Respuesta :
Answer:
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)
Answer:
These lines are intersecting at the point (4, 0, -4)
Step-by-step explanation:
We need to determine if the lines L1: x = 16 + 4t, y = 6 + 2t, z = -1 + t and
L2: x = -6 + 5t, y = -8 + 4t, z = -12 + 4t intersect, are skew, or are parallel.
The direction vector of L1 is V1 = <4, 2, 1> and that of L2 is V2 = <5, 4, 4>.
Since there no k ≠ 0 such that kV1 = V2, we conclude that the two vectors are not parallel to each other, L1 and L2 should be skew or intersecting. To check whether they are intersecting or not, we need to solve the system of equations in variables s and t, and see if there is an intersecting point (x_0, y_0, z_0).
x_0 = 16 + 4t = -6 + 5s,
y_0 = 6 + 2t = -8 + 4s,
z_0 = -1 + t = -12 + 4s.
From these, we have the following equations
16 + 4t = -6 + 5s
=> 4t + 5s = -22..........................(1)
6 + 2t = -8 + 4s
=> 2t - 4s = -14.............................(2)
-1 + t = -12 + 4s
=> t - 4s = -11.................................(3)
Eliminate t from (1) and (2), we have
3s = 6
Or s = 2
Substituting s = 2 into (2), we have
2t - 4(3) = -14
t = -3
Since (s, t) = (2, -3) also satisfies (3), we say say (s, t) = (2, -3) is the solution to the system of equations.
We can determine the intersecting points by either putting t = -3 in L1, or t = 2 in L2.
Using L1, the intersection point is (x, y, z) = (16 + 4(-3) = 4, 6 + 2(-3) = 0, -1 + (-3) = -4)
The lines intersect at (4, 0, −4)
