Find values so that the vector has the specified angle

[Calpalned]

Homework Statement

Find the values of x such that the angle between the vectors <2, 1, -1>, and <1, x, 0> is 45 degrees.

Homework Equations

<2, 1, -1> dot <1, x, 0> = 2 + x
|first vector| = (6)^0.5
|2nd vector| = (1 + x²)^1/2
Angle = cos^-1(a dot b)/(|a||b|)
Cosine of 45 is the root of two divided by two.

The Attempt at a Solution

I set (2 + x) divided by 6^0.5 and (1 + x²)^0.5 equal to 1 divided by the root of 2.
I eventually get 2√2 + x√2 = √6√1 + x²
Am I on the right track?

 

[Mark44]

Homework Statement

Find the values of x such that the angle between the vectors <2, 1, -1>, and <1, x, 0> is 45 degrees.

Homework Equations

<2, 1, -1> dot <1, x, 0> = 2 + x
|first vector| = (6)^0.5
|2nd vector| = (1 + x²)^1/2
Angle = cos^-1(a dot b)/(|a||b|)
Cosine of 45 is the root of two divided by two.

The Attempt at a Solution

I set (2 + x) divided by 6^0.5 and (1 + x²)^0.5 equal to 1 divided by the root of 2.

Or equivalently, $\frac{2 + x}{\sqrt{6}\sqrt{1 + x^2}} = \frac{\sqrt{2}}{2}$

I eventually get 2√2 + x√2 = √6√1 + x²

I'm not sure about the above, although it might be right. I get a quadratic with all integer coefficients.
Can you show what you did going from your equation above to this one?

Am I on the right track?