Q.Find the magnitude of two vectors $\vec{a}$ and $\vec{b},$ having the same magnitude and such that the angle between them is $60^\circ$ and their scalar product is $\frac{1}{2}.$

We are asked to compute

where $a = |\vec{a}|$ and $b = |\vec{b}|$. The notation $\vec{v}^2$ means $\vec{v} \cdot \vec{v} = |\vec{v}|^2$. So we are really squaring a vector difference — that is, taking its dot product with itself.

Concept & Intuition

The trick is to treat the square of a vector as its magnitude squared. Then we just expand the square of a difference, just like $(x - y)^2 = x^2 - 2xy + y^2$, but here the “multiplication” is the dot product. The denominators are scalars, so they factor out cleanly.

1. Write the square as a dot product

2. Expand using the distributive property

3. Replace $\vec{a}\cdot\vec{a} = a^2$ and $\vec{b}\cdot\vec{b} = b^2$

4. Combine into a single fraction Common denominator $a^2 b^2$:

5. Recognize the numerator The numerator is exactly $( \vec{a} - \vec{b} )^2 = a^2 - 2\vec{a}\cdot\vec{b} + b^2$. So:

6. Rewrite as a square of a vector

But this is not one of the options directly — wait, check: Option (B) is $\left( \frac{\vec{a} - \vec{b}}{ab} \right)^2$. That seems to match. However, we must be careful: the original expression had $\frac{\vec{a}}{a^2} - \frac{\vec{b}}{b^2}$, and we got $\frac{(\vec{a} - \vec{b})^2}{a^2 b^2}$. That is indeed $\left( \frac{\vec{a} - \vec{b}}{ab} \right)^2$. So why isn’t (B) the answer? Let’s check the options again carefully. …

The key idea here is that when two vectors are perpendicular, their dot product is zero. That makes the square of the magnitude of any linear combination easy to compute — you just add the squares of the scaled magnitudes, with no cross term.

Let’s walk through it.

1. Write the square of the magnitude. For any vector, $|\vec{v}|^2 = \vec{v} \cdot \vec{v}$. So

2. Expand the dot product. Using distributivity:

Since $\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{a}$, the middle terms combine:

3. Use the perpendicular condition. Because $\vec{a}$ and $\vec{b}$ are perpendicular, $\vec{a}\cdot\vec{b} = 0$. That eliminates the cross term entirely: $|\vec{a} - 2\vec{b}|^2 = |\vec{a}|^2 + 4|\vec{b}|^2.$ …

We are given three unit vectors $\vec{a}, \vec{b}, \vec{c}$ with:

• $\vec{a} \perp \vec{b}$ and $\vec{a} \perp \vec{c}$, so $\vec{a} \cdot \vec{b} = 0$ and $\vec{a} \cdot \vec{c} = 0$.
• The angle between $\vec{b}$ and $\vec{c}$ is $\frac{2\pi}{3}$, so $\vec{b} \cdot \vec{c} = |\vec{b}||\vec{c}|\cos\frac{2\pi}{3} = 1 \cdot 1 \cdot \left(-\frac12\right) = -\frac12$.

We need $\left| \vec{a} + 3\vec{b} - 4\vec{c} \right|^2$.

Concept & Intuition

When asked for the squared magnitude of a linear combination of vectors, the natural approach is to expand using the dot product:

$|\vec{v}|^2 = \vec{v} \cdot \vec{v}$.

This turns the problem into a sum of scalar products, which we can evaluate using the given dot products. No geometry beyond the dot product is needed — just careful algebra.

Step-by-step solution

1. Write the squared magnitude as a dot product

2. Expand using distributivity

3. Simplify using symmetry and given values

   • $\vec{a}\cdot\vec{a} = |\vec{a}|^2 = 1$, similarly $\vec{b}\cdot\vec{b} = 1$, $\vec{c}\cdot\vec{c} = 1$.
   • $\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{a} = 0$ and $\vec{a}\cdot\vec{c} = \vec{c}\cdot\vec{a} = 0$.
   • $\vec{b}\cdot\vec{c} = \vec{c}\cdot\vec{b} = -\frac12$.

   Substituting:

4. Compute carefully

We have three unit vectors $\vec{a}, \vec{b}, \vec{c}$ — each of magnitude 1. The only given condition is

Let’s expand this and see what it tells us about the angles between these vectors.

1. Expand each square using $|\vec{x}|^2 = \vec{x}\cdot\vec{x}$:

Since $|\vec{a}| = |\vec{b}| = 1$, this becomes $1 + 1 - 2\vec{a}\cdot\vec{b} = 2 - 2\vec{a}\cdot\vec{b}$.

Similarly,

2. Add them:

The problem says this sum equals 10, so

Hence

3. Now, each dot product is the cosine of the angle between the vectors. Since $\vec{a}, \vec{b}, \vec{c}$ are unit vectors, we have $-1 \le \vec{a}\cdot\vec{b} \le 1, \qquad -1 \le \vec{a}\cdot\vec{c} \le 1.$ …