# Orthogonal /parallel or neither

• Nov 29th 2008, 08:10 AM
Faz
Orthogonal /parallel or neither
a)Determine whether the lines (x-1)/4=(y+1)/2=(z-3)/-1 and
(x-1)/1=(y+1)/2=(z-3)/8 are orthogonal , parallel or neither.

b)Determine whether the following lines (x-3)/3 = (y+1)/-2 = (z-3)/5 contains the point (6,-3,8)

c)Find equation of the line containing the given point (5,3,4) and parallel to the vector
-> -> ->
2 i + 5 j - 8 K

plse assist
• Nov 29th 2008, 11:33 AM
Soroban
Hello, Faz!

These are very basic questions.
. . Exactly where is your difficulty?

Quote:

a) Determine whether the lines: . $\frac{x-1}{4}=\:\:\frac{y+1}{2} \:=\:\frac{z-3}{-1}\;\text{ and }\;\frac{x-1}{1}=\:\:\frac{y+1}{2}\:=\:\frac{z-3}{8}$
are orthogonal, parallel or neither.

The direction vectors of the lines are: . $\vec u \:=\:\langle 4,2,-1\rangle\:\text{ and }\:\vec v \:=\:\langle 1,2,8\rangle$

Since $\vec u$ is not a scalar multiple of $\vec v$, they are not parallel.

We find that: . $\vec u \cdot \vec v \:=\:\langle 4,2,\text{-}1\rangle\cdot\langle1,2,8\rangle \:=\:(4)(1)(+(2)(2) + (\text{-}1)(8) \:=\:0$

Therefore: . $\vec u \:\perp \;\vec v$

Quote:

b) Determine whether the line . $\frac{x-3}{3}\:=\:\frac{y+1}{-2} \:=\:\frac{z-3}{5}$ .contains the point (6,-3,8)
Do the coordinates satisfy the equation? . . . Plug them in!

Quote:

c) Find an equation of the line containing the given point (5,3,4)
and parallel to the vector: . $2\vec i + 5\vec j - 8\vec k$

We have a point $P(5,3,4)$ and the direction vector $\vec v \:=\:\langle 2,5,\text{-}8\rangle$

. . . Well?