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) only x
(b

Q: Wait, why do we use a determinant here? Can't I just calculate the cross product first and then the dot product?

You absolutely can! The determinant is simply the most elegant, compact way to represent the operation $[\vec{a} \vec{b} \vec{c}] = (\vec{a} \times \vec{b}) \cdot \vec{c}$. Using the determinant saves you from the tedious intermediate steps of calculating the cross product vector, but both methods will lead you to the exact same constant result.

Q: I got a massive expression with $x^2$ and $y$ terms. Did I make a mistake?

It is very common to feel overwhelmed by the algebra! If you see terms like $x^2$ or $y$ appearing, don't panic. Often, in these types of JEE problems, these terms are designed to cancel out perfectly. Re-check your signs during the expansion of the determinant—a single sign error is usually the culprit when variables don't vanish.

Q: Does the fact that the result is $1$ mean the vectors are unit vectors?

Not necessarily! The scalar triple product represents the volume of the parallelepiped. A volume of $1$ just means the unit volume is maintained. It doesn't imply that the individual vectors have a magnitude of $1$ or that they are orthogonal.

Q: Why does the problem ask if it depends on $x$ or $y$ if the answer is just a constant?

That is the beauty of the question! It tests your ability to look past the 'noise' of variables. In competitive exams, examiners often introduce variables to see if you will get intimidated by the algebra or if you have the confidence to trust the mathematical structure to simplify.

Q: Is there a geometric intuition for why the volume is constant?

Yes! If you look closely at the vectors, you might notice that as you change $x$ and $y$, you are essentially performing a 'shear' transformation on the parallelepiped. A shear transformation preserves the volume of a shape, which is why the final result remains invariant.

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 $x$

(B) only $y$

(C) Neither $x$ nor $y$

(D) both $x$ and $y$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

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

Usually, when we see variables like $x$ and $y$ scattered throughout the components of these vectors, our instinct is to brace for a long, grueling algebraic battle. We expect the volume of this shape to change as $x$ and $y$ shift, like a balloon being squeezed and stretched.

But today, we are going to discover something truly beautiful: the concept of **invariance**.

### The Gateway: The Determinant

To find the volume of the parallelepiped formed by these vectors, we use the scalar triple product, denoted as $[\vec{a} \vec{b} \vec{c}]$. Mathematically, this is the determinant of the matrix formed by the components of our vectors:

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

I know, looking at this matrix, you might feel a bit of anxiety. You see the $x$ and $y$ terms, and you think, 'How will these ever disappear?'

But take a deep breath. In mathematics, as in life, complexity is often just a mask for a deeper, simpler truth. We are going to peel back that mask.

### The Expansion: A Dance of Terms

Let us expand this determinant along the first row. This is our standard toolkit for handling $3 	imes 3$ matrices.

We take the first element, $1$, and multiply it by the minor determinant, then subtract the second element (which is $0$, making our lives much easier!), and finally add the third element, $-1$, multiplied by its minor.

$$
= 1 \cdot \begin{vmatrix} 1 & 1-x \ x & 1+x-y \end{vmatrix} - 0 \cdot \begin{vmatrix} x & 1-x \ y & 1+x-y \end{vmatrix} + (-1) \cdot \begin{vmatrix} x & 1 \ y & x \end{vmatrix}
$$

As we calculate the first minor, we get $1(1+x-y) - x(1-x)$. Expanding this gives us $1 + x - y - x + x^2$, which simplifies beautifully to $1 - y + x^2$.

Now, look at the third minor. We have $(-1) \cdot [x(x) - 1(y)]$, which simplifies to $-(x^2 - y)$, or $-x^2 + y$.

### The Grand Cancellation

Now, we bring it all together. This is the moment of truth. We combine our results:

$$
= (1 - y + x^2) + (-x^2 + y)
$$

Watch closely as the terms interact. The $x^2$ and $-x^2$ cancel each other out. The $-y$ and $+y$ vanish into thin air.

We are left with nothing but the constant $1$:

$$
= 1 - y + x^2 - x^2 + y = 1
$$

### The Takeaway

Isn't that breathtaking? Despite the presence of $x$ and $y$, the volume of this parallelepiped is exactly $1$, regardless of the values of $x$ and $y$.

This means the scalar triple product depends on **neither $x$ nor $y$**.

This problem teaches us a vital lesson for your JEE journey: never let the presence of variables intimidate you. Sometimes, the structure of the problem is designed to collapse into a constant.

Trust the process, keep your signs organized, and always look for the underlying symmetry. You have just mastered the art of seeing through the complexity to the **constant truth** beneath.

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

Understanding Scalar Triple Product

The Determinant Formula

Setting up the Determinant

Expanding along the First Row

Calculating the First Minor

Calculating the Third Minor

Combining the Results

Final Simplification

Conclusion and Key Takeaway

The Sigma Insight: Scalar Triple Product

Analyzing the Setup

The Gateway

The Expansion

The Grand Cancellation

The Takeaway

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