Question

find the equation of the a line passing through the point (2, 1, 3) and perpendicular to the line (x-1)/1=(y-2)/2=(z-3)/3 and x/-3=y/2=z/5

Answer 1

find the equation of the a line
passing through the point (2, 1, 3) and
perpendicular to the line

(x-1) (y-2) (z-3)
--- = ---- = ----
1 2 3

and

x y z
--- = --- = ---
-3 2 5

Answer 2

need the direction ratios of the line
using the condition of perpendicularity,
l+2m+3n=0
-3l+2m+5n=0

Answer 3

cross your vector parts to get the normal vector

Answer 4

(l,m,n) are the direction ratios of the required line

Answer 5

<1,2,3> x <-3,2,5> = normal vector right?

Answer 6

n<a,b,c> ; at point P(x,y,z)

x-Px y-Py z-Pz
----- = ----- = ----- gets your line
a b c

Answer 7

(l,m,n)=(2,-7,4)

Answer 8

line is
x-2/2=y-1/-7=z-3/4

Answer 9

i can make a matrix in my ti83, but cant figure out how to math it in lol

Answer 10

\[\left|\begin{array}c i&j&k\\1&2&3\\ -3&2&5 \end{array}\right|\]
\[(2.5-2.3)i - (1.5-(-3).3)j+(1.2-(-3.2))k\]
\[4i-14j+8k\]
\[2<2,-7,4>\]
right?

Answer 11

yay!! ... we match

Answer 12

:P

Answer 13

another way to write the line equation is:

x = 2 +2t
y = 1 -7t
z = 3 +4t

Answer 14

yes, symmetric form :)

Answer 15

thank u very much