0.3 is your answer to your question
In order to do this, you must first find the "cross product" of these vectors. To do that, we can use several methods. To simplify this first, I suggest you compute:
‹1, -1, 1› × ‹0, 1, 1›
You are interested in vectors orthogonal to the originals, which don't change when you scale them. Using 0,-1,1 is much easier than 6s and 7s.
So what methods are there to compute this? You can review them here (or presumably in your class notes or textbook):
http://en.wikipedia.org/wiki/Cross_produ...
In addition to these methods, sometimes I like to set up:
‹1, -1, 1› • ‹a, b, c› = 0
‹0, 1, 1› • ‹a, b, c› = 0
That is the dot product, and having these dot products equal zero guarantees orthogonality. You can convert that to:
a - b + c = 0
b + c = 0
This is two equations, three unknowns, so you can solve it with one free parameter:
b = -c
a = c - b = -2c
The computation, regardless of method, yields:
‹1, -1, 1› × ‹0, 1, 1› = ‹-2, -1, 1›
The above method, solving equations, works because you'd just plug in c=1 to obtain this solution. However, it is not a unit vector. There will always be two unit vectors (if you find one, then its negative will be the other of course). To find the unit vector, we need to find the magnitude of our vector:
|| ‹-2, -1, 1› || = √( (-2)² + (-1)² + (1)² ) = √( 4 + 1 + 1 ) = √6
Then we divide that vector by its magnitude to yield one solution:
‹ -2/√6 , -1/√6 , 1/√6 ›
And take the negative for the other:
‹ 2/√6 , 1/√6 , -1/√6 ›
The area of a triangle for one of the legs being 3 inches and the hypotenuse being 9 inches is 12.727 square inches
<em><u>Solution:</u></em>
Given that to find area of a triangle for one of the legs being 3 inches and the hypotenuse being 9 inches
From given information,
Let "c" = hypotenuse = 9 inches
Let "a" = length of one of the leg of triangle = 3 inches
To find: area of triangle
<u><em>The area of triangle when hypotenuse and length of one side of triangle is given:</em></u>

Where, "c" is the length of hypotenuse
"a" is the length of one side of triangle
Substituting the given values we get,


Thus area of triangle is 12.727 square inches
The GCF of 8 and 40 is 8. Hope that helps.