Question

Let
\(\vec{a}\) and \(\vec{b}\) be two vectors such that 
\(|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + 2|\vec{b}|^2,\  \vec{a} \cdot \vec{b} = 3\) and \(|\vec{a} \times \vec{b}|^2 = 75\)
Then \(|\vec{a}|^2\) is equal to _____.

Answer 1

Correct Answer: 14

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.

Answer 2

\(\because |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + 2|\vec{b}|^2\)
or \(|\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} = |\vec{a}|^2 + 2|\vec{b}|^2\)
\(∴ |\vec{b}|^2=6 …(i)\)
Now,\(|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - \left(\vec{a} \cdot \vec{b}\right)^2\)
\(75=|\vec{a}|^2⋅6−9\)
\(∴ |\vec{a}|^2=14\)
So, the correct answer is 14.