Let a = i - αj + βk, b = 3i + βj - αk and c = -αi - 2j + k, where α and β are integers. If a · b = -

Q: Why can't I just calculate the cross product first?

You are missing the components! The vectors \vec{a}, \vec{b}, and \vec{c} are incomplete. You must solve for the parameters \alpha and \beta first. Think of it as unlocking the vectors before you can use them.

Q: Wait, why do we reject the root \alpha = -\frac{1}{2}?

This is a classic JEE trap! Always read the question carefully. The problem states \alpha and \beta are integers. If you include the fraction, you are solving a different problem than the one asked. In JEE, constraints are not suggestions; they are rules.

Q: Does the order of rows in the determinant matter for the scalar triple product?

Yes! The scalar triple product (\vec{a} \times \vec{b}) \cdot \vec{c} is the determinant of the matrix with rows \vec{a}, \vec{b}, and \vec{c} in that specific order. Changing the order changes the sign, which represents the orientation of the volume. Stick to the order given!

Question

Let $\vec{a} = \hat{i} - \alpha\hat{j} + \beta\hat{k}, \vec{b} = 3\hat{i} + \beta\hat{j} - \alpha\hat{k}$ and $\vec{c} = -\alpha\hat{i} - 2\hat{j} + \hat{k}$, where $\alpha$ and $\beta$ are integers. If $\vec{a} \cdot \vec{b} = -1$ and $\vec{b} \cdot \vec{c} = 10$, then $(\vec{a} 	imes \vec{b}) \cdot \vec{c}$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To find the scalar triple product $(\vec{a} 	imes \vec{b}) \cdot \vec{c}$, we must first determine the values of the unknown variables $\alpha$ and $\beta$. We are given the dot product clues $\vec{a} \cdot \vec{b} = -1$ and $\vec{b} \cdot \vec{c} = 10$.

Using the definition of the dot product $\vec{a} \cdot \vec{b} = a_x b_x + a_y b_y + a_z b_z$, we substitute the components to get:
$$(1)(3) + (-\alpha)(\beta) + (\beta)(-\alpha) = -1$$

This simplifies to $3 - 2\alpha\beta = -1$, which yields our first vital equation:
$$\alpha\beta = 2$$

### The Quadratic Twist

Next, we apply the dot product logic to $\vec{b} \cdot \vec{c} = 10$:
$$(3)(-\alpha) + (\beta)(-2) + (-\alpha)(1) = 10$$

This simplifies to $-4\alpha - 2\beta = 10$, or more elegantly:
$$2\alpha + \beta = -5$$

We now have a system of two equations: $\alpha\beta = 2$ and $2\alpha + \beta = -5$. Substituting $\beta = \frac{2}{\alpha}$ into the second equation gives:
$$2\alpha + \frac{2}{\alpha} = -5$$

Multiplying by $\alpha$, we arrive at the quadratic equation:
$$2\alpha^2 + 5\alpha + 2 = 0$$

Factoring this expression, we find $(2\alpha + 1)(\alpha + 2) = 0$. This provides two potential values for $\alpha$: $-\frac{1}{2}$ and $-2$.

Since the problem explicitly states that $\alpha$ and $\beta$ must be **integers**, we reject the fraction and accept $\alpha = -2$. Consequently, we find $\beta = -1$.

### The Geometric Grand Finale

With our parameters $\alpha = -2$ and $\beta = -1$ confirmed, the vectors are fully defined as:
$\vec{a} = \hat{i} + 2\hat{j} - \hat{k}$
$\vec{b} = 3\hat{i} - \hat{j} + 2\hat{k}$
$\vec{c} = 2\hat{i} - 2\hat{j} + \hat{k}$

The scalar triple product is the determinant of the matrix formed by these vectors:
$$\begin{vmatrix} 1 & 2 & -1 \ 3 & -1 & 2 \ 2 & -2 & 1 \end{vmatrix}$$

Expanding along the first row, we calculate:
$$1[(-1)(1) - (2)(-2)] - 2[(3)(1) - (2)(2)] - 1[(3)(-2) - (-1)(2)]$$

This simplifies to $1(3) - 2(-1) - 1(-4)$, which equals $3 + 2 + 4 = 9$.

The volume of the parallelepiped is **9 cubic units**.

Enter Numerical Value:

Visualized Solution

Analyze the Given Vectors

Apply First Dot Product

Simplify First Equation

Apply Second Dot Product

Simplify Second Equation

Substitute $β$ into Equation 2

Form the Quadratic Equation

Solve for $α$ (Integer Constraint)

Find the Value of $β$

Define Numerical Vectors

Scalar Triple Product Concept

Set up the Determinant

Expand the Determinant

Calculate the Final Answer

The Sigma Insight: Scalar Triple Product

Analyzing the Setup

The Quadratic Twist

The Geometric Grand Finale

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