Second answer if it still helps
Answer: 25
Step-by-step explanation: got it wrong and got the answer Took 1 for the team
Answer:
Step-by-step explanation:
The normal vector to the plane <em>x</em> + 3<em>y</em> + <em>z</em> = 5 is <em>n</em> = (1, 3, 1). The line we want is parallel to this normal vector.
Scale this normal vector by any real number <em>t</em> to get the equation of the line through the point (1, 3, 1) and the origin, then translate it by the vector (1, 0, 6) to get the equation of the line we want:
(1, 0, 6) + (1, 3, 1)<em>t</em> = (1 + <em>t</em>, 3<em>t</em>, 6 + <em>t</em>)
This is the vector equation; getting the parametric form is just a matter of delineating
<em>x</em>(<em>t</em>) = 1 + <em>t</em>
<em>y</em>(<em>t</em>) = 3<em>t</em>
<em>z</em>(<em>t</em>) = 6 + <em>t</em>
Opposite sides of a parallelogram are equal lengths, so ...
... AB = CD
... 6x +30 = 2y -10
and
... BC = AD
... 2x -5 = y -35
_____
The equations can be put in standard form to get
... 3x -y = -20
... 2x -y = -30
Subtracting the second from the first, we get
... x = 10
Solving for y, we have
... y = 2x+30 = 2·10 +30 = 50
The values of x and y are x=10, y=50.
_____
AB = 6x+30 = 90
BC = 2x-5 = 15