Problem 9 - Essential Calculus Early Transcendentals

Question

Determine whether the points lie on a straight line. (a) \(A(2,4,2), \quad B(3,7,-2), \quad C(1,3,3)\) (b) \(D(0,-5,5), \quad E(1,-2,4), \quad F(3,4,2)\)

Step-by-Step Solution

Answer

(a) No, (b) Yes.

Step 1: Understand the Problem

We need to determine if the points given for each set (a) and (b) lie on a straight line in three-dimensional space.

Step 2: Use Vector Method for Collinearity

To check if three points lie on a straight line, we can use vectors. Points A, B, and C lie on a line if vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \) are parallel, meaning one is a scalar multiple of the other. This is true if the cross product of \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \) is zero.

Step 3: Find the Vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \)

For part (a), calculate \( \overrightarrow{AB} = B - A = (3-2, 7-4, -2-2) = (1, 3, -4) \) and \( \overrightarrow{AC} = C - A = (1-2, 3-4, 3-2) = (-1, -1, 1) \).

Step 4: Calculate the Cross Product for Part (a)

Compute the cross product \( \overrightarrow{AB} \times \overrightarrow{AC} \): \[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 3 & -4 \ -1 & -1 & 1 \end{vmatrix} \] This simplifies to \( \mathbf{i}(3 \times 1 - (-4) \times (-1)) - \mathbf{j}(1 \times 1 - (-4) \times (-1)) + \mathbf{k}(1 \times (-1) - 3 \times (-1)) \) = \(3 + 4, -1 - 4, -1 + 3 \) = \(7, -5, 2\). Since this vector is not zero, points A, B, and C do not lie on the same line.

Step 5: Repeat Steps for Part (b)

Find vectors \( \overrightarrow{DE} = E - D = (1-0, -2+5, 4-5) = (1, 3, -1) \) and \( \overrightarrow{DF} = F - D = (3-0, 4+5, 2-5) = (3, 9, -3) \).

Step 6: Calculate the Cross Product for Part (b)

Compute \( \overrightarrow{DE} \times \overrightarrow{DF} \): \[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 3 & -1 \ 3 & 9 & -3 \end{vmatrix} \] This simplifies to \( \mathbf{i}(3 \times (-3) - (-1) \times 9) - \mathbf{j}(1 \times (-3) - (-1) \times 3) + \mathbf{k}(1 \times 9 - 3 \times 3) \) = \((-9 + 9), (3 - 3), (9 - 9) \) = \(0, 0, 0\). Since this vector is zero, points D, E, and F lie on the same line.

Key Concepts

Vector Cross Product

The vector cross product is an essential tool in 3D geometry that helps us determine the relationship between two vectors. When you take the cross product of two vectors, you essentially get another vector that is perpendicular to both of the initial vectors. This new vector's direction is given by the right-hand rule.
To find the cross product of two vectors \(\mathbf{a} \times \mathbf{b}\), you arrange the components of the vectors in a 3x3 determinant like this:

• The first row consists of the unit vectors \(\mathbf{\hat{i}}, \mathbf{\hat{j}}, \mathbf{\hat{k}}\).
• The second row has the components of vector \(\mathbf{a}\).
• The third row contains the components of vector \(\mathbf{b}\).

To solve the determinant, you can apply cofactor expansion across the first row. Each component of the resulting vector can be calculated separately.
More precisely, if you have vectors \(\mathbf{a} = (a_1, a_2, a_3)\) and \(\mathbf{b} = (b_1, b_2, b_3)\), their cross product is given by:\[\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{\hat{i}} & \mathbf{\hat{j}} & \mathbf{\hat{k}} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix}\]This results in:\[(a_2b_3 - a_3b_2)\mathbf{\hat{i}} - (a_1b_3 - a_3b_1)\mathbf{\hat{j}} + (a_1b_2 - a_2b_1)\mathbf{\hat{k}}\]

Scalar Multiples of Vectors

Understanding scalar multiples is critical when exploring collinearity in vectors. A vector \(\mathbf{b}\) is a scalar multiple of another vector \(\mathbf{a}\) if there exists a scalar \(k\) such that \(\mathbf{b} = k \times \mathbf{a}\). This means that the two vectors are parallel and lie along the same line or direction.
Scalar multiplication is simple; each component of a vector is multiplied by the same scalar value. If you have a vector \(\mathbf{v} = (v_1, v_2, v_3)\) and a scalar \(k\), the scalar multiple \(\mathbf{w} = k\mathbf{v}\) is calculated as follows:

• For the \(x\)-component: \(w_1 = k \times v_1\)
• For the \(y\)-component: \(w_2 = k \times v_2\)
• For the \(z\)-component: \(w_3 = k \times v_3\)

If two vectors are scalar multiples, they will appear to overlap when plotted, underscoring their parallelism. Knowing when vectors are scalar multiples allows us to confirm if points are collinear.

Vector Method for Collinearity

In 3D geometry, the vector method for determining collinearity involves checking the cross product result of vectors created from three points. If you have three points \(A\), \(B\), and \(C\), you form vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\).
To check for collinearity, calculate the cross product \(\overrightarrow{AB} \times \overrightarrow{AC}\). If the result is the zero vector \((0, 0, 0)\), this indicates that the vectors are parallel, and hence, the points are collinear, lying on the same straight line.
Here's a simple step-by-step guide:

• Find vector \(\overrightarrow{AB}\) as \(B - A\).
• Find vector \(\overrightarrow{AC}\) as \(C - A\).
• Calculate the cross product \(\overrightarrow{AB} \times \overrightarrow{AC}\).
• If the cross product equals zero, the points are collinear.

This method conclusively tells us whether three points in 3D space lie along a single straight line. It's an invaluable technique that combines the geometric properties of vectors with algebraic computations.