Question

let p(1,3,2) and let L be the line with parametric equation x=2-t,y=-1+2t, z=3+t use the cross product to find the distance between P to L

Answer 1

I have no idea, but I tried to find the plane that passes thru the line and used point and plane distance formula.

Answer 2

Look at your parametric equation.
Can you put it into a form\[\frac{ x-x _{1} }{ a }=\frac{ y-y _{1} }{ b }=\frac{ z-z _{1} }{ c }=t\]

Answer 3

hello?

Answer 4

yup, \[\frac{ x-2 }{ -1 }=\frac{ y+1 }{ 2 }=\frac{ z-3 }{ }\]

Answer 5

1

Answer 6

@science0229

Answer 7

right!

Answer 8

Now, in that form, we know that the vector form is v=<a,b,c>, or <-1,2,1>

Answer 9

Wait, do you know the symmetrical line equation?

Answer 10

ys

Answer 11

ok. then does this make sense to you so far?

Answer 12

yes

Answer 13

Now, the formula for the distance between plane 'n' and line 'v' is known as \[\frac{ n*v }{ |n| }\]
* should be a dot.

Answer 14

is n= (1,3,2)

Answer 15

yes

Answer 16

\[\frac{ 7 }{ \sqrt{14}}\]

Answer 17

"Now, the formula for the distance between plane 'n' and line 'v' is known as"
n is not the plane it is a point

Answer 18

the answer is correct

Answer 19

no, \[\it is \frac{ \sqrt{66} }{ 3}\]

Answer 20

it is that

Answer 21

Wait. Oops.
I thought p meant plane :(

Answer 22

P is the point

Answer 23

I think the idea is this