Question

If the triple scalar product \((\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{c}\) is equal to zero, then (i) \(a=0\), or \(b=0\), or \(c=0\) or (ii) two of the vectors are parallel, or (iii) the three vectors lie in the same plane (they are said to be coplanar). Show that the vectors $$ 2 \boldsymbol{i}-\boldsymbol{j}+\boldsymbol{k}, 3 \boldsymbol{i}-4 \boldsymbol{j}+5 \boldsymbol{k}, \boldsymbol{i}+2 \boldsymbol{j}-3 \boldsymbol{k} $$ are coplanar.

Short answer

Answer: Yes, the vectors are coplanar since their triple scalar product is equal to zero.

Step-by-step solution

Find the cross product of the first two vectors.

Let the first two vectors be represented as \(\boldsymbol{a} = (2, -1, 1)\) and \(\boldsymbol{b} = (3, -4, 5)\). Calculate their cross product \(\boldsymbol{a} \times \boldsymbol{b}\) using the formula $$ \boldsymbol{a} \times \boldsymbol{b} = \begin{pmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{pmatrix} $$

Evaluate the determinant.

Expanding the determinant, we have: $$ \boldsymbol{a} \times \boldsymbol{b} = \hat{i}(a_2b_3 - a_3b_2) - \hat{j}(a_1b_3 - a_3b_1) + \hat{k}(a_1b_2 - a_2b_1) $$ After substituting the components of \(\boldsymbol{a}\) and \(\boldsymbol{b}\), we get: $$ \boldsymbol{a} \times \boldsymbol{b} = \hat{i}((-1)(5) - (1)(-4)) - \hat{j}((2)(5) - (1)(3)) + \hat{k}((2)(-4) - (-1)(3)) $$

Simplify the cross product.

Now, simplify the cross product: $$ \boldsymbol{a} \times \boldsymbol{b} = \hat{i}(5+4) - \hat{j}(10-3) + \hat{k}(-8+3) \\ \boldsymbol{a} \times \boldsymbol{b} = 9\hat{i} - 7\hat{j} - 5\hat{k} $$

Calculate the dot product with the third vector.

Let the third vector be represented as \(\boldsymbol{c} = (1, 2, -3)\). Calculate the dot product \((\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{c}\) as follows: $$ (\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{c} = (9\hat{i} - 7\hat{j} - 5\hat{k}) \cdot (1\hat{i} + 2\hat{j} - 3\hat{k}) \\ = 9(1) - 7(2) - 5(-3) $$

Simplify the dot product.

Finally, simplify the dot product: $$ (\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{c} = 9 - 14 + 15 \\ (\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{c} = 0 $$ Since the triple scalar product is equal to zero, the vectors are coplanar.

Key concepts

Coplanar Vectors

Vectors are called **coplanar** if they all lie in the same plane, even if they are in a three-dimensional space. In simpler terms, if you have three vectors and you can imagine them on a flat sheet of paper without any of them pointing out of the paper, they are coplanar. For this particular scenario, the vectors do not necessitate being parallel; they only need to exist together within the same plane.
The condition for coplanarity can be mathematically expressed using the triple scalar product. If for three vectors \( \boldsymbol{a} \), \( \boldsymbol{b} \), and \( \boldsymbol{c} \), the triple scalar product \((\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{c}\) equals zero, then these vectors are coplanar.
This exercise demonstrates that the vectors:

• \(2 \boldsymbol{i} - \boldsymbol{j} + \boldsymbol{k}\)
• \(3 \boldsymbol{i} - 4 \boldsymbol{j} + 5 \boldsymbol{k}\)
• \(\boldsymbol{i} + 2 \boldsymbol{j} - 3 \boldsymbol{k}\)

are coplanar because their calculated triple scalar product is zero.

Cross Product

The **cross product** is an operation used in vector calculus to find a vector that is perpendicular to two given vectors. For vectors \( \boldsymbol{a} = (a_1, a_2, a_3) \) and \( \boldsymbol{b} = (b_1, b_2, b_3) \), the cross product \( \boldsymbol{a} \times \boldsymbol{b} \) is computed by determining the determinant of a 3x3 matrix.
The formula for the cross product is:\[ \boldsymbol{a} \times \boldsymbol{b} = \begin{pmatrix} \hat{i} & \hat{j} & \hat{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{pmatrix} \]
This involves finding the determinants of the cofactors multiplied by the unit vectors \(\hat{i}, \hat{j},\) and \(\hat{k}\). For instance, we found the cross product to be:

• \(9\hat{i} - 7\hat{j} - 5\hat{k}\)

This cross product provides a vector orthogonal to the initial pair, pointing in a different direction in the space.

Dot Product

The **dot product** is a scalar value calculated as the sum of the products of the corresponding components of two vectors. It measures the extent to which the vectors point in the same direction.
For vectors \( \boldsymbol{p} = (p_1, p_2, p_3) \) and \( \boldsymbol{q} = (q_1, q_2, q_3) \), the dot product \( \boldsymbol{p} \cdot \boldsymbol{q} \) is given by:\[ \boldsymbol{p} \cdot \boldsymbol{q} = p_1 q_1 + p_2 q_2 + p_3 q_3 \]
In the example, the dot product \((\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{c} \) for the vector \(\boldsymbol{c} = (1, 2, -3)\) resulted in `9(1) - 7(2) - 5(-3)`, which simplifies to zero. This zero result is critical because it confirms the vectors are coplanar. It indicates that the vector obtained from the cross product is orthogonal to the third vector.

Triple Scalar Product

The **triple scalar product** involves three vectors and combines both the cross and dot product operations. It is a measure of the "volume" of the parallelepiped formed by the three vectors. Mathematically, for vectors \( \boldsymbol{a}, \boldsymbol{b},\) and \(\boldsymbol{c}\), it is expressed as:\[(\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{c} \]
The result is a scalar that represents the volume of the parallelepiped spanned by the vectors. If this volume is zero, it indicates that the three vectors are coplanar, since they must lie on a flat surface, unable to form a three-dimensional shape.
In practical terms, the triple scalar product can be calculated as the determinant of a 3x3 matrix formed by the components of the vectors. In the given example, knowing that this product is zero confirms coplanarity, as no "thickness" exists for the shape.