## anonymous 3 years ago Find the coordinates of the point where the vector line $$\vec{r} = (6 + \frac{7}{9}d)\vec{i} + (3 + \frac{4}{9}d)\vec{j} + (2 - \frac{4}{9}d)\vec{k}$$ intersects the plane 3x - 7y + z = 5. I'd like an explanation or better yet, a guide through it rather than just a plain answer.

1. anonymous

holy good luck with that!

2. anonymous

-_-, thanks lol

3. anonymous

you know that: x=6+7/9d, y=3+4/9d, and z=2-4/9d ?

4. anonymous

Yes. I actually think I might have an answer since I asked this yesterday. Could you check it?

5. anonymous

I sence De Moivre's Theorum comming on here =)

6. anonymous

d = -54/11, so (24/11, 9/11, 46/11) Is this right?

7. anonymous

In your question you've mentioned i..... Is that i as in i, or i as in the Square root of -1???

8. anonymous

i as in unit vector $$\vec{i}$$

9. anonymous

How so?

10. anonymous

Of course. Thank you :)

11. anonymous

that 7/9 or what, cant see the problem clearly

12. anonymous

7/9

13. anonymous

maybe I'm wrong in sth

14. anonymous

d= -54/11

15. anonymous

Here's my work if it helps... $\vec{r} = (6 + \frac{7}{9}d)\vec{i} + (3 + \frac{4}{9}d)\vec{j} + (2 - \frac{4}{9}d)\vec{k}$$3(6 + \frac79d) - 7(3 + \frac49d) + (2 - \frac49d) = 5$$18 + \frac{21}{9}d - 21 - \frac{28}{9}d + 2 - \frac49d = 5$$-\frac{11}9d = 6$$d = 6 \times -\frac9{11}$$d = -\frac{54}{11}$

16. anonymous

lol, that's right, sr

17. anonymous

lol ok, thanks :)

18. anonymous

So just plug in to find coordinate. Is this cal 3?