Question

−1

x−5
​
= 3
y+3
​
= −2
z−9
​
−(x−5)=3(y+3)=−2(z−9)
x+5= 3
y+3
​
= −2
z−9
​
−1
x+5
​
= 3
y−3
​
= −2
z+9
​
−(x+5)=3(y−3)=−2(z+9).
​
(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 zplane point of intersection with xet-plane

Answer 1

To find the points at which the line intersects the coordinate planes, we can set one variable at a time to zero and solve for the other two. Let's start with the xy-plane, which corresponds to z = 0.

1. Intersection with the xy-plane (z = 0):

Substitute z = 0 into the equations from part (a):

-1x + 5 = 3y + 3

x - 5 = -3y - 3

Solve for x and y:

x = 3y + 3 + 5

x = 3y + 8

So, the line intersects the xy-plane at the point (8, 0, 0).

2. Intersection with the yz-plane (x = 0):

Substitute x = 0 into the equations from part (a):

-0 + 5 = 3y + 3

5 = 3y + 3

Solve for y:

3y = 5 - 3

3y = 2

y = 2/3

So, the line intersects the yz-plane at the point (0, 2/3, 0).

3. Intersection with the xz-plane (y = 0):

Substitute y = 0 into the equations from part (a):

-1x + 5 = -2z - 9

x - 5 = -2z - 9

Solve for x and z:

x = -2z - 9 + 5

x = -2z - 4

So, the line intersects the xz-plane at the point (-4, 0, 0).

These are the points of intersection of the line with the coordinate planes:

- xy-plane: (8, 0, 0)

- yz-plane: (0, 2/3, 0)

- xz-plane: (-4, 0, 0)

Explanation: