Given Information:
We are given two vectors:
1. v1 = ai^ + bj^ + ck^ which makes an angle of 60 degrees with vector v2.
2. v2 = i^ + j^ + √2k^

To Find:
The number of values of theta between 0 and 360 degrees.

Explanation:
To find the number of values of theta between 0 and 360 degrees, we need to understand the relationship between the given vectors v1 and v2. Let's break down the problem step by step.

Step 1: Finding the Magnitude of v1 and v2
The magnitude of a vector can be found using the formula:
|m| = √(a^2 + b^2 + c^2)

Applying this formula to v1 and v2, we get:
|m1| = √(a^2 + b^2 + c^2)
|m2| = √(1^2 + 1^2 + (√2)^2)
|m2| = √(1 + 1 + 2)
|m2| = √4
|m2| = 2

Step 2: Finding the Dot Product of v1 and v2
The dot product of two vectors can be found using the formula:
v1 · v2 = |v1| |v2| cos(theta)

Here, theta is the angle between the two vectors.

Let's substitute the values of v1, v2, and theta into the formula:
v1 · v2 = |v1| |v2| cos(60)
v1 · v2 = |v1| |v2| (1/2)

We know that the dot product of two vectors is given by:
v1 · v2 = a1a2 + b1b2 + c1c2

Let's equate the two expressions for v1 · v2:
a1a2 + b1b2 + c1c2 = |v1| |v2| (1/2)

Step 3: Simplifying the Equation
Since v1 · v2 = a1a2 + b1b2 + c1c2, we can rearrange the equation as follows:
a1a2 + b1b2 + c1c2 = (a^2 + b^2 + c^2) * (1/2)

Expanding the equation, we get:
a1a2 + b1b2 + c1c2 = (a^2/2) + (b^2/2) + (c^2/2)

Comparing the coefficients of the variables, we have:
a1a2 = a^2/2
b1b2 = b^2/2
c1c2 = c^2/2

From these equations, we can conclude that a1a2 = a^2/2, b1b2 = b^2/2, and c1c2 = c^2/2.

Step 4: Finding the Values of Theta
Since a1a2 =