Show that the line with parametric equations x = 6 + 8t, y = −5 + t, z = 2 + 3t does not intersect the plane with equation 2x − y − 5z − 2 = 0.

Hey there! Let's see if these two entities, the line and the plane, can be friends or if they're meant to go their separate ways.

To determine if they intersect, we need to find the values of 't' that satisfy both the line and the plane equations. So, let's substitute the equations of the line into the equation of the plane:

2(6 + 8t) − (−5 + t) − 5(2 + 3t) − 2 = 0

Now, let's simplify and solve for 't':

12 + 16t + 5 - t - 10 - 15t - 2 = 0
16t - t - 15t + 12 + 5 - 10 - 2 = 0
0t + 5 = 0

Uh-oh! 0t + 5 = 0 means that 't' can be any real number! This indicates that there are infinitely many values of 't' that satisfy both equations.

So, it seems like the line and the plane are destined to meet and intersect somewhere, my friend. They're not going their separate ways after all!

I hope that clarifies things for you! Let me know if there's anything else I can help you with.

To determine whether the line and the plane intersect, we need to determine if there is a point that satisfies both the line's parametric equations and the equation of the plane.

First, let's substitute the parametric equations of the line into the equation of the plane:

2x - y - 5z - 2 = 0

Substituting x = 6 + 8t, y = -5 + t, and z = 2 + 3t:

2(6 + 8t) - (-5 + t) - 5(2 + 3t) - 2 = 0

Simplifying:

12 + 16t + 5 - t - 10 - 15t - 2 = 0

Combining like terms:

-15t + 15 = 0

Dividing both sides by 15:

- t + 1 = 0

Simplifying:

t = 1

Substituting t = 1 back into the parametric equations of the line:

x = 6 + 8(1) = 14
y = -5 + (1) = -4
z = 2 + 3(1) = 5

Therefore, the point that satisfies both the line and the plane is (14, -4, 5) with t = 1.

Since we found a point of intersection, the line with parametric equations x = 6 + 8t, y = −5 + t, z = 2 + 3t does intersect the plane with equation 2x − y − 5z − 2 = 0.

To determine whether the line intersects the plane or not, we need to find a point of intersection between the line and the plane.

Given a line with parametric equations:
x = 6 + 8t
y = −5 + t
z = 2 + 3t

And a plane with the equation:
2x − y − 5z − 2 = 0

To find the point of intersection, substitute the parametric equations into the equation of the plane:

2(6 + 8t) − (−5 + t) − 5(2 + 3t) − 2 = 0

Simplify:

12 + 16t + 5 - t - 10 - 15t - 2 = 0
-16t - 3 = 0

Solve for t:

-16t = 3
t = -3/16

Now substitute this value of t into the parametric equations to find the corresponding values of x, y, and z:

x = 6 + 8(-3/16) = 6 - 12/16 = 6 - 3/4 = 21/4
y = -5 - 3/16 = -80/16 - 3/16 = -83/16
z = 2 + 3(-3/16) = 2 - 9/16 = 23/16

Therefore, the point of intersection is (21/4, -83/16, 23/16).

Since we found a point of intersection, the line intersects the plane. Therefore, the statement that the line does not intersect the plane is incorrect.

well let's just substitute ...

2(6+8t) - (-5+t) - 5(2+3t) = 0
12 + 16t + 5 -t - 10 - 15t = 0
7 = 0

This was false when I went to school , and I don't think it has changed since then.

So, since we end up with a contradiction, (7=0), there is no solution, and the line does not intersect with the plane.

or

direction of the line is <8,1,3>
normal to the plane is <2, -1, -5>

<8,1,3> dot <2, -1, -5>
= 16 -1 - 15 = 0

thus the line and the normal are perpendicular, or
the line is parallel to the plane.
However, my second method would not be complete, since our line could lie on the plane itself. So we still have to perform the calculation of my first method to show it does NOT lie on the plane