If a = 1/√10 (3i + k) and b = 1/7 (2i + 3j - 6k), then the value of (2a - b) · [(a × b) × (a + 2b)]

Q: Why do we waste time checking magnitudes and dot products? Can't we just expand the cross product directly?

You *could*, but that is the path of the brute-force warrior, not the strategist. In JEE Advanced, the numbers are rarely random. Checking properties first reveals the 'skeleton' of the problem, turning a 10-minute calculation into a 30-second observation. Always look for the 'hidden' simplicity before you start the heavy lifting.

Q: I always get the Vector Triple Product formula mixed up. Is it $(\vec{u} \cdot \vec{w})\vec{v} - (\vec{v} \cdot \vec{w})\vec{u}$ or something else?

It is easy to confuse! Remember the 'BAC-CAB' rule for $(\vec{a} \times \vec{b}) \times \vec{c} = \vec{b}(\vec{a} \cdot \vec{c}) - \vec{a}(\vec{b} \cdot \vec{c})$. The key is that the vector *outside* the parenthesis in the cross product (the one being crossed last) is the one that gets subtracted. Keep this mnemonic in your toolkit, and you will never doubt the formula again.

Q: Wait, why did the final answer come out to be negative? Can a dot product be negative?

Absolutely! A dot product is just a scalar. It represents the projection of one vector onto another, scaled by their magnitudes. A negative result simply means the angle between the two vectors is obtuse (greater than 90 degrees). In our case, it confirms the geometric orientation of the resulting vectors.

Question

If $\vec{a} = \frac{1}{\sqrt{10}}(3\hat{i} + \hat{k})$ and $\vec{b} = \frac{1}{7}(2\hat{i} + 3\hat{j} - 6\hat{k})$, then the value of $(2\vec{a} - \vec{b}) \cdot [(\vec{a} 	imes \vec{b}) 	imes (\vec{a} + 2\vec{b})]$ is

Options:

(A) $-3$

(B) $5$

(C) $3$

(D) $-5$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

My dear student, take a deep breath. When you look at an expression like $(2\vec{a} - \vec{b}) \cdot [(\vec{a} 	imes \vec{b}) 	imes (\vec{a} + 2\vec{b})]$, it is natural to feel a surge of anxiety. It looks like a chaotic mess of brackets, cross products, and dot products.

But here is the secret of the JEE Advanced topper: the complexity is often a mask. The problem is not asking you to perform a massive, tedious calculation; it is inviting you to a dance of symmetry and properties. Let us peel back the layers together.

### The Detective Work

Before we touch the main expression, we must understand our players. We are given $\vec{a} = \frac{1}{\sqrt{10}}(3\hat{i} + \hat{k})$ and $\vec{b} = \frac{1}{7}(2\hat{i} + 3\hat{j} - 6\hat{k})$.

Let us calculate their magnitudes. For $\vec{a}$, we have:

$$|\vec{a}|^2 = \frac{1}{10}(3^2 + 1^2) = \frac{10}{10} = 1$$

It is a unit vector! Now, for $\vec{b}$, we have:

$$|\vec{b}|^2 = \frac{1}{49}(2^2 + 3^2 + (-6)^2) = \frac{4+9+36}{49} = \frac{49}{49} = 1$$

Another unit vector! This is not a coincidence. Now, check their interaction via the dot product:

$$\vec{a} \cdot \vec{b} = \frac{1}{7\sqrt{10}}(3(2) + 0(3) + 1(-6)) = \frac{1}{7\sqrt{10}}(6 - 6) = 0$$

They are orthogonal! We have just discovered that $\vec{a}$ and $\vec{b}$ are **orthonormal vectors**. This is our 'skeleton key' that will unlock the entire problem.

### The Vector Triple Product

Now, let us turn our attention to the 'monster' inside the square brackets: $(\vec{a} 	imes \vec{b}) 	imes (\vec{a} + 2\vec{b})$. This is a classic Vector Triple Product.

We use the identity $(\vec{u} 	imes \vec{v}) 	imes \vec{w} = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{v} \cdot \vec{w})\vec{u}$. Mapping our variables as $\vec{u} = \vec{a}$, $\vec{v} = \vec{b}$, and $\vec{w} = \vec{a} + 2\vec{b}$, we get:

$$[\vec{a} \cdot (\vec{a} + 2\vec{b})]\vec{b} - [\vec{b} \cdot (\vec{a} + 2\vec{b})]\vec{a}$$

Let us evaluate those dot products. The first term is:

$$\vec{a} \cdot (\vec{a} + 2\vec{b}) = \vec{a} \cdot \vec{a} + 2(\vec{a} \cdot \vec{b}) = 1 + 0 = 1$$

The second term is:

$$\vec{b} \cdot (\vec{a} + 2\vec{b}) = \vec{b} \cdot \vec{a} + 2(\vec{b} \cdot \vec{b}) = 0 + 2(1) = 2$$

Suddenly, our massive triple product has collapsed into the simple expression: $\vec{b} - 2\vec{a}$.

### The Final Act

We are almost at the finish line. We need to evaluate $(2\vec{a} - \vec{b}) \cdot (\vec{b} - 2\vec{a})$.

Look closely at these two vectors. The second vector, $\vec{b} - 2\vec{a}$, is exactly the negative of the first vector, $2\vec{a} - \vec{b}$. So, we are calculating the dot product of a vector with its own negative:

$$(2\vec{a} - \vec{b}) \cdot [-(2\vec{a} - \vec{b})] = -|2\vec{a} - \vec{b}|^2$$

Now, we expand the magnitude squared:

$$|2\vec{a} - \vec{b}|^2 = (2\vec{a} - \vec{b}) \cdot (2\vec{a} - \vec{b}) = 4|\vec{a}|^2 + |\vec{b}|^2 - 4(\vec{a} \cdot \vec{b})$$

Plugging in our known values:

$$4(1) + 1 - 4(0) = 5$$

Therefore, our final result is **$-5$**. We have conquered the monster! Remember, in physics and mathematics, the most complex-looking problems are often just simple truths dressed in complicated clothing.

Animated Solution for Mathematics - Vector Algebra: If $a = \frac{1}{10} (3 \hat{i} + \hat{k})$ and $b = \frac{1}{7} (2 \hat{i} + 3 \hat{j} - 6 \hat{k})$ , then the value of $(2 a - b) \cdot [(a \times b) \times (a + 2 b)]$ is

Select Answer:

Visualized Solution

Given Vectors

Calculating Magnitudes

Checking Orthogonality

The Target Expression

Vector Triple Product Formula

Expanding the VTP

Evaluating Dot Products (Part 1)

Evaluating Dot Products (Part 2)

Simplified VTP

Final Dot Product Setup

Calculating the Final Value

The Sigma Insight: Vector Triple Product

Analyzing the Setup

The Detective Work

The Vector Triple Product

The Final Act

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Animated Solution for Mathematics - Vector Algebra: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a=10​1​(3i^+k^) and b=71​(2i^+3j^​−6k^), then the value of (2a−b)⋅[(a×b)×(a+2b)] is