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# Advanced Math/Lines and planes in space

Question

vectors
QUESTION: Consider the line L :(x+ 10)/(-6)=(y-16)/(-2)= (z+ 3)/6

and the plane pi: 3x -3y + 2z = 4.

(a) Explain why L and pi are parallel.
(b) Of all the lines L_2 inside which are parallel to L, find a vector equation for the one which is closest to L.

Private:   No
Subject:   linear algebra
Question:       vectors

Consider the line L :(x+ 10)/(-6)=(y-16)/(-2)= (z+ 3)/6

and the plane pi: 3x -3y + 2z = 4.

(a) Explain why L and pi are parallel.
(b) Of all the lines L_2 inside which are parallel to L, find a vector equation for the one which is closest to L.

(x+ 10)/(-6)=(y-16)/(-2)= (z+ 3)/6

a) L: x = -6t + 10;  y = -2t + 16;   z = 6t - 3

DV = <-6,-2,6>

NV to pi = <3,-3,2>

Dot those.

........................................
b) Pick a point on  L, such as P(10,16,-3)

Find the point Q IN pi that is the result of 'dropping the perpendicular' to pi from P.  That means, along the DV of pi.

Q = P + s NV
Q = (10,16,-3) + s<3,-3,2>

But Q in pi, so Q must satisfy the equation of pi.  That will let you solve for s and thus find the coordinates of Q.  Use that, plus the DV for L, to get your vector equation.

[an error occurred while processing this directive]---------- FOLLOW-UP ----------

QUESTION: I'm not quite sure by what you mean by this: "But Q in pi, so Q must satisfy the equation of pi.  That will let you solve for s and thus find the coordinates of Q."

What i have done so far is:

Q = (0,0,2)
Q = (10,16,-3) + s<3,-3,2>
(0,0,2) = (10,16,-3) + s<3,-3,2>
(-10,-16,5) = s<3,-2,3>

i don't know if this is the right way of doing the problem. Am i doing it right? I undrstand the part saying that when i find the point Q that i have to dot prodcut it afterwards with the DV of L. Thanks for any help.

QUESTION: I'm not quite sure by what you mean by this: "But Q in pi, so Q must satisfy the equation of pi.  That will let you solve for s and thus find the coordinates of Q."

Q is in pi, so Q must satisfy the equation of pi, 3x -3y + 2z = 4.

and  x = 10 + 3s,  y = 16 - 3s,  z = -3 + 2s

Substitute and solve for s.  Use that value to write Q, the nearest point to the given point on L, then use this Q and the DV of L to write your equation.

What i have done so far is:

Q = (0,0,2)   <<<<<<<<<<<<< what's this?
Q = (10,16,-3) + s<3,-3,2>
(0,0,2) = (10,16,-3) + s<3,-3,2>
(-10,-16,5) = s<3,-2,3>

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