Let a = i + j + k and b = j - k. If c is a vector such that a × c = b and a · c = 3, then a · (b × c

Q: Why do we use the Vector Triple Product identity? Can't we just solve for $\vec{c}$ component by component?

You absolutely can! Solving for $\vec{c}$ is a robust, reliable method. However, in JEE, time is a currency. The identity $\vec{a} \times (\vec{a} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{a} - (\vec{a} \cdot \vec{a})\vec{c}$ is a powerful tool that bypasses the need to find individual components if you only need the final scalar value.

Q: Is the scalar triple product always equal to the determinant of the components?

Yes, exactly! The scalar triple product $[\vec{a} \vec{b} \vec{c}]$ represents the volume of the parallelepiped formed by the three vectors. It is equivalent to the determinant of the matrix formed by the components of $\vec{a}$, $\vec{b}$, and $\vec{c}$. This geometric interpretation is why cyclic permutations don't change the value—rotating the vectors doesn't change the volume!

Q: Wait, why did the shortcut result in $-|\vec{b}|^2$?

That is the beauty of vector algebra! By using the cyclic property, we transformed the expression into $\vec{b} \cdot (\vec{c} \times \vec{a})$. Since $\vec{a} \times \vec{c} = \vec{b}$, then $\vec{c} \times \vec{a} = -\vec{b}$. Substituting this gives $\vec{b} \cdot (-\vec{b})$, which is $-|\vec{b}|^2$. It turns a complex algebraic problem into a simple magnitude calculation.

Question

Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{j} - \hat{k}$. If $\vec{c}$ is a vector such that $\vec{a} 	imes \vec{c} = \vec{b}$ and $\vec{a} \cdot \vec{c} = 3$, then $\vec{a} \cdot (\vec{b} 	imes \vec{c})$ is equal to :

Options:

(A) -2

(B) -6

(C) 6

(D) 2

Sigma Percentile · Accepted Answer

### The Architecture of Space: A Vector Odyssey

Welcome, future engineer. Today, we are not just solving a vector problem; we are exploring the architecture of three-dimensional space.

When you look at vectors $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{j} - \hat{k}$, do not just see numbers. See arrows piercing through the Cartesian grid.

Vector $\vec{a}$ is a diagonal reaching out into the first octant, while $\vec{b}$ lies flat in the $yz$-plane. We are tasked with finding the value of the scalar triple product $\vec{a} \cdot (\vec{b} 	imes \vec{c})$, given that $\vec{a} 	imes \vec{c} = \vec{b}$ and $\vec{a} \cdot \vec{c} = 3$.

This is a classic JEE challenge—it tests not just your ability to calculate, but your ability to choose the most elegant path.

### Phase 1: The Brute Force Approach

Imagine you are a detective. You have a mystery vector $\vec{c}$. You know two things about it: its cross product with $\vec{a}$ is $\vec{b}$, and its dot product with $\vec{a}$ is $3$.

The most straightforward way to solve this is to find $\vec{c}$ explicitly. To do this, we use the **Vector Triple Product identity**. We take the cross product of $\vec{a}$ with the equation $\vec{a} 	imes \vec{c} = \vec{b}$.

This gives us:
$$\vec{a} 	imes (\vec{a} 	imes \vec{c}) = \vec{a} 	imes \vec{b}$$

Applying the **BAC-CAB rule**, we expand the left side:
$$(\vec{a} \cdot \vec{c})\vec{a} - (\vec{a} \cdot \vec{a})\vec{c} = \vec{a} 	imes \vec{b}$$

Here, we substitute the knowns: $\vec{a} \cdot \vec{c} = 3$ and $\vec{a} \cdot \vec{a} = |\vec{a}|^2 = 1^2 + 1^2 + 1^2 = 3$. The equation simplifies beautifully to:
$$3\vec{a} - 3\vec{c} = \vec{a} 	imes \vec{b}$$

By calculating $\vec{a} 	imes \vec{b}$ using the determinant method, we find it equals $-2\hat{i} + \hat{j} + \hat{k}$. Solving for $\vec{c}$ yields:
$$\vec{c} = \frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}$$

With $\vec{c}$ in hand, the final dot product is just a matter of arithmetic, leading us to **$-2$**. This is the path of the reliable worker, and it works every time.

### Phase 2: The Ninja Shortcut

But wait! In the high-pressure environment of the JEE Advanced exam, time is your most precious resource. Is there a faster way? Absolutely.

Look at the target expression: $\vec{a} \cdot (\vec{b} 	imes \vec{c})$. This is a scalar triple product, denoted as $[\vec{a} \vec{b} \vec{c}]$.

One of the most powerful properties of this product is **cyclic permutation**:
$$\vec{a} \cdot (\vec{b} 	imes \vec{c}) = \vec{b} \cdot (\vec{c} 	imes \vec{a})$$

Now, look at our given condition: $\vec{a} 	imes \vec{c} = \vec{b}$. This implies that:
$$\vec{c} 	imes \vec{a} = -(\vec{a} 	imes \vec{c}) = -\vec{b}$$

Substitute this into our permuted expression:
$$\vec{b} \cdot (-\vec{b}) = -|\vec{b}|^2$$

We don't even need to find $\vec{c}$! We just need the magnitude of $\vec{b}$. Since $\vec{b} = \hat{j} - \hat{k}$, its magnitude squared is:
$$|\vec{b}|^2 = 0^2 + 1^2 + (-1)^2 = 2$$

Therefore, the answer is **$-2$**. In ten seconds, we have bypassed the algebra and arrived at the truth. This is the difference between knowing the math and mastering it.

Animated Solution for Mathematics - Vector Algebra: Let $a = \hat{i} + \hat{j} + \hat{k}$ and $b = \hat{j} - \hat{k}$ . If $c$ is a vector such that $a \times c = b$ and $a \cdot c = 3$ , then $a \cdot (b \times c)$ is equal to :

Select Answer:

Visualized Solution

Given Vectors $a$ and $b$

The Conditions for Vector $c$

Target Expression: $a \cdot (b \times c)$

Calculating $a \times b$

Using Vector Triple Product Identity

Calculating Magnitude Squared $∣ a ∣^{2}$

Substituting Values into the VTP Equation

Solving for Vector $c$

Final Dot Product Calculation

The Shortcut (Ninja Technique)

The Sigma Insight: Vector Triple Product

The Architecture of Space

Phase 1

Phase 2

Similar Questions

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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; } Let a=i^+j^​+k^ and b=j^​−k^. If c is a vector such that a×c=b and a⋅c=3, then a⋅(b×c) is equal to :

Select Answer:

Visualized Solution

Given 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

The Conditions for Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

Target Expression: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a⋅(b×c)

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

Using Vector Triple Product Identity

Calculating Magnitude Squared .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣a∣2