Given information:
- Vector a + Vector b = 2i cap
- Vector a - Vector b = 4j cap

To find: The angle between vector a and vector b.

Approach:
We can solve this problem using vector addition and subtraction properties.

1. Expressing vector a and vector b in terms of their components:
We are given that vector a + vector b = 2i cap. This can be written as:
a1i cap + a2j cap + a3k cap + b1i cap + b2j cap + b3k cap = 2i cap

From this equation, we can equate the i cap components:
a1 + b1 = 2 ...(1)

Similarly, we are given that vector a - vector b = 4j cap. This can be written as:
a1i cap + a2j cap + a3k cap - b1i cap - b2j cap - b3k cap = 4j cap

From this equation, we can equate the j cap components:
a2 - b2 = 4 ...(2)

2. Solving the equations:
From equation (1), we can express b1 in terms of a1:
b1 = 2 - a1

Substituting this value in equation (2), we get:
a2 - (2 - a1) = 4
a2 - 2 + a1 = 4
a2 + a1 = 6 ...(3)

3. Finding the dot product between vector a and vector b:
The dot product between vector a and vector b is given by:
a · b = |a| |b| cos(theta)

We can express vector a and vector b in terms of their components:
a = a1i cap + a2j cap + a3k cap
b = b1i cap + b2j cap + b3k cap

The dot product can be calculated as:
a · b = (a1i cap + a2j cap + a3k cap) · (b1i cap + b2j cap + b3k cap)
= a1b1 + a2b2 + a3b3

4. Calculating the dot product:
Using equation (1), we can express b1 in terms of a1:
b1 = 2 - a1

Substituting this value in the dot product equation, we get:
a · b = a1(2 - a1) + a2b2 + a3b3
= 2a1 - a1^2 + a2b2 + a3b3

5. Using the dot product to find the angle:
We know that a · b = |a| |b| cos(theta)
Therefore, we can write:
2a1 - a1^2 + a2b2 + a3b3 = |a| |b| cos(theta)

From equation (3), we know that a2 + a1 = 6. We can express a2 in terms of a1:
a2 = 6 - a1

Substituting this value in the equation above, we get:
2a1 - a1^2 + (6 - a1)b2 + a3b3 =