Vectors

**Simplicity** · June 7th 2008, 10:18 AM

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 $C$ .

**bobak** · June 7th 2008, 10:41 AM

you forgot to post a question.

also this problem looks similar to the this one

Bobak

**o_O** · June 7th 2008, 10:46 AM

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.

**Simplicity** · June 7th 2008, 10:47 AM

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 $C$ .'

**Simplicity** · June 7th 2008, 10:48 AM

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?

**Moo** · June 7th 2008, 10:54 AM

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

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