5.4: The Cross Product (2024)

Definition: cross product

Let \(\vecs u=⟨u_1,u_2,u_3⟩\) and \(\vecs v=⟨v_1,v_2,v_3⟩.\) Then, the cross product \(\vecs u×\vecs v\) is vector

\[\begin{align} \vecs u×\vecs v &= (u_2v_3−u_3v_2)\hat{\imath}−(u_1v_3−u_3v_1) \hat{\jmath}+(u_1v_2−u_2v_1)\hat{k} \nonumber \\[5pt] &=⟨u_2v_3−u_3v_2,−(u_1v_3−u_3v_1),u_1v_2−u_2v_1⟩. \label{cross}\end{align}\]

Example \(\PageIndex{1}\): Finding a Cross Product

Let \(\vecs p=⟨−1,2,5⟩\) and \(\vecs q=⟨4,0,−3⟩\) (Figure \(\PageIndex{1}\)). Find \(\vecs p×\vecs q\).

Solution

Substitute the components of the vectors into Equation \ref{cross}:

\[\begin{align*} \vecs p×\vecs q&=⟨−1,2,5⟩×⟨4,0,−3⟩ \\[5pt] &= ⟨p_2q_3−p_3q_2,p_1q_3−p_3q_1,p_1q_2−p_2q_1⟩ \\[5pt] &= ⟨2(−3)−5(0),−(−1)(−3)+5(4),(−1)(0)−2(4)⟩ \\[5pt] &= ⟨−6,17,−8⟩.\end{align*}\]

Exercise \(\PageIndex{1}\)

Find \(\vecs p×\vecs q\) for \(\vecs p=⟨5,1,2⟩\) and \(\vecs q=⟨−2,0,1⟩.\) Express the answer using standard unit vectors.

Hint

Use the formula \(\vecs u×\vecs v=(u_2v_3−u_3v_2)\hat{\imath}−(u_1v_3−u_3v_1)\hat{\jmath}+(u_1v_2−u_2v_1)\hat{k}.\)

Answer

\(\hat{\imath}−9\hat{\jmath}+2\hat{k}\)

Example \(\PageIndex{2}\): Anticommutativity of the Cross Product

Let \(\vecs u=⟨0,2,1⟩\) and \(\vecs v=⟨3,−1,0⟩\). Calculate \(\vecs u×\vecs v\) and \(\vecs v×\vecs u\) and graph them.

Solution

We have

\(\vecs u×\vecs v=⟨(0+1),−(0−3),(0−6)⟩=⟨1,3,−6⟩\)

\(\vecs v×\vecs u=⟨(−1−0),−(3−0),(6−0)⟩=⟨−1,−3,6⟩.\)

We see that, in this case, \(\vecs u×\vecs v=−(\vecs v×\vecs u)\) (Figure \(\PageIndex{4}\)). We prove this in general later in this section.

Exercise \(\PageIndex{2}\)

Suppose vectors \(\vecs u\) and \(\vecs v\) lie in the \(xy\)-plane (the \(z\)-component of each vector is zero). Now suppose the \(x\)- and \(y\)-components of \(\vecs u\) and the \(y\)-component of \(\vecs v\) are all positive, whereas the \(x\)-component of \(\vecs v\) is negative. Assuming the coordinate axes are oriented in the usual positions, in which direction does \(\vecs u×\vecs v\) point?

Hint

Remember the right-hand rule (Figure \(\PageIndex{2}\)).

Answer

Up (the positive \(z\)-direction)

Example \(\PageIndex{3}\): Cross Product of Standard Unit Vectors

Find \(\hat{\imath} ×(\hat{\jmath}×\hat{k})\).

Solution

We know that \(\hat{\jmath}×\hat{k}=\hat{\imath}\). Therefore, \(\hat{\imath}×(\hat{\jmath}×\hat{k})=\hat{\imath}×\hat{\imath}=0.\)

Exercise \(\PageIndex{3}\)

Find \((\hat{\imath}×\hat{\jmath})×(\hat{k}×\hat{\imath}).\)

Hint

Remember the right-hand rule (Figure \(\PageIndex{2}\)).

Answer

\(−\hat{\imath}\)

Properties of the Cross Product

Let \(\vecs u,\vecs v,\) and \(\vecs w\) be vectors in space, and let \(c\) be a scalar.

1. Anticommutative property: \[\vecs u×\vecs v=−(\vecs v×\vecs u)\]
2. Distributive property: \[\vecs u×(\vecs v+\vecs w)=\vecs u×\vecs v+\vecs u×\vecs w\]
3. Multiplication by a constant: \[c(\vecs u×\vecs v)=(c\vecs u)×\vecs v=\vecs u×(c\vecs v)\]
4. Cross product of the zero vector: \[\vecs u×\vecs 0=\vecs 0×\vecs u=\vecs 0\]
5. Cross product of a vector with itself: \[\vecs v×\vecs v=\vecs 0\]
6. Scalar triple product: \[\vecs u⋅(\vecs v×\vecs w)=(\vecs u×\vecs v)⋅\vecs w\]

Proof

For property \(i\), we want to show \(\vecs u×\vecs v=−(\vecs v×\vecs u).\) We have

\[\begin{align*} \vecs u×\vecs v &=⟨u_1,u_2,u_3⟩×⟨v_1,v_2,v_3⟩ \\[5pt] &=⟨u_2v_3−u_3v_2,−u_1v_3+u_3v_1,u_1v_2−u_2v_1⟩ \\[5pt] &=−⟨u_3v_2−u_2v_3,−u_3v_1+u_1v_3,u_2v_1−u_1v_2⟩ \\[5pt] &=−⟨v_1,v_2,v_3⟩×⟨u_1,u_2,u_3⟩\\[5pt] &=−(\vecs v×\vecs u).\end{align*}\]

Unlike most operations we’ve seen, the cross product is not commutative. This makes sense if we think about the right-hand rule.

For property \(iv\)., this follows directly from the definition of the cross product. We have

\[\vecs u × \vecs 0=⟨u_2(0)−u_3(0),−(u_2(0)−u_3(0)),u1(0)−u_2(0)⟩=⟨0,0,0⟩=\vecs 0. \]

Then, by property i., \(\vecs 0×\vecs u=\vecs 0\) as well. Remember that the dot product of a vector and the zero vector is the scalar \(0\), whereas the cross product of a vector with the zero vector is the vector \(\vecs 0\).

Property \(vi\). looks like the associative property, but note the change in operations:

\[\begin{align*} \vecs u⋅(\vecs v×\vecs w)&=u⋅⟨v_2w_3−v_3w_2,−v_1w_3+v_3w_1,v_1w_2−v_2w_1⟩ \\[5pt] &= u_1(v_2w_3−v_3w_2)+u_2(−v_1w_3+v_3w_1)+u_3(v_1w_2−v_2w_1) \\[5pt] &=u_1v_2w_3−u_1v_3w_2−u_2v_1w_3+u_2v_3w_1+u_3v_1w_2−u_3v_2w_1 \\[5pt] &=(u_2v_3−u_3v_2)w_1+(u_3v_1−u_1v_3)w_2+(u_1v_2−u_2v_1)w_3 \\[5pt] &=⟨u_2v_3−u_3v_2,u_3v_1−u_1v_3,u_1v_2−u_2v_1⟩⋅⟨w_1,w_2,w_3⟩ =(\vecs u×\vecs v)⋅\vecs w.\end{align*}\]

\(\square\)

Example \(\PageIndex{4}\): Using the Properties of the Cross Product

Use the cross product properties to calculate \((2\hat{\imath}×3\hat{\jmath})×\hat{\jmath}.\)

Solution

\[\begin{align*} (2\hat{\imath}×3 \hat{\jmath})×\hat{\jmath}&=2(\hat{\imath}×3\hat{\jmath})×\hat{\jmath} \\[5pt] &=2(3)(\hat{\imath}×\hat{\jmath})×\hat{\jmath} \\[5pt] &=(6\hat{k})×\hat{\jmath} \\[5pt] &=6(\hat{k}×\hat{\jmath}) \\[5pt] &=6(−\hat{\imath})=−6\hat{\imath}. \end{align*}\]

Exercise \(\PageIndex{4}\)

Use the properties of the cross product to calculate \((\hat{\imath}×\hat{k})×(\hat{k}×\hat{\jmath}).\)

Hint

\(\vecs u×\vecs v=−(\vecs v×\vecs u)\)

Answer

\(−\hat{k}\)

Magnitude of the Cross Product

Let \(\vecs u\) and \(\vecs v\) be vectors, and let \(θ\) be the angle between them. Then, \(‖\vecs u×\vecs v‖=‖\vecs u‖⋅‖\vecs v‖⋅\sin θ.\)

Proof

Let \(\vecs u=⟨u_1,u_2,u_3⟩\) and \(\vecs v=⟨v_1,v_2,v_3⟩\) be vectors, and let \(θ\) denote the angle between them. Then

\[ \begin{align*} ‖\vecs u×\vecs v‖^2&=(u_2v_3−u_3v_2)^2+(u_3v_1−u_1v_3)^2+(u_1v_2−u_2v_1)^2 \\[5pt] &=u^2_2v^2_3−2u_2u_3v_2v_3+u^2_3v^2_2+u^2_3v^2_1−2u_1u_3v_1v_3+u^2_1v^2_3+u^2_1v^2_2−2u_1u_2v_1v_2+u^2_2v^2_1 \\[5pt] &=u^2_1v^2_1+u^2_1v^2_2+u^2_1v^2_3+u^2_2v^2_1+u^2_2v^2_2+u^2_2v^2_3+u^2_3v^2_1+u^2_3v^2_2+u^2_3v^2_3−(u^2_1v^2_1+u^2_2v^2_2+u^2_3v^2_3+2u_1u_2v_1v_2+2u_1u_3v_1v_3+2u_2u_3v_2v_3) \\[5pt] &=(u^2_1+u^2_2+u^2_3)(v^2_1+v^2_2+v^2_3)−(u_1v_1+u_2v_2+u_3v_3)^2 \\[5pt] &=‖\vecs u‖^2‖\vecs v‖^2−(\vecs u⋅\vecs v)^2 \\[5pt] &=‖\vecs u‖^2‖\vecs v‖^2−‖\vecs u‖^2‖\vecs v‖^2 \cos^2θ \\[5pt] &=‖\vecs u‖^2‖\vecs v‖^2(1−\cos^2θ) \\[5pt] &=‖\vecs u‖^2‖\vecs v‖^2(\sin^2θ). \end{align*} \]

Taking square roots and noting that \(\sqrt{\sin^2θ}=\sinθ\) for \(0≤θ≤180°,\) we have the desired result:

\[‖\vecs u×\vecs v‖=‖\vecs u‖‖\vecs v‖ \sin θ.\]

□

Example \(\PageIndex{5}\): Calculating the Cross Product

Use Note to find the magnitude of the cross product of \(\vecs u=⟨0,4,0⟩\) and \(\vecs v=⟨0,0,−3⟩\).

Solution

We have

\[\begin{align*} ‖\vecs u×\vecs v‖ &= ‖\vecs u‖⋅‖\vecs v‖⋅\sinθ \\[5pt] &=\sqrt{0^2+4^2+0^2}⋅\sqrt{0^2+0^2+(−3)^2}⋅\sin{\dfrac{π}{2}} \\[5pt] &=4(3)(1)=12 \end{align*}\]

Exercise \(\PageIndex{5}\)

Use Note to find the magnitude of \(\vecs u×\vecs v\), where \(\vecs u=⟨−8,0,0⟩\) and \(\vecs v=⟨0,2,0⟩\).

Hint

Vectors \(\vecs u\) and \(\vecs v\) are orthogonal.

Answer

16

Example \(\PageIndex{6}\): Using Expansion Along the First Row to Compute a \(3×3\) Determinant

Evaluate the determinant \(\begin{bmatrix}2&5&−1\\−1&1&3\\−2&3&4\end{bmatrix}\).

Solution

We have

\[\begin{align*} \begin{bmatrix}2&5&−1\\−1&1&3\\−2&3&4\end{bmatrix} &=2\begin{bmatrix}1&3\\3&4\end{bmatrix}−5\begin{bmatrix}−1&3\\−2&4\end{bmatrix}−1\begin{bmatrix}−1&1\\−2&3\end{bmatrix} \\[5pt] &=2(4−9)−5(−4+6)−1(−3+2) \\[5pt] &= 2(−5)−5(2)−1(−1)=−10−10+1 \\[5pt] &=−19 \end{align*}\]

Exercise \(\PageIndex{6}\)

Evaluate the determinant \(\begin{bmatrix}1&−2&−1\\3&2&−3\\1&5&4\end{bmatrix}\).

Hint

Expand along the first row. Don’t forget the second term is negative!

Answer

40

Rule: Cross Product Calculated by a Determinant

Let \(\vecs u=⟨u_1,u_2,u_3⟩\) and \(\vecs v=⟨v_1,v_2,v_3⟩\) be vectors. Then the cross product \(\vecs u×\vecs v\) is given by

\[\vecs u×\vecs v=\begin{bmatrix}\hat{\imath}&\hat{\jmath}&\hat{k}\\u_1&u_2&u_3\\v_1&v_2&v_3\end{bmatrix}=\begin{bmatrix}u_2&u_3\\v_2&v_3\end{bmatrix}\hat{\imath}−\begin{bmatrix}u_1&u_3\\v_1&v_3\end{bmatrix}\hat{\jmath}+\begin{bmatrix}u_1&u_2\\v_1&v_2\end{bmatrix}\hat{k}.\]

Example \(\PageIndex{7}\): Using Determinant Notation to find \(\vecs p×\vecs q\)

Let \(\vecs p=⟨−1,2,5⟩\) and \(\vecs q=⟨4,0,−3⟩\). Find \(\vecs p×\vecs q\).

Solution

We set up our determinant by putting the standard unit vectors across the first row, the components of \(\vecs u\) in the second row, and the components of \(\vecs v\) in the third row. Then, we have

\[\begin{align*} \vecs p×\vecs q&=\begin{bmatrix}\hat{\imath}&\hat{\jmath}&\hat{k}\\−1&2&5\\4&0&−3\end{bmatrix}=\begin{bmatrix}2&0\\5&−3\end{bmatrix}\hat{\imath}−\begin{bmatrix}−1&5\\4&−3\end{bmatrix}\hat{\jmath}+\begin{bmatrix}−1&2\\4&0\end{bmatrix}\hat{k} \\[5pt] &= (−6−0)\hat{\imath}−(3−20)\hat{\jmath}+(0−8)\hat{k} \\[5pt] &=−6\hat{\imath}+17\hat{\jmath}−8\hat{k}.\end{align*}\]

Notice that this answer confirms the calculation of the cross product in Example \(\PageIndex{1}\).

Exercise \(\PageIndex{7}\)

Use determinant notation to find \(\vecs a×\vecs b,\) where \(\vecs a=⟨8,2,3⟩\) and \(\vecs b=⟨−1,0,4⟩.\)

Hint

Calculate the determinant \(\begin{bmatrix}\hat{\imath}&\hat{\jmath}&\hat{k}\\8&2&3\\−1&0&4\end{bmatrix}\).

See Also

Cross Product (vector Product) - Definition, Formula and Properties

Answer

\(8\hat{\imath}−35\hat{\jmath}+2\hat{k}\)

Example \(\PageIndex{8}\): Finding a Unit Vector Orthogonal to Two Given Vectors

Let \(\vecs a=⟨5,2,−1⟩\) and \(\vecs b=⟨0,−1,4⟩\). Find a unit vector orthogonal to both \(\vecs a\) and \(\vecs b\).

Solution

The cross product \(\vecs a×\vecs b\) is orthogonal to both vectors \(\vecs a\) and \(\vecs b\). We can calculate it with a determinant:

\[\begin{align*} \vecs a×\vecs b &=\begin{bmatrix}\hat{\imath}&\hat{\jmath}&\hat{k}\\5&2&−1\\0&−1&4\end{bmatrix}=\begin{bmatrix}2&−1\\−14\end{bmatrix}\hat{\imath}−\begin{bmatrix}5&−1\\0&4\end{bmatrix}\hat{\jmath}+\begin{bmatrix}5&2\\0&−1\end{bmatrix}\hat{k} \\[5pt] &=(8−1)\hat{\imath}−(20−0)\hat{\jmath}+(−5−0)\hat{k} \\[5pt] &=7\hat{\imath}−20\hat{\jmath}−5\hat{k}.\end{align*} \]

Normalize this vector to find a unit vector in the same direction:

\(\|\vecs a×\vecs b\|=\sqrt{(7)^2+(−20)^2+(−5)^2}=\sqrt{474}\).

Thus, \(⟨\dfrac{7}{\sqrt{474}},\dfrac{−20}{\sqrt{474}},\dfrac{−5}{\sqrt{474}}⟩\) is a unit vector orthogonal to \(\vecs a\) and \(\vecs b\).

Simplified, this vector becomes \(⟨\dfrac{7\sqrt{474}}{474},\dfrac{−10\sqrt{474}}{237},\dfrac{−5\sqrt{474}}{474}⟩\).

Exercise \(\PageIndex{8}\)

Find a unit vector orthogonal to both \(\vecs a\) and \(\vecs b\), where \(\vecs a=⟨4,0,3⟩\) and \(\vecs b=⟨1,1,4⟩.\)

Hint

Normalize the cross product.

Answer

\(⟨\dfrac{−3}{\sqrt{194}},\dfrac{−13}{\sqrt{194}},\dfrac{4}{\sqrt{194}}⟩\) or, simplified as \(⟨\dfrac{−3\sqrt{194}}{194},\dfrac{−13\sqrt{194}}{194},\dfrac{2\sqrt{194}}{97}⟩\)

Area of a Parallelogram

If we locate vectors \(\vecs u\) and \(\vecs v\) such that they form adjacent sides of a parallelogram, then the area of the parallelogram is given by \(‖\vecs u×\vecs v‖\) (Figure \(\PageIndex{5}\)).

Proof

We show that the magnitude of the cross product is equal to the base times height of the parallelogram.

\[\begin{align*} \text{Area of a parallelogram} &= \text{base} × \text{height} \\[5pt] &=‖\vecs u‖(‖\vecs v‖\sin θ) \\[5pt] &=‖\vecs u×\vecs v‖ \end{align*}\]

□

Example \(\PageIndex{9}\): Finding the Area of a Triangle

Let \(P=(1,0,0),Q=(0,1,0),\) and \(R=(0,0,1)\) be the vertices of a triangle (Figure \(\PageIndex{6}\)). Find its area.

Solution

We have \(\vecd{PQ}=⟨0−1,1−0,0−0⟩=⟨−1,1,0⟩\) and \(\vecd{PR}=⟨0−1,0−0,1−0⟩=⟨−1,0,1⟩\). The area of the parallelogram with adjacent sides \(\vecd{PQ}\) and \(\vecd{PR}\) is given by \(∥\vecd{PQ}×\vecd{PR}∥\):

\[ \begin{align*} \vecd{PQ} \times \vecd{PR} &= \begin{bmatrix}\hat{\imath}&\hat{\jmath}&\hat{k}\\−1&1&0\\−1&0&1\end{bmatrix} \\[5pt] &=(1−0)\hat{\imath}−(−1−0)\hat{\jmath}+(0−(−1))\hat{k} \\[5pt] &=\hat{\imath}+\hat{\jmath}+\hat{k} \\[10pt] ∥\vecd{PQ}×\vecd{PR}∥ &=∥⟨1,1,1⟩∥ \\[5pt] &=\sqrt{1^2+1^2+1^2} \\[5pt] &=\sqrt{3}. \end{align*} \]

The area of \(ΔPQR\) is half the area of the parallelogram or \(\sqrt{3}/2 \, \text{units}^2\).

Exercise \(\PageIndex{9}\)

Find the area of the parallelogram \( PQRS\) with vertices \( P(1,1,0)\), \(Q(7,1,0)\), \(R(9,4,2)\), and \( S(3,4,2)\).

Hint

Sketch the parallelogram and identify two vectors that form adjacent sides of the parallelogram.

Answer

\(6\sqrt{13}\, \text{units}^2\)

The Triple Scalar Product

Because the cross product of two vectors is a vector, it is possible to combine the dot product and the cross product. The dot product of a vector with the cross product of two other vectors is called the triple scalar product because the result is a scalar.

When we create a matrix from three vectors, we must be careful about the order in which we list the vectors. If we list them in a matrix in one order and then rearrange the rows, the absolute value of the determinant remains unchanged. However, each time two rows switch places, the determinant changes sign:

\(\begin{bmatrix}a_1&a_2&a_3\\b_1&b_2&b_3\\c_1&c_2&c_3\end{bmatrix}=d \begin{bmatrix}a_1&a_2&a_3\\b_1&b_2&b_3\\c_1&c_2&c_3\end{bmatrix}=−d \begin{bmatrix}a_1&a_2&a_3\\b_1&b_2&b_3\\c_1&c_2&c_3\end{bmatrix}=d \begin{bmatrix}a_1&a_2&a_3\\b_1&b_2&b_3\\c_1&c_2&c_3\end{bmatrix}=−d\)

Verifying this fact is straightforward, but rather messy. Let’s take a look at this with an example:

\[ \begin{align} \begin{bmatrix}1&2&1\\−2&0&3\\4&1&−1\end{bmatrix} &=\begin{bmatrix}0&3\\1&−1\end{bmatrix}−2\begin{bmatrix}−2&3\\4&−1\end{bmatrix}+\begin{bmatrix}−2&0\\4&1\end{bmatrix} \\[5pt] &=(0−3)−2(2−12)+(−2−0) \\[5pt] &=−3+20−2=15. \end{align} \]

Switching the top two rows we have

\[ \begin{align} \begin{bmatrix}−2&0&3\\1&2&1\\4&1&−1\end{bmatrix} &=-2\begin{bmatrix}2&1\\1&−1\end{bmatrix}+3\begin{bmatrix}1&2\\4&1\end{bmatrix} \\[5pt] &=−2(−2−1)+3(1−8)\\[5pt] &=6−21=−15. \end{align} \]

Rearranging vectors in the triple products is equivalent to reordering the rows in the matrix of the determinant. Let \(\vecs u=u_1\hat{\imath}+u_2\hat{\jmath}+u_3\hat{k}, \vecs v=v_1\hat{\imath}+v_2\hat{\jmath}+v_3\hat{k},\) and \(\vecs w=w_1\hat{\imath}+w_2\hat{\jmath}+w_3\hat{k}.\) Applying Note, we have

\[\vecs u⋅(\vecs v×\vecs w)=\begin{bmatrix}u_1&u_2&u_3\\v_1&v_2&v_3\\w_1&w_2&w_3\end{bmatrix} \]

and

\[\vecs u⋅(\vecs w×\vecs v)=\begin{bmatrix}u_1&u_2&u_3\\w_1&w_2&w_3\\v_1&v_2&v_3\end{bmatrix}.\]

We can obtain the determinant for calculating \(\vecs u⋅(\vecs w×\vecs v)\) by switching the bottom two rows of \(\vecs u⋅(\vecs v×\vecs w).\) Therefore, \(\vecs u⋅(\vecs v×\vecs w)=−\vecs u⋅(\vecs w×\vecs v).\)

Following this reasoning and exploring the different ways we can interchange variables in the triple scalar product lead to the following identities:

\[\vecs u⋅(\vecs v×\vecs w)=−\vecs u⋅(\vecs w×\vecs v)\]

\[\vecs u⋅(\vecs v×\vecs w)=\vecs v⋅(\vecs w×\vecs u)=\vecs w⋅(\vecs u×\vecs v).\]

Let \(\vecs u\) and \(\vecs v\) be two vectors in standard position. If \(\vecs u\) and \(\vecs v\) are not scalar multiples of each other, then these vectors form adjacent sides of a parallelogram. We saw in Note that the area of this parallelogram is \(‖\vecs u×\vecs v‖\). Now suppose we add a third vector \(\vecs w\) that does not lie in the same plane as \(\vecs u\) and \(\vecs v\) but still shares the same initial point. Then these vectors form three edges of a parallelepiped, a three-dimensional prism with six faces that are each parallelograms, as shown in Figure. The volume of this prism is the product of the figure’s height and the area of its base. The triple scalar product of \(\vecs u,\vecs v,\) and \(\vecs w\) provides a simple method for calculating the volume of the parallelepiped defined by these vectors.

Volume of a Parallelepiped

The volume of a parallelepiped with adjacent edges given by the vectors \(\vecs u,\vecs v\), and \(\vecs w\) is the absolute value of the triple scalar product (Figure \(\PageIndex{7}\)):

\[V=||\vecs u⋅(\vecs v×\vecs w)||.\]

Note that, as the name indicates, the triple scalar product produces a scalar. The volume formula just presented uses the absolute value of a scalar quantity.

Example \(\PageIndex{11}\): Calculating the Volume of a Parallelepiped

Let \(\vecs u=⟨−1,−2,1⟩,\vecs v=⟨4,3,2⟩,\) and \(\vecs w=⟨0,−5,−2⟩\). Find the volume of the parallelepiped with adjacent edges \(\vecs u,\vecs v\), and \(\vecs w\) (Figure \(\PageIndex{8}\)).

Solution

We have

\[\begin{align*} \vecs u⋅(\vecs v×\vecs w)&=\begin{bmatrix}−1&−2&1\\4&3&2\\0&−5&−2\end{bmatrix} \\[5pt] &= (−1)\begin{bmatrix}3&2\\−5&−2\end{bmatrix}+2\begin{bmatrix}4&2\\0&−2\end{bmatrix}+\begin{bmatrix}4&3\\0&−5\end{bmatrix} \\[5pt] &=(−1)(−6+10)+2(−8−0)+(−20−0) \\[5pt] &=−4−16−20 \\[5pt] &=−40.\end{align*}\]

Thus, the volume of the parallelepiped is \(|−40|=40\) units³

Example \(\PageIndex{12}\): Using the Triple Scalar Product

Use the triple scalar product to show that vectors \(\vecs u=⟨2,0,5⟩,\vecs v=⟨2,2,4⟩\), and \(\vecs w=⟨1,−1,3⟩\) are coplanar—that is, show that these vectors lie in the same plane.

Solution

Start by calculating the triple scalar product to find the volume of the parallelepiped defined by \(\vecs u,\vecs v,\) and \(\vecs w\):

\(\vecs u⋅(\vecs v×\vecs w)=\begin{bmatrix}2&0&5\\2&2&4\\1&−1&3\end{bmatrix}\)

\(=[2(2)(3)+(0)(4)(1)+5(2)(−1)]−[5(2)(1)+(2)(4)(−1)+(0)(2)(3)]\)

\(=2−2\)

\(=0\).

The volume of the parallelepiped is \(0\) units³, so one of the dimensions must be zero. Therefore, the three vectors all lie in the same plane.

Exercise \(\PageIndex{12}\)

Are the vectors \(\vecs a=\hat{\imath}+\hat{\jmath}−\hat{k}, \vecs b=\hat{\imath}−\hat{\jmath}+\hat{k},\) and \(\vecs c=\hat{\imath}+\hat{\jmath}+\hat{k}\) coplanar?

Hint

Calculate the triple scalar product.

Answer

No, the triple scalar product is \(−4≠0,\) so the three vectors form the adjacent edges of a parallelepiped. They are not coplanar.

Example \(\PageIndex{13}\): Finding an Orthogonal Vector

Only a single plane can pass through any set of three noncolinear points. Find a vector orthogonal to the plane containing points \(P=(9,−3,−2),Q=(1,3,0),\) and \(R=(−2,5,0).\)

Solution

The plane must contain vectors \(\vecd{PQ}\) and \(\vecd{QR}\):

\(\vecd{PQ}=⟨1−9,3−(−3),0−(−2)⟩=⟨−8,6,2⟩\)

\(\vecd{QR}=⟨−2−1,5−3,0−0⟩=⟨−3,2,0⟩.\)

The cross product \(\vecd{PQ}×\vecd{QR}\) produces a vector orthogonal to both \(\vecd{PQ}\) and \(\vecd{QR}\). Therefore, the cross product is orthogonal to the plane that contains these two vectors:

\(\vecd{PQ}×\vecd{QR}=\begin{bmatrix}\hat{\imath}&\hat{\jmath}&\hat{k}\\−8&6&2\\−3&2&0\end{bmatrix}\)

\(=0\hat{\imath}−6\hat{\jmath}−16\hat{k}−(−18\hat{k}+4\hat{\imath}+0\hat{\jmath})\)

\(=−4\hat{\imath}−6\hat{\jmath}+2\hat{k}.\)

Definition: Torque

Torque, \(\vecs \tau\) (the Greek letter tau), measures the tendency of a force to produce rotation about an axis of rotation. Let \(\vecs r\) be a vector with an initial point located on the axis of rotation and with a terminal point located at the point where the force is applied, and let vector \(\vecs F\) represent the force. Then torque is equal to the cross product of \(r\) and \(F\):

\[\vecs \tau=\vecs r×\vecs F.\]

See Figure \(\PageIndex{9}\).

Example \(\PageIndex{14}\): Evaluating Torque

A bolt is tightened by applying a force of \(6\) N to a 0.15-m wrench (Figure \(\PageIndex{10}\)). The angle between the wrench and the force vector is \(40°\). Find the magnitude of the torque about the center of the bolt. Round the answer to two decimal places.

Solution:

Substitute the given information into the equation defining torque:

\[ \begin{align*} ‖\vecs τ‖&=\|\vecs r×\vecs F\| \\[5pt] &=‖\vecs r‖∥\vecs F∥\sinθ \\[5pt] &=(0.15\,\text{m})(6\,\text{N})\sin 40° \\[5pt] &≈0.58\,\text{N⋅m.} \end{align*}\]

Exercise \(\PageIndex{14}\)

Calculate the force required to produce \(15\) N⋅m torque at an angle of \(30º\) from a \(150\)-cm rod.

Hint

\(‖\vecs τ‖=15\) N⋅m and \(‖\vecs r‖=1.5\) m

Answer

\(20\) N