Let a, b, c be three vectors mutually perpendicular to each other and have same magnitude. If a vect

Q: Wait, why can't we just assume $\vec{r} = x\vec{a} + y\vec{b} + z\vec{c}$ and solve for $x, y, z$?

You absolutely can! That is a perfectly valid approach. However, the Vector Triple Product method is often faster and more elegant for JEE Advanced problems because it allows you to manipulate the entire expression without needing to solve for individual components until the very end. It reveals the underlying symmetry of the problem.

Q: Why does the term $(\vec{a} \cdot \vec{r})\vec{a} + (\vec{b} \cdot \vec{r})\vec{b} + (\vec{c} \cdot \vec{r})\vec{c}$ equal $k^2\vec{r}$?

This is a fundamental property of orthogonal bases. Any vector $\vec{r}$ can be written as $\vec{r} = \frac{(\vec{r} \cdot \vec{a})}{|\vec{a}|^2}\vec{a} + \frac{(\vec{r} \cdot \vec{b})}{|\vec{b}|^2}\vec{b} + \frac{(\vec{r} \cdot \vec{c})}{|\vec{c}|^2}\vec{c}$. Since $|\vec{a}| = |\vec{b}| = |\vec{c}| = k$, the denominators are all $k^2$. Multiplying both sides by $k^2$ gives you exactly that expression.

Q: Is it always safe to divide by $k^2$?

Yes, in this context. Since $k$ is the magnitude of the vectors $\vec{a}, \vec{b}, \vec{c}$, and these are given as non-zero vectors (implied by the fact that they form a basis), $k$ must be greater than zero. Therefore, $k^2$ is non-zero, and we can safely divide by it.

Question

Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors mutually perpendicular to each other and have same magnitude. If a vector $\vec{r}$ satisfies $\vec{a} 	imes \{(\vec{r} - \vec{b}) 	imes \vec{a}\} + \vec{b} 	imes \{(\vec{r} - \vec{c}) 	imes \vec{b}\} + \vec{c} 	imes \{(\vec{r} - \vec{a}) 	imes \vec{c}\} = \vec{0}$, then $\vec{r}$ is equal to :

Options:

(A) $\frac{1}{3}(\vec{a} + \vec{b} + \vec{c})$

(B) $\frac{1}{3}(2\vec{a} + \vec{b} - \vec{c})$

(C) $\frac{1}{2}(\vec{a} + \vec{b} + \vec{c})$

(D) $\frac{1}{2}(\vec{a} + \vec{b} + 2\vec{c})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, empty 3D space. You have three vectors, $\vec{a}, \vec{b}, \vec{c}$, which are mutually perpendicular and share the same magnitude, $k$.

Think of them as the $x, y, z$ axes of your own personal coordinate system. Because they are perpendicular, their dot products with each other are zero—a beautiful, simplifying fact.

We are given a complex equation:

$$ \vec{a} 	imes \{(\vec{r} - \vec{b}) 	imes \vec{a}\} + \vec{b} 	imes \{(\vec{r} - \vec{c}) 	imes \vec{b}\} + \vec{c} 	imes \{(\vec{r} - \vec{a}) 	imes \vec{c}\} = \vec{0} $$

It looks intimidating, but every complex vector equation is just a puzzle waiting for the right tool.

### The Vector Triple Product: Our Swiss Army Knife

To dismantle this, we need the Vector Triple Product (VTP) identity:

$$ \vec{x} 	imes (\vec{y} 	imes \vec{z}) = (\vec{x} \cdot \vec{z})\vec{y} - (\vec{x} \cdot \vec{y})\vec{z} $$

This is the key that unlocks the nested cross products. Let us apply this to the first term: $\vec{a} 	imes \{(\vec{r} - \vec{b}) 	imes \vec{a}\}$.

According to our VTP rule, this becomes:

$$ (\vec{a} \cdot \vec{a})(\vec{r} - \vec{b}) - (\vec{a} \cdot (\vec{r} - \vec{b}))\vec{a} $$

Now, watch the magic of orthogonality. We know $\vec{a} \cdot \vec{a} = k^2$ and $\vec{a} \cdot \vec{b} = 0$. The expression simplifies beautifully to:

$$ k^2(\vec{r} - \vec{b}) - (\vec{a} \cdot \vec{r})\vec{a} $$

### The Power of Symmetry

We do not need to repeat this grueling process for the other two terms. By symmetry, the second term, $\vec{b} 	imes \{(\vec{r} - \vec{c}) 	imes \vec{b}\}$, must simplify to:

$$ k^2(\vec{r} - \vec{c}) - (\vec{b} \cdot \vec{r})\vec{b} $$

Similarly, the third term, $\vec{c} 	imes \{(\vec{r} - \vec{a}) 	imes \vec{c}\}$, becomes:

$$ k^2(\vec{r} - \vec{a}) - (\vec{c} \cdot \vec{r})\vec{c} $$

We have successfully broken down the monster into manageable pieces.

### The Grand Assembly

Now, let us sum these three simplified terms and set them equal to $\vec{0}$:

$$ 3k^2\vec{r} - k^2(\vec{a} + \vec{b} + \vec{c}) - [(\vec{a} \cdot \vec{r})\vec{a} + (\vec{b} \cdot \vec{r})\vec{b} + (\vec{c} \cdot \vec{r})\vec{c}] = \vec{0} $$

Look closely at the term in the square bracket. This is the resolution of vector $\vec{r}$ along the orthogonal basis $\vec{a}, \vec{b}, \vec{c}$.

In any orthogonal basis, the vector $\vec{r}$ is represented as:

$$ \vec{r} = \frac{\vec{r} \cdot \vec{a}}{|\vec{a}|^2}\vec{a} + \frac{\vec{r} \cdot \vec{b}}{|\vec{b}|^2}\vec{b} + \frac{\vec{r} \cdot \vec{c}}{|\vec{c}|^2}\vec{c} $$

Multiplying by $k^2$, we find that the entire bracket is simply $k^2\vec{r}$.

### The Final Victory

Substituting this back into our equation, we get:

$$ 3k^2\vec{r} - k^2(\vec{a} + \vec{b} + \vec{c}) - k^2\vec{r} = \vec{0} $$

This simplifies to:

$$ 2k^2\vec{r} = k^2(\vec{a} + \vec{b} + \vec{c}) $$

Dividing both sides by $2k^2$, we arrive at the elegant solution:

$$ \vec{r} = \frac{1}{2}(\vec{a} + \vec{b} + \vec{c}) $$

We have navigated the complexity and found the simple, beautiful truth hidden within. Keep practicing, keep visualizing, and remember that every vector problem is just a path to a deeper understanding of the geometry of our world.

Animated Solution for Mathematics - Vector Algebra: Let $a, b, c$ be three vectors mutually perpendicular to each other and have same magnitude. If a vector $r$ satisfies $a \times {(r - b) \times a} + b \times {(r - c) \times b} + c \times {(r - a) \times c} = 0$ , then $r$ is equal to :

Select Answer:

Visualized Solution

Visualizing the Orthogonal Basis

The Vector Triple Product Tool

Expanding the First Term

Simplifying the First Term

Expanding Remaining Terms

Summing the Expansions

Vector Resolution Concept

Final Substitution

Solving for $r$

The Sigma Insight: Vector Triple Product

Analyzing the Setup

The Vector Triple Product

The Power of Symmetry

The Grand Assembly

The Final Victory

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