A and B, respectively. Determine the length...

We have to find the distance between the points A and B
which are the points of intersection of the line of the line r= (-8,-6,-1) + s(2,2,1),
and the xz and yz-coordinate planes respectively.

The
equation of the xz plane is y = 0 and that of the yz plane is x =
0.

r= (-8,-6,-1) + s(2,2,1) can be written
as

(x + 8)/2 = (y + 6)/2 = (z + 1)/1 =
t

A point on the line can be expressed as (2t - 8 , 2t - 6,
t - 1)

The point A is where the line intersects y =
0

=> 2t - 6 =
0

=> t = 3

The required
point A is (-2, 0, 2)

Similarly for point B, the equation
of the plane is x = 0.

=> 2t - 8 =
0

=> t = 4

The point B
is (0 , 2, 3)

The distance between the points A and B is
sqrt [(x1 - x2)^2+ (y1 - y2)^2 + (z1 - z2)^2]

=> D =
sqrt [ ( 0 + 2)^2 + (2 - 0)^2 +(3 - 2)^2]

=> D =
sqrt (4 + 4 + 1)

=> D = sqrt
9

=> D =
3

The length of the line segment AB = 3
units.