Question

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.

Answer 1

holy good luck with that!

Answer 2

-_-, thanks lol

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

i as in unit vector \(\vec{i}\)

Answer 9

How so?

Answer 10

Of course. Thank you :)

Answer 11

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

Answer 12

7/9

Answer 13

maybe I'm wrong in sth

Answer 14

d= -54/11

Answer 15

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

Answer 16

lol, that's right, sr

Answer 17

lol ok, thanks :)

Answer 18

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