Let a, b and c be three non zero vectors such that b · c = 0 and a × (b × c) = (b - c)/2. If d be a

Q: Wait, why can we just compare the coefficients of $\vec{b}$ and $\vec{c}$? Is that always allowed?

Great question! We can do this because $\vec{b}$ and $\vec{c}$ are non-zero and orthogonal, which makes them linearly independent. In a vector space, if a linear combination of linearly independent vectors equals another, their coefficients must match exactly. It is the bedrock of vector algebra!

Q: I got stuck at $(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d})$. How do I even start expanding that?

Don't let the notation intimidate you! Think of $(\vec{a} \times \vec{b})$ as a single vector, let's call it $\vec{V}$. Then you have $\vec{V} \cdot (\vec{c} \times \vec{d})$. This is just a scalar triple product. You can rotate the dot and cross operators cyclically or use the property that allows you to move the cross product to the other side. It is all about rearranging the pieces to reveal the hidden structure.

Q: Why did the $\vec{d}$ vector disappear in the final step?

It didn't disappear—it was 'absorbed' into the dot product! When we expanded $\vec{b} \times (\vec{c} \times \vec{d})$, we got $(\vec{b} \cdot \vec{d})\vec{c} - (\vec{b} \cdot \vec{c})\vec{d}$. Since $\vec{b} \cdot \vec{c} = 0$, the term with $\vec{d}$ vanished, leaving us with a scalar $(\vec{b} \cdot \vec{d})$ multiplying $\vec{c}$. It is a beautiful simplification that rewards your patience.

Question

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non zero vectors such that $\vec{b} \cdot \vec{c} = 0$ and $\vec{a} 	imes (\vec{b} 	imes \vec{c}) = \frac{\vec{b} - \vec{c}}{2}$. If $\vec{d}$ be a vector such that $\vec{b} \cdot \vec{d} = \vec{a} \cdot \vec{b}$, then $(\vec{a} 	imes \vec{b}) \cdot (\vec{c} 	imes \vec{d})$ is equal to

Options:

(A) $\frac{3}{4}$

(B) $\frac{1}{2}$

(C) $-\frac{1}{4}$

(D) $\frac{1}{4}$

Sigma Percentile · Accepted Answer

### The Dance of Vectors: Unlocking the Triple Product

My dear student, welcome to the arena. Today, we are not just solving a problem; we are choreographing a dance between vectors.

When you look at an expression like $\vec{a} 	imes (\vec{b} 	imes \vec{c}) = \frac{\vec{b} - \vec{c}}{2}$, do not see a wall of symbols. See a map. This equation is telling us exactly how these vectors are oriented in space.

Let us peel back the layers together.

### Phase 1: The BAC-CAB Revelation

We begin with the Vector Triple Product. You have likely memorized the identity:

$$\vec{a} 	imes (\vec{b} 	imes \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$$

But have you ever stopped to appreciate its elegance? It takes a cross product—a vector—and transforms it into a linear combination of the vectors $\vec{b}$ and $\vec{c}$.

We are given that $\vec{a} 	imes (\vec{b} 	imes \vec{c}) = \frac{1}{2}\vec{b} - \frac{1}{2}\vec{c}$. By equating our expansion to this given condition, we get:

$$(\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} = \frac{1}{2}\vec{b} - \frac{1}{2}\vec{c}$$

Because $\vec{b}$ and $\vec{c}$ are orthogonal and non-zero, they are linearly independent. This is our golden ticket!

It allows us to equate the coefficients directly. We instantly find that $\vec{a} \cdot \vec{c} = \frac{1}{2}$ and $\vec{a} \cdot \vec{b} = \frac{1}{2}$. Just like that, the complexity begins to dissolve.

### Phase 2: The Mystery of Vector $\vec{d}$

The problem introduces a new player: vector $\vec{d}$. We are told $\vec{b} \cdot \vec{d} = \vec{a} \cdot \vec{b}$.

Since we just discovered that $\vec{a} \cdot \vec{b} = \frac{1}{2}$, we now know that $\vec{b} \cdot \vec{d} = \frac{1}{2}$. Keep this value close; it is the key to the final lock.

### Phase 3: The Grand Finale

Now, we face the target: $(\vec{a} 	imes \vec{b}) \cdot (\vec{c} 	imes \vec{d})$. This is a scalar quadruple product.

It looks terrifying, doesn't it? But remember, in JEE physics and math, we rarely fight monsters head-on. We use strategy.

We can treat $(\vec{a} 	imes \vec{b})$ as a single vector and use the property of the scalar triple product to shift the cross product:

$$(\vec{a} 	imes \vec{b}) \cdot (\vec{c} 	imes \vec{d}) = \vec{a} \cdot (\vec{b} 	imes (\vec{c} 	imes \vec{d}))$$

Now, look at the inner term: $\vec{b} 	imes (\vec{c} 	imes \vec{d})$. We apply the BAC-CAB rule again:

$$\vec{b} 	imes (\vec{c} 	imes \vec{d}) = (\vec{b} \cdot \vec{d})\vec{c} - (\vec{b} \cdot \vec{c})\vec{d}$$

Here is the moment of triumph. We know $\vec{b} \cdot \vec{c} = 0$. The entire second term vanishes into thin air!

We are left with:

$$\vec{a} \cdot ((\vec{b} \cdot \vec{d})\vec{c}) = (\vec{b} \cdot \vec{d})(\vec{a} \cdot \vec{c})$$

Substitute our known values:

$$\left( \frac{1}{2} ight) 	imes \left( \frac{1}{2} ight) = \frac{1}{4}$$

There it is. The chaos has settled into a simple, beautiful fraction. The final answer is **$\frac{1}{4}$**.

You didn't just solve a problem; you navigated the geometry of space. Keep this confidence with you—you are ready for the next challenge.

Animated Solution for Mathematics - Vector Algebra: Let $a, b$ and $c$ be three non zero vectors such that $b \cdot c = 0$ and $a \times (b \times c) = \frac{b - c}{2}$ . If $d$ be a vector such that $b \cdot d = a \cdot b$ , then $(a \times b) \cdot (c \times d)$ is equal to

Select Answer:

Visualized Solution

Visualizing the Orthogonal Vectors

The Vector Triple Product Condition

Expanding the Vector Triple Product

Equating the Expressions

Comparing Coefficients of $b$

Comparing Coefficients of $c$

Introducing Vector $d$

Evaluating $b \cdot d$

The Target Expression

Interchanging Dot and Cross

Expanding the Inner VTP

Applying Orthogonality

Simplifying the Target Expression

Final Numerical Substitution

The Sigma Insight: Vector Triple Product

The Dance of Vectors

Phase 1

Phase 2

Phase 3

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