If a and b are vectors in space given by a = (i - 2j)/√5 and b = (2i + j + 3k)/√14, then find the va

Q: Why do we calculate the magnitude and dot product first? Isn't it faster to just plug the vectors into the cross product?

That is a classic trap! If you try to compute the cross product directly, you will be dealing with messy components and square roots, which is a recipe for calculation errors. In JEE Advanced, whenever you see square roots in vector components, it is almost always a hint to check for unit vectors or orthogonality. By checking these properties first, you turn a nightmare calculation into a simple algebraic identity.

Q: I always forget the vector triple product identity. Is there a way to derive it or remember it easily?

The identity $(\vec{u} \times \vec{v}) \times \vec{w} = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{v} \cdot \vec{w})\vec{u}$ is often called the 'BAC-CAB' rule, though that applies to $\vec{a} \times (\vec{b} \times \vec{c})$. For this specific form, remember that the result must be a linear combination of $\vec{v}$ and $\vec{u}$. The coefficients are simply the dot products of the 'outer' vector $\vec{w}$ with the 'inner' vectors. It is a powerful tool that collapses complex cross products into simple vector additions.

Q: Why does the expression $(2\vec{a} + \vec{b}) \cdot (2\vec{a} + \vec{b})$ simplify to $|2\vec{a} + \vec{b}|^2$? Can I just square the coefficients?

Be very careful! You cannot simply square the coefficients like $(2\vec{a})^2 + (\vec{b})^2$. Because these are vectors, the square of a sum is governed by the dot product: $(\vec{x} + \vec{y}) \cdot (\vec{x} + \vec{y}) = |\vec{x}|^2 + |\vec{y}|^2 + 2(\vec{x} \cdot \vec{y})$. You must include the cross-term $2(\vec{x} \cdot \vec{y})$, which in this case happens to be zero, but you must always account for it!

Question

If $\vec{a}$ and $\vec{b}$ are vectors in space given by $\vec{a} = \frac{\hat{i} - 2\hat{j}}{\sqrt{5}}$ and $\vec{b} = \frac{2\hat{i} + \hat{j} + 3\hat{k}}{\sqrt{14}}$, then find the value of $(2\vec{a} + \vec{b}) \cdot [(\vec{a} 	imes \vec{b}) 	imes (\vec{a} - 2\vec{b})]$.

Sigma Percentile · Accepted Answer

### The Hidden Elegance of Vector Algebra

Welcome, fellow traveler on the JEE journey. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of square roots and cross products.

Many students see an expression like $(2\vec{a} + \vec{b}) \cdot [(\vec{a} 	imes \vec{b}) 	imes (\vec{a} - 2\vec{b})]$ and immediately panic, reaching for the determinant method to compute the cross product. But stop. Take a breath. In the world of JEE Advanced, complexity is often a mask for hidden simplicity.

### Phase 1: The Inspection

Before we touch the main expression, let us look at our building blocks: $\vec{a} = \frac{\hat{i} - 2\hat{j}}{\sqrt{5}}$ and $\vec{b} = \frac{2\hat{i} + \hat{j} + 3\hat{k}}{\sqrt{14}}$. Do you see the square roots? They are not there to make your life difficult; they are there to normalize the vectors.

Let us calculate the magnitude of $\vec{a}$ squared: 
$$|\vec{a}|^2 = \frac{1^2 + (-2)^2}{(\sqrt{5})^2} = \frac{1+4}{5} = 1$$

Similarly, for $\vec{b}$, we find: 
$$|\vec{b}|^2 = \frac{2^2 + 1^2 + 3^2}{(\sqrt{14})^2} = \frac{4+1+9}{14} = 1$$
Both are unit vectors!

But wait, there is more. Let us check their dot product: 
$$\vec{a} \cdot \vec{b} = \frac{(1)(2) + (-2)(1) + (0)(3)}{\sqrt{5} \cdot \sqrt{14}} = \frac{2-2}{\sqrt{70}} = 0$$
They are orthogonal! This is the 'Aha!' moment. We are not working with random vectors; we are working with an orthonormal-like pair. This discovery is our key to the kingdom.

### Phase 2: The Triple Product Identity

Now, let us face the beast: the vector triple product $(\vec{a} 	imes \vec{b}) 	imes (\vec{a} - 2\vec{b})$. We invoke the vector triple product identity: 
$$(\vec{u} 	imes \vec{v}) 	imes \vec{w} = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{v} \cdot \vec{w})\vec{u}$$

Here, let $\vec{u} = \vec{a}$, $\vec{v} = \vec{b}$, and $\vec{w} = (\vec{a} - 2\vec{b})$. Substituting these into the identity, we get: 
$$[\vec{a} \cdot (\vec{a} - 2\vec{b})]\vec{b} - [\vec{b} \cdot (\vec{a} - 2\vec{b})]\vec{a}$$

This looks intimidating, but remember our findings from Phase 1. The first term is $\vec{a} \cdot \vec{a} - 2(\vec{a} \cdot \vec{b}) = |\vec{a}|^2 - 0 = 1$. The second term is $\vec{b} \cdot \vec{a} - 2(\vec{b} \cdot \vec{b}) = 0 - 2|\vec{b}|^2 = -2$.

Suddenly, the entire triple product collapses into: 
$$(1)\vec{b} - (-2)\vec{a} = 2\vec{a} + \vec{b}$$

### Phase 3: The Final Collapse

Look at what we have achieved. The original expression was $(2\vec{a} + \vec{b}) \cdot [(\vec{a} 	imes \vec{b}) 	imes (\vec{a} - 2\vec{b})]$. We just proved that the bracketed term is exactly $2\vec{a} + \vec{b}$.

So, the expression is simply $(2\vec{a} + \vec{b}) \cdot (2\vec{a} + \vec{b})$, which is $|2\vec{a} + \vec{b}|^2$. Expanding this using the dot product property, we get: 
$$4|\vec{a}|^2 + |\vec{b}|^2 + 4(\vec{a} \cdot \vec{b})$$

Substituting our known values: 
$$4(1) + 1 + 4(0) = 5$$

### Conclusion

We started with a terrifying expression and ended with a simple integer. This is the beauty of vector algebra. It is not about brute-forcing calculations; it is about recognizing the underlying structure.

You have the tools, you have the identity, and now you have the mindset. The final answer is **5**. Go forth and conquer.

Animated Solution for Mathematics - Vector Algebra: If $a$ and $b$ are vectors in space given by $a = \frac{i ^ - 2 j ^}{5}$ and $b = \frac{2 i ^ + j ^ + 3 k ^}{14}$ , then find the value of $(2 a + b) \cdot [(a \times b) \times (a - 2 b)]$ .

Enter Numerical Value:

Visualized Solution

Visualizing Vectors $a$ and $b$

Magnitude of $a$

Magnitude of $b$

Dot Product $a \cdot b$

Vector Triple Product Identity

Applying the Identity

Simplifying the First Dot Product

Simplifying the Second Dot Product

Reconstructing the Triple Product

The Final Expression

Expanding $∣2 a + b ∣^{2}$

Atomic Compute: Final Value

The Sigma Insight: Vector Triple Product

The Hidden Elegance of Vector Algebra

Phase 1

Phase 2

Phase 3

Conclusion

Similar Questions

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Enter Numerical Value:

Visualized Solution

Visualizing Vectors .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Magnitude of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Magnitude of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Dot Product .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a⋅b