Question

Multivariable Calculus help

Answer 1

number 38

Answer 2

@agent0smith @thephysicsman @ageta

Answer 3

Be back in an hour, gotta catch the bus before it stops running for the night.

Answer 4

Q is dots not a line?

Answer 5

@legomyego180 were you able to find the equation of the line L or plane p ?

Answer 6

I'll come back to this if i have time later

Answer 7

Problem 38)
Parametric equation of a line passes through \(P_1(4,-3,1),P_2(2,-2,3)\)

direction vector: \(\vec d= (-2,1,2)\), for this step, just take \(P_2-P_1\)

So, the parametric equation is
\(x=4-2t\\y=-3+t\\z=1+2t\)

To get it, you just pick a point ( I pick P1\) and combine with \(\vec d\)

Answer 8

Now, equation of a plane contains \(Q_1(2,0,-4),Q_2(1,2,3),Q_2(-1,2,1)\)

Find normal vector n by take cross product of \(Q_1Q_2,~and~Q_1Q_3\). You do this step

I got \(\vec n= (-4,-16,4)\)

hence, the equation of the plane is

\(-4(x-2)-16y+4(z+4)=0\)

\(-4x+8-16y+4z+16=0\) \(\rightarrow\color{red}{ -4x-16y+4z=-24}\)

Answer 9

Now, you plug the values x, y, z of the line to equation of the plane, you get

\(-4(4-2t)-16(-3+t)+4(1+2t) =-24\) to solve for t.

After getting value of t, plug back to parametric equation of the line to find x, y, z.

Good luck

Answer 10

To my calculation, you don't have any t satisfies the equation. hence, the line and the plane don't intersect. dat sit