Question

Consider the cube in the first octant with sides \(O P, O Q\) and \(O R\) of length 1 , along the \(x\) -axis, \(y\) -axis and \(z\) -axis, respectively, where \(O(0,0,0)\) is the origin. Let \(S\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)\) 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 \(O T\). If \(\vec{p}=\overrightarrow{S P}, \vec{q}=\overrightarrow{S Q}, \vec{r}=\overrightarrow{S R}\) and \(\vec{t}=\overrightarrow{S T}\), then the value of \(|(\vec{p} \times \vec{q}) \times(\vec{r} \times \vec{t})|\) is

Short answer

The value of \(|(\vec{p} \times \vec{q}) \times(\vec{r} \times \vec{t})|\) is 0.

Step-by-step solution

Identify the position vectors of points P, Q, R, and T

Since the cube is in the first octant with sides OP, OQ, and OR of length 1 along the x-axis, y-axis, and z-axis respectively, the coordinates of points P, Q, and R are P(1,0,0), Q(0,1,0), and R(0,0,1). The vertex T opposite to O has coordinates T(1,1,1). To find the position vectors of P, Q, R, and T with respect to S, we subtract S' coordinates from each point's coordinates: \(\vec{p} = P - S = (1 - \frac{1}{2}, 0 - \frac{1}{2}, 0 - \frac{1}{2})\), \(\vec{q} = Q - S = (0 - \frac{1}{2}, 1 - \frac{1}{2}, 0 - \frac{1}{2})\), \(\vec{r} = R - S = (0 - \frac{1}{2}, 0 - \frac{1}{2}, 1 - \frac{1}{2})\), and \(\vec{t} = T - S = (1 - \frac{1}{2}, 1 - \frac{1}{2}, 1 - \frac{1}{2})\).

Calculate the vectors \(\vec{p}\), \(\vec{q}\), \(\vec{r}\), and \(\vec{t}\)

Using the coordinates found in Step 1, we find the vectors \(\vec{p} = (\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2})\), \(\vec{q} = (-\frac{1}{2}, \frac{1}{2}, -\frac{1}{2})\), \(\vec{r} = (-\frac{1}{2}, -\frac{1}{2}, \frac{1}{2})\), and \(\vec{t} = (\frac{1}{2}, \frac{1}{2}, \frac{1}{2})\).

Compute the cross products \(\vec{p} \times \vec{q}\) and \(\vec{r} \times \vec{t}\)

To find the cross products, we can use the determinant of a matrix with the unit vectors \(i, j, k\) in the first row, the components of the first vector in the second row, and the components of the second vector in the third row: \(\vec{p} \times \vec{q} = \begin{vmatrix} i & j & k \ \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} \ -\frac{1}{2} & \frac{1}{2} & -\frac{1}{2} \ \end{vmatrix}\) and \(\vec{r} \times \vec{t} = \begin{vmatrix} i & j & k \ -\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \ \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \end{vmatrix}\). Calculate these determinants to get the cross products.

Calculate the double cross product

The double cross product \((\vec{p} \times \vec{q}) \times (\vec{r} \times \vec{t})\) can again be calculated using the determinant method as in Step 3. Use the components of the cross products obtained in Step 3 to compute this.

Find the magnitude of the double cross product

The magnitude of a vector \(\vec{v}\) is given by \(|\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}\). Use this formula to calculate the magnitude of the double cross product obtained in Step 4.

Key concepts

Cross Product

In vector algebra, the cross product is a binary operation on two vectors in three-dimensional space. It results in a third vector that is perpendicular to the plane containing the original vectors. Deemed useful for determining the area, torque, and direction, the cross product has significant physical representations.

The cross product of two vectors \(\vec{a}\) and \(\vec{b}\) is denoted \(\vec{a} \times \vec{b}\) and is calculated using the determinant of a matrix with the unit vectors \(i, j, k\) along the top row, and the components of each vector beneath them. Formally, if \(\vec{a} = (a_x, a_y, a_z)\) and \(\vec{b} = (b_x, b_y, b_z)\), then:
\[\vec{a} \times \vec{b} = \begin{vmatrix} i & j & k \ a_x & a_y & a_z \ b_x & b_y & b_z \end{vmatrix}\]
For example, in our cube problem from the textbook, to find the cross product of \(\vec{p}\) and \(\vec{q}\), we set up a determinant with \(\vec{p}\)'s and \(\vec{q}\)'s components and calculate. The use of the cross product tells us about the area spanned by two sides of the cube and the orientation of the resulting vector in space. It is important to remember that the cross product is not commutative, meaning \(\vec{a} \times \vec{b} eq \vec{b} \times \vec{a}\) and actually \(\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})\).

Position Vectors

Position vectors serve as foundational elements in vector algebra. A position vector denotes the location of a point in space relative to an origin, which is typically the point \(O(0,0,0)\) in a three-dimensional coordinate system. In the context of the exercise from the textbook, \(\vec{p}\), \(\vec{q}\), \(\vec{r}\), and \(\vec{t}\) are position vectors expressing the locations of points P, Q, R, and T with respect to point S, the center of the cube.

To find a position vector such as \(\vec{p}\), one simply subtracts the coordinates of the starting point S from point P, resulting in \(\vec{p} = P - S\). This subtraction yields the directed line segment from S to P and provides the components of the position vector. The concept of position vectors is fundamental in various applications, ranging from physics to engineering, as it allows for the expression of points in space in a standardized mathematical form.

Magnitude of Vectors

The magnitude of a vector is a measure of its length and is a crucial concept in vector algebra. Mathematically, the magnitude is defined as the square root of the sum of the squares of its components. For any vector \(\vec{v} = (v_x, v_y, v_z)\), the magnitude is denoted and calculated as follows:
\[|\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}\]
Using this formula, one can compute the magnitude of the cross product vectors found in the textbook exercise. The concept of magnitude becomes particularly important when calculating the length of a vector, which in the context of the exercise refers to the size of the diagonal spanned by two vectors in the cross product. The magnitude is always a non-negative value and gives a sense of scale and distance in vector analysis.

Through understanding how to properly find the magnitude, students can interpret physical quantities like force, velocity, and displacement that are often represented as vectors in physics. It can also help in normalizing vectors, resulting in a unit vector which is essential for direction-related calculations without considering the scale.