# Vectors

• Jun 7th 2008, 10:18 AM
Simplicity
Vectors
Relative to a fixed origin, $O,$ the line $l$ has the equation:

$r = (\b i + 7\b j - 5 \b k) + \lambda(3 \b i - \b j + 2 \b k)$

The point $C$ lies on $l$ and is such that $OC$ is perpendicular to $l$.

EDIT: The question is - Find the coordinates of $C$.
• Jun 7th 2008, 10:41 AM
bobak
you forgot to post a question.

also this problem looks similar to the this one

Bobak
• Jun 7th 2008, 10:46 AM
o_O
Let this point be $C(a,b,c)$ and $\vec{OC} = (a,b,c)$

Since C is on the line, we can see that: $a = 1 + 3\lambda, b = 7 - \lambda, c = -5 + 2\lambda$

Since $\vec{OC} \perp (3, -1, 2)$, then $\vec{OC} \cdot (-3,-1,2) = 0$:

$-3a - b + 2c = 0$

Convert a,b, c into terms of lambda and solve for it. Then, plug it in back to the equation of your line to get $\vec{r}$ which should correspond to your point as it originates form the origin.
• Jun 7th 2008, 10:47 AM
Simplicity
Quote:

Originally Posted by bobak
you forgot to post a question.

also this problem looks similar to the this one

Bobak

Ooop, here it is...

'Find the coordinates of $C$.'
• Jun 7th 2008, 10:48 AM
Simplicity
Quote:

Originally Posted by o_O
Let this point be $C(a,b,c)$ and $\vec{OC} = (a,b,c)$

Since C is on the line, we can see that: $a = 1 + 3\lambda, b = 7 - \lambda, c = -5 + 2\lambda$

Since $\vec{OC} \perp (3, -1, 2)$, then $\vec{OC} \cdot (-3,-1,2) = 0$:

$-3a - b + 2c = 0$

Convert a,b, c into terms of lambda and solve for it. Then, plug it in back to the equation of your line to get $\vec{r}$ which should correspond to your point as it originates form the origin.

So, if it is perpendicular it is equated to zero. What about if it was parallel?
• Jun 7th 2008, 10:54 AM
Moo
Hello,

Quote:

Originally Posted by Air
So, if it is perpendicular it is equated to zero. What about if it was parallel?

The two vectors would be proportional (with a common ratio), because they have the same direction :)