Find vector that is orthogonal to (-1,2,2)

[jwpaine]

Find vector that is orthogonal to (-1,2,2)

I begin by doing the cross-product of (-1,2,2) x (a,b,c) = (2c - 2b)i + (-c - 2a)j + (-b - 2a)k => (2c - 2b, -c - 2a, -b - 2a) which, when values for a,b and c are substituted in, should yield a new vector that is perpendicular to the original vector.

Yet, for instance, when I substitute {a = 2, b = 3, c = 1} => (2c - 2b, -c - 2a, -b - 2a) => (-4, -5, 7) which, when doing the dot product of this vector with the original: (-1,2,2) . (-4, -5, 7) does not equal 0, thus these two vectors are not perpendicular. I have used this method before, why is it not working in this case? Perhaps I made a simple mistake?

Thanks,
John

 

[jwpaine]

I of course could just do the dot product to begin with:

(-1,2,2)(a,b,c) = 0
-a + 2b + 2c = 0

Let a = 2, b = 3

-2 + 4 + 2c = 0 ==> c = -1

Thus (-1,2,2) is perp to (2,3,-1)

That works, and is simpler, but what did I do wrong before with the cross product?

Thanks,
John

 

[Pklarreich]

I of course could just do the dot product to begin with:

(-1,2,2)(a,b,c) = 0
-a + 2b + 2c = 0

Let a = 2, b = 3

-2 + 4 + 2c = 0 ==> c = -1

Thus (-1,2,2) is perp to (2,3,-1)

That works, and is simpler, but what did I do wrong before with the cross product?

Thanks,
John

Simple. You blew the arithmetic. I get, for the cross-product:

i j k
-1 2 2
a b c

= (2c - 2b, 2a + c, - b - 2a)

Now take a = 2, b = 3, c = 1, as you did, and get:

(2 - 6, 4 + 1, -3 - 4)
(-4, 5,-7)

Now do:
(-4, 5,-7) dot (-1,2,2) = 4 + 10 - 14 = 0
and it works.