Problem with vectors 2

1. Jan 7, 2013

MMCS

Three vectors

a = xi - 3j + 9K
b = -3i + xj - 10K
c = 9i - 10j + xk

Find the largest value of x for which the magnitude of the resultant is equal to 17.
I am given the correct answer of 8.564 but i dont know how to get that.

I have no working out to show for this as i dont know where to start, but some advice would be helpful

Thanks

2. Jan 7, 2013

haruspex

How do you work out the resultant of two vectors?

3. Jan 7, 2013

HallsofIvy

Staff Emeritus
I'm not sure that mentioning the "resultant of two vectors" will help here at all- except, of course, to point out that it is just as easy to find the resultant of three vectors!

MMCS, do you understand that the "resultant" of any number of vectors is just their sum? Do you understand that you can sum vectors "component wise"- that is just by adding the i, j, and k components separately? What do you get for the resultant of these three vectors? (There will, of course, be an "x" in each component.)

What is the magnitude of that resultant? That will be a forumula with terms involving "x". Set that equal to 17 and solve for x.

4. Jan 7, 2013

MMCS

ok so once i complete the addition of i, j and k and then square the brackets and squared 17 to eliminate square root i get

(x^2 + 12x + 36) + ( X^2 + 14x + 49) + (x^2 - 2x + 1) = 289

x^6 + 24x + 86 = 289

Is this correct? can this be solved?

5. Jan 7, 2013

scurty

You summed the j hat components wrong. Also, $x^2 + x^2 + x^2 = 3 x^2$

6. Jan 7, 2013

haruspex

You shouldn't have an x^6 there. try that last step again.

7. Jan 7, 2013

MMCS

Oh right i never noticed that, good spot, so now i have,

3x^2 - 16x + 206

8. Jan 7, 2013

scurty

Yes. Assuming you didn't subtract the 289 yet.

9. Jan 7, 2013

MMCS

Thats it, finally got it, Thanks for you help!!