Answer:
3) the lines are coincident
4) (9/11, 1/11, 0) +t(-5/11, 8/11, 1)
Step-by-step explanation:
You want the point of intersection of the given lines, and the line of intersection of the given planes.
3) Lines
Writing two equations in s and t that make L1 have the same point as L2, we have ...
2 -s = -3 +2t ⇒ s +2t = 5 . . . . . . . . . . equating x-values
5 -2s = -5 +4t ⇒ 2s +4t = 10 . . . . . . . equating y-values
By inspection, these are dependent equations such that s = 5-2t. This means every point on L1 is also a point on L2.
The lines are coincident.
4) Planes
Solving the equations for x and y, as in the second part of the attachment, we find:
x = 9/11 -5/11z
y = 1/11 +8/11z
This means we can use z = t and write the vector equation of the line as ...
L = (9/11, 1/11, 0) +t(-5/11, 8/11, 1)
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Additional comment
When we say the equations are dependent "by inspection", it means we can see that multiplying the first equation by 2 gives the second equation. Or, writing the second equation in standard form (removing common factors from the coefficients) makes it identical to the first equation.
In problem 4, we can write the vector equation of the line without fractions as ...
L = (4, -5, -7) +t(-5, 8, 11)
The integer-valued point can be found by combining the two equations to give 8x +5y = 7. We recognize that a solution is x=2·7+5n, y=-3·7-8n, and for n=-2, the values are (x, y) = (4, -5). The corresponding value of z is 4+2(-5)-1 = -7.
