Let a = i - k, b = xi + j + (1 - x)k and c = yi + xj + (1 + x - y)k. Then [a b c] depends on
(a) onl

Q: Wait, why do we use a determinant to find the volume of a parallelepiped?

Great question! The scalar triple product, defined as $\vec{a} \cdot (\vec{b} \times \vec{c})$, is geometrically equivalent to the volume of a parallelepiped with edges $\vec{a}$, $\vec{b}$, and $\vec{c}$. The determinant is simply the most efficient algebraic way to compute this product. It encapsulates the cross product and the dot product in one elegant structure.

Q: Can we just use cross products instead of the determinant?

Absolutely! You could calculate $\vec{b} \times \vec{c}$ first and then take the dot product with $\vec{a}$. However, the determinant method is generally faster and less prone to sign errors during the intermediate steps. Both methods will lead you to the same result—a constant value of 1.

Q: Why did the variables $x$ and $y$ cancel out? Is that a coincidence?

It might feel like magic, but it is not a coincidence! In many JEE problems, you will encounter expressions that look like they depend on variables, but the underlying geometry is invariant. Here, the vectors are constructed such that the volume of the parallelepiped they form is constant, regardless of the specific values of $x$ and $y$.

Question

Let $\vec{a} = \hat{i} - \hat{k}, \vec{b} = x\hat{i} + \hat{j} + (1 - x)\hat{k}$ and $\vec{c} = y\hat{i} + x\hat{j} + (1 + x - y)\hat{k}$. Then $[\vec{a} \vec{b} \vec{c}]$ depends on

Options:

(A) only $y$

(B) only $x$

(C) both $x$ and $y$

(D) neither $x$ nor $y$

Sigma Percentile · Accepted Answer

### The Geometry of Invariance: A Vector Journey

Imagine you are standing in a three-dimensional coordinate system. You have three vectors, $\vec{a}$, $\vec{b}$, and $\vec{c}$, acting as the edges of a parallelepiped.

In the world of JEE physics and mathematics, the scalar triple product, denoted as $[\vec{a} \vec{b} \vec{c}]$, is not just a dry calculation; it is a measure of spatial capacity. It tells us the volume of that parallelepiped.

Today, we are going to explore why this volume, despite the presence of variables $x$ and $y$, remains perfectly constant.

### The Determinant: Our Mathematical Lens

To find the volume, we turn to the determinant. The determinant is the language of linear transformations and volumes.

When we arrange the components of our vectors into a $3 	imes 3$ matrix, we are essentially creating a map of how these vectors span space. Our vectors are given as:

$\vec{a} = \hat{i} - \hat{k}$

$\vec{b} = x\hat{i} + \hat{j} + (1 - x)\hat{k}$

$\vec{c} = y\hat{i} + x\hat{j} + (1 + x - y)\hat{k}$

By extracting the coefficients, we build our matrix:

$$[\vec{a} \vec{b} \vec{c}] = \begin{vmatrix} 1 & 0 & -1 \ x & 1 & 1-x \ y & x & 1+x-y \end{vmatrix}$$

### The Dance of Expansion

Now, let us expand this determinant along the first row. This is a strategic choice because the second element is zero, which simplifies our work significantly.

We break it down into three parts:

1. The first term: $1 \cdot \begin{vmatrix} 1 & 1-x \ x & 1+x-y \end{vmatrix}$

2. The second term: $-0 \cdot \begin{vmatrix} x & 1-x \ y & 1+x-y \end{vmatrix} = 0$

3. The third term: $+ (-1) \cdot \begin{vmatrix} x & 1 \ y & x \end{vmatrix}$

As we calculate these, watch how the algebra unfolds. The first minor gives us $1(1+x-y) - x(1-x) = 1 + x - y - x + x^2 = 1 - y + x^2$.

The third minor gives us $x^2 - y$. When we combine these with the coefficients from the first row, we get:

$$[\vec{a} \vec{b} \vec{c}] = (1 - y + x^2) - (x^2 - y)$$

### The Beauty of Cancellation

Look closely at the expression $(1 - y + x^2) - (x^2 - y)$. This is the moment of truth.

We distribute the negative sign: $1 - y + x^2 - x^2 + y$. The $x^2$ terms cancel out, and the $y$ terms cancel out.

We are left with exactly **$1$**.

This result is profound. It tells us that no matter what values you choose for $x$ and $y$, the volume of the parallelepiped formed by these vectors is always **$1$**.

The variables were merely a distraction, a test of your ability to see through the algebraic noise to the geometric truth. The scalar triple product is independent of both $x$ and $y$.

Keep this in mind as you tackle future problems: sometimes, the most complex-looking expressions hide the simplest, most elegant constants.

Animated Solution for Mathematics - Vector Algebra: Let $a = \hat{i} - \hat{k}, b = x \hat{i} + \hat{j} + (1 - x) \hat{k}$ and $c = y \hat{i} + x \hat{j} + (1 + x - y) \hat{k}$ . Then $[a b c]$ depends on

Select Answer:

Visualized Solution

Given Vectors $a, b, c$

Scalar Triple Product Formula

Setting up the Determinant

Expanding: First Term

Expanding: Second Term

Expanding: Third Term

Writing the Expanded Expression

Simplifying the First Bracket

Simplifying the Second Bracket

Combining All Terms

Final Conclusion

The Sigma Insight: Scalar Triple Product

The Geometry of Invariance

The Determinant

The Dance of Expansion

The Beauty of Cancellation

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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^−k^,b=xi^+j^​+(1−x)k^ and c=yi^+xj^​+(1+x−y)k^. Then [abc] depends on