# Math Help - Vectors

1. ## 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 $C$.

2. you forgot to post a question.

also this problem looks similar to the this one

Bobak

3. 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.

4. 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 $C$.'

5. 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?

6. Hello,

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