Types Of Vectors

To solve this problem, we need to find the angle between the vectors and , and then evaluate .

First, we calculate the dot product of the given vectors. The dot product formula for two vectors and is given by:

The vectors we are dealing with are:

Compute , knowing that the angle between and is , which gives:

Compute :

Compute :

1. 
   After simplifying:
   

Compute :

1. 
   Calculating further:
   

Finally, use the dot product to find :

Simplify and evaluate :

Therefore, the answer is

90+27

.

Given the conditions:
|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + 2|\vec{b}|^2 , \vec{a} \cdot \vec{b} = 3 , and |\vec{a} \times \vec{b}|^2 = 75 , we need to find |\vec{a}|^2 .

Start by expanding |\vec{a}+\vec{b}|^2 :
|\vec{a}+\vec{b}|^2 = (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = |\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 .
Given |\vec{a}+\vec{b}|^2 = |\vec{a}|^2 + 2|\vec{b}|^2 , equate the expressions:
|\vec{a}|^2 + 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 = |\vec{a}|^2 + 2|\vec{b}|^2 .
Simplifying gives:
2(\vec{a} \cdot \vec{b}) = |\vec{b}|^2 .
We know \vec{a} \cdot \vec{b} = 3 , so:
2 \times 3 = |\vec{b}|^2 .
Thus, |\vec{b}|^2 = 6 .

Next, use |\vec{a} \times \vec{b}|^2 = 75 :
The magnitude of the cross product is given by:
|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2 .
Substitute the known values:
75 = |\vec{a}|^2 \times 6 - 3^2 .
Simplifying:
75 = 6|\vec{a}|^2 - 9 .
Add 9 to both sides:
84 = 6|\vec{a}|^2 .
Divide by 6:
|\vec{a}|^2 = 14 .

The value falls within the expected range of 14,14. Hence, |\vec{a}|^2 = 14 .