Further Maths Vectors Help

The line l has equations (x-1)/2 = (y+2)/3 = (z-7)/5. The line n has equation 4x - y - z = 8.

(i) Show that l is parrallel to n but does not lie on n.

(ii) The point A(1, -2, 7) is on l. Write down the vector equation of the line through A whihc is perpendicular to n. Hence find the position vector of the point on n which is closest to A

No idea where to start I have never seen a line equation in that form before.

Reply 1

the bear

20

Original post by amarooDan

The line l has equations (x-1)/2 = (y+2)/3 = (z-7)/5. The line n has equation 4x - y - z = 8.

(i) Show that l is parrallel to n but does not lie on n.

(ii) The point A(1, -2, 7) is on l. Write down the vector equation of the line through A whihc is perpendicular to n. Hence find the position vector of the point on n which is closest to A

No idea where to start I have never seen a line equation in that form before.

this first line has direction vector 2i + 3j + 5k and contains the point ( 1, -2, 7 ).

4x - y - z = 8 looks more like the equation of a plane.

Reply 2

Can'tDecideAName

8

For part (i) you would have to show that there is no point on the line that is also in the plane and that the planes normal is perpendicular to the direction vector of the line.

Reply 3

mathcool

9

Hello , when it is said a line, it means line (although the form of the equation makes it look like a plane).

How does a line can be formed?

say a line (named line l1) is given, <r> = <a> (point vector) + t (parameter) x (a direction vector) (normal vector form of a line).

now, say, you are asked to form an equation of the line (named l2) through point vector <c>, perpendicular to l1.

How would you form it?

Take a general point <r> on l2, so the (direction) vector (<r> - <c>) is on the line l2, and since l1 and l2 are perpendicular,

(<r> - <c>) (dot product with) = 0, i.e. direction vector l2 dot product with , direction vector of l1, is zero.

i.e. <r> (dot product with) = <c> (dot product with) = p (real number)

Now, this equation is of a line, although it looks like a plane. (infact a plane, because multiple lines can go thorough <a> that are perpendicular to l1). However, all those lines will form a plane (whose normal will be .) because those lines are perpendicular to l1.

However, on a specific instance of a given <r> (on l2, or in the plane), the equation becomes a line (i.e. subset of the plane).

Now approach the question with this understanding.

(edited 1 month ago)

Reply 4

chavvo

7

Original post by mathcool