Q/ Find the parametric equations of the line joining the points P1( 1, 1, -1) and P2( 1, 2, 0)?

using the usual rules, you get

(x-1)/(1-1) = (y-1)/(2-1) = (z+1)/(0+1)

so, what do you need to do to make it work here?

To find the parametric equations of the line joining two points, we need to determine the direction vector and a point on that line.

Let's start by finding the direction vector:

Direction vector = P2 - P1
= (1, 2, 0) - (1, 1, -1)
= (1 - 1, 2 - 1, 0 - (-1))
= (0, 1, 1)

Now, let's choose one of the given points, P1, as our reference point. We can use P1(1, 1, -1).

Finally, we can write the parametric equations of the line using the reference point and the direction vector:

x = 1 + 0t
y = 1 + t
z = -1 + t

Therefore, the parametric equations of the line joining the points P1(1, 1, -1) and P2(1, 2, 0) are:
x = 1
y = 1 + t
z = -1 + t

Where t is a parameter that represents any real number.

To find the parametric equations of the line joining two points in 3D space, you need to consider the components of each point and find the equations that describe the movement along each axis.

Let's consider the points P1(1, 1, -1) and P2(1, 2, 0).

For the x-axis, both points have the same x-coordinate, which means that the movement along the x-axis is constant. Therefore, the equation for the x-component is:

x = 1

For the y-axis, the y-components of the two points are different, so there is a change along the y-axis. We can express this change as a parameter t, and the equation becomes:

y = 1 + t

For the z-axis, the z-components of the two points are also different, so there is a change along the z-axis. We can similarly express this change as a parameter t, and the equation becomes:

z = -1 + t

Putting it all together, we have the parametric equations of the line joining the points P1(1, 1, -1) and P2(1, 2, 0) as:

x = 1
y = 1 + t
z = -1 + t

These equations describe the line that connects the two given points in 3D space, where the parameter t represents the movement along the line.