Consider the cube in the first octant with sides OP, OQ and OR of length 1, along the x-axis, y-axis

Q: Why can't I just calculate the cross products directly? It seems straightforward.

While calculating the cross products directly is mathematically valid, it is a significant trap in competitive exams like JEE. You would have to compute two separate cross products, then a third one, involving multiple determinants and a high probability of sign errors. The vector quadruple product identity is the 'JEE shortcut' that transforms a tedious, error-prone calculation into an elegant algebraic reduction. Always look for these identities to save time and ensure accuracy.

Q: How do I know which vector to treat as $\vec{C}$ in the identity?

The identity $(\vec{A} \times \vec{B}) \times \vec{C} = (\vec{A} \cdot \vec{C})\vec{B} - (\vec{B} \cdot \vec{C})\vec{A}$ is designed for a cross product of two vectors, where the second term is itself a cross product. By defining $\vec{C} = (\vec{r} \times \vec{t})$, you treat the entire second parenthesis as a single vector. This allows you to apply the identity directly, simplifying the expression into a linear combination of $\vec{p}$ and $\vec{q}$.

Q: Is there a geometric interpretation of the scalar triple product $[\vec{p} \vec{r} \vec{t}]$?

Absolutely! The scalar triple product $[\vec{a} \vec{b} \vec{c}]$ represents the signed volume of the parallelepiped formed by vectors $\vec{a}, \vec{b},$ and $\vec{c}$. In this problem, it serves as a powerful tool to project the vectors onto the normal of the plane formed by $\vec{r}$ and $\vec{t}$. It effectively collapses the 3D complexity into a simple scalar value, which is the key to solving this problem efficiently.

Question

Consider the cube in the first octant with sides $OP, OQ$ and $OR$ of length $1$, along the $x$-axis, $y$-axis and $z$-axis, respectively, where $O(0, 0, 0)$ is the origin. Let $S\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}ight)$ be the centre of the cube and $T$ be the vertex of the cube opposite to the origin $O$ such that $S$ lies on the diagonal $OT$. If $\vec{p} = \vec{SP}, \vec{q} = \vec{SQ}, \vec{r} = \vec{SR}$ and $\vec{t} = \vec{ST}$, then the value of $|(\vec{p} 	imes \vec{q}) 	imes (\vec{r} 	imes \vec{t})|$ is ________.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a 3D coordinate system, looking at a cube perfectly aligned with the axes in the first octant. The origin $O(0,0,0)$ is at your feet, and the cube extends to the point $T(1,1,1)$.

We are given the center of the cube, $S(\frac{1}{2}, \frac{1}{2}, \frac{1}{2})$, and four vectors radiating from this center: $\vec{p} = \vec{SP}$, $\vec{q} = \vec{SQ}$, $\vec{r} = \vec{SR}$, and $\vec{t} = \vec{ST}$. Our mission is to find the magnitude of the vector quadruple product $|(\vec{p} 	imes \vec{q}) 	imes (\vec{r} 	imes \vec{t})|$.

### Defining the Vectors

To proceed, we determine the components by subtracting the coordinates of $S$ from the vertices $P(1,0,0)$, $Q(0,1,0)$, $R(0,0,1)$, and $T(1,1,1)$.

$$\vec{p} = P - S = \frac{1}{2}(\hat{i} - \hat{j} - \hat{k})$$
$$\vec{q} = Q - S = \frac{1}{2}(-\hat{i} + \hat{j} - \hat{k})$$
$$\vec{r} = R - S = \frac{1}{2}(-\hat{i} - \hat{j} + \hat{k})$$
$$\vec{t} = T - S = \frac{1}{2}(\hat{i} + \hat{j} + \hat{k})$$

Notice the symmetry here; each vector is a combination of $\hat{i}$, $\hat{j}$, and $\hat{k}$ with coefficients of $\pm \frac{1}{2}$. This symmetry is our best friend in simplifying the upcoming calculations.

### The Power of the Quadruple Product Identity

We face the expression $|(\vec{p} 	imes \vec{q}) 	imes (\vec{r} 	imes \vec{t})|$. Instead of calculating cross products directly, we use the vector quadruple product identity: $(\vec{A} 	imes \vec{B}) 	imes \vec{C} = (\vec{A} \cdot \vec{C})\vec{B} - (\vec{B} \cdot \vec{C})\vec{A}$.

Let $\vec{C} = \vec{r} 	imes \vec{t}$. Our expression transforms into:

$$(\vec{p} 	imes \vec{q}) 	imes \vec{C} = (\vec{p} \cdot \vec{C})\vec{q} - (\vec{q} \cdot \vec{C})\vec{p}$$

This simplifies to the scalar triple products: $[\vec{p} \vec{r} \vec{t}]\vec{q} - [\vec{q} \vec{r} \vec{t}]\vec{p}$. The problem is now reduced to calculating two scalar triple products.

### The Calculation

First, we compute $\vec{r} 	imes \vec{t}$:

$$\vec{r} 	imes \vec{t} = \frac{1}{4} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & -1 & 1 \ 1 & 1 & 1 \end{vmatrix} = \frac{1}{4} (-2\hat{i} + 2\hat{j} + 0\hat{k}) = \frac{1}{2}(-\hat{i} + \hat{j})$$

Next, we evaluate the scalar triple products:

$$[\vec{p} \vec{r} \vec{t}] = \vec{p} \cdot (\vec{r} 	imes \vec{t}) = \frac{1}{2}(\hat{i} - \hat{j} - \hat{k}) \cdot \frac{1}{2}(-\hat{i} + \hat{j}) = \frac{1}{4}(-1 - 1) = -\frac{1}{2}$$
$$[\vec{q} \vec{r} \vec{t}] = \vec{q} \cdot (\vec{r} 	imes \vec{t}) = \frac{1}{2}(-\hat{i} + \hat{j} - \hat{k}) \cdot \frac{1}{2}(-\hat{i} + \hat{j}) = \frac{1}{4}(1 + 1) = \frac{1}{2}$$

Substituting these values back into our identity yields:

$$-\frac{1}{2}\vec{q} - \frac{1}{2}\vec{p} = -\frac{1}{2}(\vec{p} + \vec{q})$$

### Final Calculation

Finally, we compute the sum $\vec{p} + \vec{q}$:

$$\vec{p} + \vec{q} = \frac{1}{2}(\hat{i} - \hat{j} - \hat{k}) + \frac{1}{2}(-\hat{i} + \hat{j} - \hat{k}) = \frac{1}{2}(0\hat{i} + 0\hat{j} - 2\hat{k}) = -\hat{k}$$

Our resulting vector is $-\frac{1}{2}(-\hat{k}) = \frac{1}{2}\hat{k}$. The magnitude of this vector is:

$$|\frac{1}{2}\hat{k}| = \mathbf{0.5}$$

Enter Numerical Value:

Visualized Solution

Cube in the First Octant

Locating the Center $S$

Defining the Position Vectors

Calculating Vector Components

Vector Quadruple Product Expansion

Applying the Expansion

Calculating $r \times t$

Evaluating $[p r t]$

Evaluating $[q r t]$

Substituting Back

Calculating $p + q$

Final Magnitude

The Sigma Insight: Vector Triple Product

Analyzing the Setup

Defining the Vectors

The Power of the Quadruple Product Identity

The Calculation

Final Calculation

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