Question

Use cross products to determine whether the points \(A, B,\) and C are collinear. $$A(-3,-2,1), B(1,4,7), \text { and } C(4,10,14)$$

Short answer

Answer: No, the points A, B, and C are not collinear.

Step-by-step solution

Find vectors \(\vec{AB}\) and \(\vec{AC}\)

First of all, we need to find the vectors formed between the points given. We start by finding out the vector formed between points A and B as follows: $$\vec{AB} = B - A = (1 - (-3), 4 - (-2), 7 - 1)=(4, 6, 6) $$ Similarly, we can find the vector between points A and C: $$\vec{AC} = C - A = (4 - (-3), 10 - (-2), 14 - 1)=(7, 12, 13) $$ Now that we have these two vectors, we are ready to find the cross product.

Compute the cross product of vectors \(\vec{AB}\) and \(\vec{AC}\)

The cross product of two vectors \(\vec{AB}\) and \(\vec{AC}\) is calculated as follows: $$\vec{AB} \times \vec{AC} = \left| \begin{array}{ccc} \hat i & \hat j & \hat k\\ 4 & 6 & 6\\ 7 & 12 & 13 \end{array} \right| $$ Now, we can solve the determinant: $$\vec{AB} \times \vec{AC} = \left| \begin{array}{cc} 6 & 6 \\ 12 & 13 \end{array} \right| \hat i - \left| \begin{array}{cc} 4 & 6 \\ 7 & 13 \end{array} \right| \hat j + \left| \begin{array}{cc} 4 & 6 \\ 7 & 12 \end{array} \right| \hat k$$ Computing the determinants, we get: $$\vec{AB} \times \vec{AC} = (6(13) - 6(12))\hat i - (4(13)-6(7))\hat j + (4(12)-6(7))\hat k = (6)\hat i + (2)\hat j + (2)\hat k$$ We got a nonzero vector as the cross product so the points are not collinear.

Final Step: Conclusion

Since the cross product of vectors \(\vec{AB}\) and \(\vec{AC}\) is nonzero, the points A, B, and C are not collinear.

Key concepts

Vectors

A vector is a mathematical object that has both a magnitude and a direction. In a three-dimensional space, we express vectors using coordinates. For example, if we have two points A and B, with coordinates \(A(x_1, y_1, z_1)\) and \(B(x_2, y_2, z_2)\), the vector from A to B is given by \(\vec{AB} = (x_2 - x_1, y_2 - y_1, z_2 - z_1)\).
Vectors are essential in physics and engineering because they help describe quantities that have directions, like force and velocity.

• Vectors can be added together to find a resultant vector.
• Vectors can be multiplied by scalars (real numbers) to change their magnitudes.
• Vectors allow us to solve problems related to geometry and physics.

In this exercise, we find vectors \( \vec{AB} \) and \( \vec{AC} \) to later determine if the points A, B, and C are collinear.

Cross Product

The cross product is an operation on two vectors in three-dimensional space that results in another vector. This new vector is perpendicular to the plane formed by the original two vectors. The cross product is useful for finding the normal vector, which helps in determining concepts like torque and angular momentum.
To compute the cross product, we use the determinant of a matrix that includes the unit vectors \(\hat{i}, \hat{j}, \hat{k}\) and the components of the vectors:\[\vec{AB} \times \vec{AC} = \left| \begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \4 & 6 & 6 \7 & 12 & 13 \end{array} \right|\] The cross product has applications such as:

• Determining if two vectors are parallel (cross product is zero).
• Finding the area of parallelograms spanned by two vectors.
• Determining orientation of vectors.

In this problem, a nonzero cross product means the vectors \( \vec{AB} \) and \( \vec{AC} \) are not parallel, indicating that points A, B, and C are not collinear.

Determinants

Determinants provide a scalar value that can be calculated from the elements of a square matrix. They are incredibly helpful in evaluating properties of linear transformations, solving systems of linear equations, and in geometric interpretations.
When we compute the cross product with vectors \( \vec{AB} \) and \( \vec{AC} \), we use a 3x3 determinant.
The determinant helps us find the new vector by calculating the values using the following smaller 2x2 determinants:\[\vec{AB} \times \vec{AC} = \left| \begin{array}{cc} 6 & 6 \ 12 & 13 \end{array} \right| \hat{i} - \left| \begin{array}{cc} 4 & 6 \ 7 & 13 \end{array} \right| \hat{j} + \left| \begin{array}{cc} 4 & 6 \ 7 & 12 \end{array} \right| \hat{k}\]Calculating determinants:

• For a 2x2 matrix \( \begin{pmatrix} a & b \ c & d \end{pmatrix} \), the determinant is \(ad - bc\).
• Using determinants helps find solutions in algebraic equations.

In this exercise, the non-zero value of the cross product's determinant confirms the points are not collinear.