Determine whether the lines L1:x=16+4t, y=6+2t, z=−1+t and L2:x=−6+5ty=−8+4tz=−12+4t intersect, are skew, or are parallel. If they intersect, determine the point of intersection; if not leave the remaining answer blanks empty

The point of interception is (x y z)

(4 0 -4)

Step-by-step explanation:

L1: x=16+4t, y=6+2t, z=−1+t

y=mx+c, this is the equation of a line, where m is the slope and c is the intercept

Then, The slope of the line 1 is

L1={4 2 1)

Also, for line 2

L2: x=−6+5t. y=−8+4t. z=−12+4t

The slope of the line is

L2=(5 4 4)

Since those two vectors are not proportional to each other, L1 and L2 should be skew or intersecting. To check whether they are intersecting or not, we need to

check the existence of (t0, s0) such that.

x=x. y=y. z=z

Therefore

x=16+4t1=-6+5t2,

4t1-5t2=-22. Equation 1

y= 6+2t1=-8+4t2,

2t1-4t2=-14. Equation 2

z= -1+t1=-12+4t1,

t1-4t2=-11. Equation 3

Solving equation 3 and 2 simultaneously

Divide equation 2 by 2

Then,

t1-2t2=-7. Equation 2

t1-4t2=-11. Equation 3

Subtract equation 3 from 2

2t2=4

Then, t2=4/2

t2=2

from equation 3

t1-4t2=-11. t2=2

Then, t1=-11+4t2

t1=-11+4×2

t1=-11+8=-3

t1=-3

Let check for this in the equation 1

4t1-5t2=-22. t1=-3 and t2=2

4(-3)-5(2)

-12-10=-22

Then the line intercept

The point of interception

Now,

For x

x=16+4t1=-6+5t2,

t1=-3. t2=2

x=16+4(-3)=16-12

x=4

For y,

y=6+2t1=-8+4t2,

t1=-3 and t2=2

y=6+2(-3)

y=0

For z,

z=-1+t1=-12+4t1,

t1=-3 and t2=2

z=−1+t1

z=−1-3

z=-4

The point of interception is (x y z)

(4 0 -4)

(x, y)→(x − 8, y − 7)

step-by-step explanation:

enjoy!

it is a. 4.2 because 7.6-3.4 gets the remaining length of 4.2. i might be wrong but i am pretty sure this is the answer.