Problem 56 - University Calculus: Early Transcendentals

Question

Determine whether the given points are coplanar. $$A(0,0,4), \quad B(6,2,0), \quad C(2,-1,1), \quad D(-3,-4,3)$$

Step-by-Step Solution

Answer

The points are not coplanar.

Step 1: Recognize the Points

The four points given are: \( A(0,0,4) \), \( B(6,2,0) \), \( C(2,-1,1) \), and \( D(-3,-4,3) \). We need to determine if these points lie on the same plane.

Step 2: Calculate Vectors between the Points

Compute vectors \( \overrightarrow{AB} \), \( \overrightarrow{AC} \), and \( \overrightarrow{AD} \).- \( \overrightarrow{AB} = B - A = (6-0, 2-0, 0-4) = (6, 2, -4) \)- \( \overrightarrow{AC} = C - A = (2-0, -1-0, 1-4) = (2, -1, -3) \)- \( \overrightarrow{AD} = D - A = (-3-0, -4-0, 3-4) = (-3, -4, -1) \)

Step 3: Compute the Scalar Triple Product

The scalar triple product of vectors \( \overrightarrow{AB} \), \( \overrightarrow{AC} \), and \( \overrightarrow{AD} \) is given by \( \overrightarrow{AB} \cdot (\overrightarrow{AC} \times \overrightarrow{AD}) \). Calculate it using the determinant:\[\begin{vmatrix} 6 & 2 & -4 \ 2 & -1 & -3 \ -3 & -4 & -1\end{vmatrix}\]First, find the cross product \( \overrightarrow{AC} \times \overrightarrow{AD} \):\[\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & -1 & -3 \ -3 & -4 & -1\end{vmatrix}\]= \( \mathbf{i}((-1)(-1) - (-3)(-4)) - \mathbf{j}(2(-1) - (-3)(-3)) + \mathbf{k}(2(-4) - (-1)(-3)) \) = \( \mathbf{i}(1 + 12) - \mathbf{j}(-2 + 9) + \mathbf{k}(-8 + 3) \) = \( 13\mathbf{i} - 7\mathbf{j} - 5\mathbf{k} \).Next, calculate the dot product \( \overrightarrow{AB} \cdot \) \((13, -7, -5)\):\[6(13) + 2(-7) + (-4)(-5) = 78 - 14 + 20 = 84\].

Step 4: Determine Coplanarity

Since the scalar triple product is 84 (not zero), the points are not coplanar. If the scalar triple product were zero, the points would have been coplanar.

Key Concepts

Vector Operations

Vectors are essential in understanding positions and movements in space. A vector represents a quantity with both magnitude and direction.
In our context, we are interested in vectors connecting certain points. For example, to calculate the vector \( \overrightarrow{AB} \), we subtract the coordinates of point \( A \) from point \( B \). This gives us a new vector that starts from \( A \) and points towards \( B \).
Here are some essential operations you might perform with vectors:

• Addition: For vectors \( \mathbf{u} = (u_1, u_2, u_3) \) and \( \mathbf{v} = (v_1, v_2, v_3) \), \( \mathbf{u} + \mathbf{v} = (u_1+v_1, u_2+v_2, u_3+v_3) \).
• Subtraction: Similarly, \( \mathbf{u} - \mathbf{v} \) involves subtracting each component respectively.
• Scalar Multiplication: Scaling a vector \( \mathbf{u} \) by a scalar \( a \) gives \( a\mathbf{u} = (au_1, au_2, au_3) \).

These operations allow us to explore relationships between points in a 3-dimensional space.

Scalar Triple Product

The scalar triple product is a useful tool when it comes to checking the coplanarity of three vectors. It involves calculating both a cross product and a dot product.
Given three vectors \( \mathbf{u} \), \( \mathbf{v} \), and \( \mathbf{w} \), the scalar triple product is formed by: \( \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) \).
This process paints a geometric picture as such:

• Determinant Interpretation: The scalar triple product can be seen as the determinant of a 3x3 matrix. The vectors form the matrix rows, with the result being a measure of the volume of the parallelepiped framed by these vectors.
• Coplanarity Check: If the scalar triple product is zero, it indicates that the vectors (and thus points derived from these vectors) are coplanar, lying on the same plane.

In our example, calculating the scalar triple product results in 84. Since it is non-zero, points \( A \), \( B \), \( C \), and \( D \) are not coplanar.

Cross Product

The cross product is a vector operation that results in a vector perpendicular to the initial two vectors in three-dimensional space. This operation is denoted by the symbol \( \times \).
For vectors \( \mathbf{a} = (a_1, a_2, a_3) \) and \( \mathbf{b} = (b_1, b_2, b_3) \), the cross product \( \mathbf{a} \times \mathbf{b} \) is calculated as:
\[\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix}\]Here is a simple breakdown of its calculation:

• The resulting vector is calculated using \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) components from the determinant. It's a formulaic approach to capturing geometric insights.
• These computed vectors provide the normal direction, which can be used in assessing the volume or orientation of a space or plane.

Cross products are not commutative, meaning \( \mathbf{a} \times \mathbf{b} eq \mathbf{b} \times \mathbf{a} \), providing a hint about directionality.

Dot Product

The dot product, or scalar product, results in a scalar when given two vectors. This operation tells us about the angle between the vectors and their alignment.
For vectors \( \mathbf{a} = (a_1, a_2, a_3) \) and \( \mathbf{b} = (b_1, b_2, b_3) \), the dot product is calculated via:
\( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \).
In practical terms:

• Magnitude Relation: Determine if vectors form a right angle (orthogonal) by checking if the dot product equals zero.
• Projection Insight: By considering the dot product in conjunction with magnitudes, it can be used to project one vector onto another, providing insights about shadowing or alignment.

In the context of our exercise, the dot product was used in the scalar triple product to determine if the computed cross product vector aligns or misaligns with the third vector, influencing the coplanarity condition.