Question

Let E be the point where the line l intersects the plane p. Find, in the scalar parametric equation for the line, the value of the parameter which corresponds to the point E and hence find the co-ordinates of this point.

Answer 1

"scalar parametric"?
In 3D, there is no scalar equation of a line because no unique normal.
Do you just mean parametric (x=x_0+at, etc)?

Answer 2

yes that's the one.
Also i forgot to include the information i already worked out for the questions before this one which i should probably specify!

equation of the plane -x-y+4z=3
equation of normal vector to the plane n=-i-j+4k
scalar parametric equations of a line perpendicular to the plane which passes through the point (-1,1,1) ... (x=-1-t),(y=1-t),(z=1+4t)

Answer 3

the line (x=-1-t), (y=1-t), (z=1+4t) is the line i'm talking about in the question

Answer 4

given a line and a plane, solve for t that will yield point of intersection
line:
\[x = x_0+\alpha t\]
\[y=y_0+\beta t\]
\[z =z_0 +\gamma t\]
plane:
\[ax+by+cz+d = 0\]
by plugging in line equations into plane equation you can find point that is both in plane and on line
then solving for t
\[t = \frac{-(ax_0+by_0+cz_0+d)}{a \alpha +b \beta +c \gamma}\]

Answer 5

thankyou! but what does aα+bβ+cγ correspond to?

Answer 6

they are all scalar coefficients
a,b,c refer to plane coefficients
alpha,beta,gamma refer to line coefficients

Answer 7

oh in your case, since line is perpendicular
a = alpha
b = beta
c = gamma

Answer 8

that is fantastic and so helpful, thank you! you've saved me! would you mind helping with another question if you've got time?

Answer 9

sure i can take a look at it

Answer 10

line 1:
\[x=1+2r\]
\[y=1+2r\]
\[z=-3+r\]

line 2:
\[x=-2\]
\[y=-s\]
\[z=2+s\]

Find the unit vector nˆ with negative i component which is perpendicular to
both line1 and line2

i only know how to take the cross product of ordinary vectors to get a unit vector, i'm not sure what to do with parametric equations?

Answer 11

ok for parametric equations , look at directional vectors which are the "t" coefficients
if the dot product of 2 vectors equals 0 then they are perpendicular
line 1: 2i+2j+k
line 2: 0i-j+k
let the unit vector be , ai +bj +ck
where a <0 and a^2 +b^2 +c^2 = 1 (unit vector)

now apply dot product
2a+2b+c = 0
0 -b +c = 0

find (a,b,c) that satisfies all conditions

Answer 12

i have to get off OS now...here is the solution
\[<a,b,c> = <-\frac{3\sqrt{17}}{17},\frac{2\sqrt{17}}{17},\frac{2\sqrt{17}}{17}>\]

Answer 13

Thanks so much I really was stuck and you have helped a great deal I am in your debt !

Answer 14

your very welcome