2.4 The cross product (Page 4/16)

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Using expansion along the first row to compute a $3 \times 3$ Determinant

Evaluate the determinant $| \begin{matrix} 2 & 5 & −1 \\ −1 & 1 & 3 \\ −2 & 3 & 4 \end{matrix} | .$

We have

\begin{matrix} | \begin{matrix} 2 & 5 & −1 \\ −1 & 1 & 3 \\ −2 & 3 & 4 \end{matrix} | & = 2 | \begin{matrix} 1 & 3 \\ 3 & 4 \end{matrix} | - 5 | \begin{matrix} −1 & 3 \\ −2 & 4 \end{matrix} | - 1 | \begin{matrix} −1 & 1 \\ −2 & 3 \end{matrix} | \\ = 2 (4 - 9) - 5 (−4 + 6) - 1 (−3 + 2) \\ = 2 (−5) - 5 (2) - 1 (−1) = −10 - 10 + 1 \\ = −19. \end{matrix}

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Evaluate the determinant $| \begin{matrix} 1 & −2 & −1 \\ 3 & 2 & −3 \\ 1 & 5 & 4 \end{matrix} | .$

$40$

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Technically, determinants are defined only in terms of arrays of real numbers. However, the determinant notation provides a useful mnemonic device for the cross product formula.

Rule: cross product calculated by a determinant

Let $u = ⟨ u_{1}, u_{2}, u_{3} ⟩$ and $v = ⟨ v_{1}, v_{2}, v_{3} ⟩$ be vectors. Then the cross product $u \times v$ is given by

u \times v = | \begin{matrix} i & j & k \\ u_{1} & u_{2} & u_{3} \\ v_{1} & v_{2} & v_{3} \end{matrix} | = | \begin{matrix} u_{2} & u_{3} \\ v_{2} & v_{3} \end{matrix} | i - | \begin{matrix} u_{1} & u_{3} \\ v_{1} & v_{3} \end{matrix} | j + | \begin{matrix} u_{1} & u_{2} \\ v_{1} & v_{2} \end{matrix} | k .

Using determinant notation to find $p \times q$

Let $p = ⟨ −1, 2, 5 ⟩$ and $q = ⟨ 4, 0, −3 ⟩ .$ Find $p \times q .$

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

\begin{matrix} p \times q & = | \begin{matrix} i & j & k \\ −1 & 2 & 5 \\ 4 & 0 & −3 \end{matrix} | = | \begin{matrix} 2 & 5 \\ 0 & −3 \end{matrix} | i - | \begin{matrix} −1 & 5 \\ 4 & −3 \end{matrix} | j + | \begin{matrix} −1 & 2 \\ 4 & 0 \end{matrix} | k \\ = (−6 - 0) i - (3 - 20) j + (0 - 8) k \\ = −6 i + 17 j - 8 k . \end{matrix}

Notice that this answer confirms the calculation of the cross product in [link] .

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Use determinant notation to find $a \times b,$ where $a = ⟨ 8, 2, 3 ⟩$ and $b = ⟨ −1, 0, 4 ⟩ .$

$8 i - 35 j + 2 k$

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Using the cross product

The cross product is very useful for several types of calculations, including finding a vector orthogonal to two given vectors, computing areas of triangles and parallelograms, and even determining the volume of the three-dimensional geometric shape made of parallelograms known as a parallelepiped . The following examples illustrate these calculations.

Finding a unit vector orthogonal to two given vectors

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

The cross product $a \times b$ is orthogonal to both vectors $a$ and $b .$ We can calculate it with a determinant:

\begin{matrix} a \times b & = | \begin{matrix} i & j & k \\ 5 & 2 & −1 \\ 0 & −1 & 4 \end{matrix} | = | \begin{matrix} 2 & −1 \\ −1 & 4 \end{matrix} | i - | \begin{matrix} 5 & −1 \\ 0 & 4 \end{matrix} | j + | \begin{matrix} 5 & 2 \\ 0 & −1 \end{matrix} | k \\ = (8 - 1) i - (20 - 0) j + (−5 - 0) k \\ = 7 i - 20 j - 5 k . \end{matrix}

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

‖ a \times b ‖ = \sqrt{{(7)}^{2} + {(−20)}^{2} + {(−5)}^{2}} = \sqrt{474} .

Thus, $⟨ \frac{7}{\sqrt{474}}, \frac{−20}{\sqrt{474}}, \frac{−5}{\sqrt{474}} ⟩$ is a unit vector orthogonal to $a$ and $b .$

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Find a unit vector orthogonal to both $a$ and $b,$ where $a = ⟨ 4, 0, 3 ⟩$ and $b = ⟨ 1, 1, 4 ⟩ .$

$⟨ \frac{−3}{\sqrt{194}}, \frac{−13}{\sqrt{194}}, \frac{4}{\sqrt{194}} ⟩$

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To use the cross product for calculating areas, we state and prove the following theorem.

Area of a parallelogram

If we locate vectors $u$ and $v$ such that they form adjacent sides of a parallelogram, then the area of the parallelogram is given by $‖ u \times v ‖$ ( [link] ).

This figure is a parallelogram. One side is represented with a vector labeled “v.” The second side, the base, has the same initial point as vector v and is labeled “u.” The angle between u and v is theta. Also, a perpendicular line segment is drawn from the terminal point of v to vector u. It is labeled “|v|sin(theta).” — The parallelogram with adjacent sides $u$ and $v$ has base $‖ u ‖$ and height $‖ v ‖ sin θ .$

Proof

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

\begin{matrix} Area of a parallelogram & = base \times height \\ = ‖ u ‖ (‖ v ‖ sin θ) \\ = ‖ u \times v ‖ \end{matrix}

□

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 ( [link] ). Find its area.

This figure is the 3-dimensional coordinate system. It has a triangle drawn in the first octant. The vertices of the triangle are points P(1, 0, 0); Q(0, 1, 0); and R(0, 0, 1). — Finding the area of a triangle by using the cross product.

We have $\vec{P Q} = ⟨ 0 - 1, 1 - 0, 0 - 0 ⟩ = ⟨ −1, 1, 0 ⟩$ and $\vec{P R} = ⟨ 0 - 1, 0 - 0, 1 - 0 ⟩ = ⟨ −1, 0, 1 ⟩ .$ The area of the parallelogram with adjacent sides $\vec{P Q}$ and $\vec{P R}$ is given by $‖ \vec{P Q} \times \vec{P R} ‖ :$

\begin{matrix} \vec{P Q} \times \vec{P R} & = & | \begin{matrix} i & j & k \\ −1 & 1 & 0 \\ −1 & 0 & 1 \end{matrix} | = (1 - 0) i - (−1 - 0) j + (0 - (−1)) k = i + j + k \\ ‖ \vec{P Q} \times \vec{P R} ‖ & = & ‖ ⟨ 1, 1, 1 ⟩ ‖ = \sqrt{1^{2} + 1^{2} + 1^{2}} = \sqrt{3} . \end{matrix}

The area of $Δ P Q R$ is half the area of the parallelogram, or $\sqrt{3} / 2 .$

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

$6 \sqrt{13}$

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Practice Key Terms 6

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Source: OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2

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2.4 The cross product (Page 4/16)

Using expansion along the first row to compute a 3 × 3 Determinant

Rule: cross product calculated by a determinant

Using determinant notation to find p × q

Using the cross product

Finding a unit vector orthogonal to two given vectors

Area of a parallelogram

Proof

Finding the area of a triangle

Read also:

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Using expansion along the first row to compute a $3 \times 3$ Determinant

Using determinant notation to find $p \times q$