Q.Classify the following measures as scalars and vectors.

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.

Option (B): $\left( \dfrac{\vec{a} - \vec{b}}{ab} \right)^2$ — yes, that’s exactly what we have. But wait: the problem might intend the square of a vector, not the scalar square? Actually, the notation $(\cdots)^2$ in the question means the scalar (dot) square, so our result is a scalar. Option (B) is also a scalar square. So (B) seems correct.

But let’s verify the other options to be sure. Option (C) is $\left( \dfrac{b\vec{a} - a\vec{b}}{ab} \right)^2$. That expands to $\frac{b^2 a^2 - 2ab(\vec{a}\cdot\vec{b}) + a^2 b^2}{a^2 b^2} = \frac{2a^2 b^2 - 2ab(\vec{a}\cdot\vec{b})}{a^2 b^2}$, which is not the same unless $a = b$. So (C) is different.

Option (A) and (D) also differ. So (B) is correct.

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:

4. Plug in the given magnitudes. $|\vec{a}| = 8$ and $|\vec{b}| = 3$, so

5. Take the square root. Magnitude is non-negative, so

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

The two cross terms each give $+6$ because $-12 \times (-\frac12) = 6$.

5. Thus

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

The sum of two numbers each at most 1 can be at most 2, and at least $-2$. But here the sum is $-3$, which is less than $-2$. This is impossible.

Since the premise itself is contradictory, neither Statement I nor Statement II can be true for any actual triple of unit vectors. Therefore both are false.