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

Evaluate the determinant | 2 5 −1 −1 1 3 −2 3 4 | .

We have

| 2 5 −1 −1 1 3 −2 3 4 | = 2 | 1 3 3 4 | 5 | −1 3 −2 4 | 1 | −1 1 −2 3 | = 2 ( 4 9 ) 5 ( −4 + 6 ) 1 ( −3 + 2 ) = 2 ( −5 ) 5 ( 2 ) 1 ( −1 ) = −10 10 + 1 = −19.
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Evaluate the determinant | 1 −2 −1 3 2 −3 1 5 4 | .

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 × v is given by

u × v = | i j k u 1 u 2 u 3 v 1 v 2 v 3 | = | u 2 u 3 v 2 v 3 | i | u 1 u 3 v 1 v 3 | j + | u 1 u 2 v 1 v 2 | k .

Using determinant notation to find p × q

Let p = −1 , 2 , 5 and q = 4 , 0 , −3 . Find p × 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

p × q = | i j k −1 2 5 4 0 −3 | = | 2 5 0 −3 | i | −1 5 4 −3 | j + | −1 2 4 0 | k = ( −6 0 ) i ( 3 20 ) j + ( 0 8 ) k = −6 i + 17 j 8 k .

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

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Use determinant notation to find a × 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 × b is orthogonal to both vectors a and b . We can calculate it with a determinant:

a × b = | i j k 5 2 −1 0 −1 4 | = | 2 −1 −1 4 | i | 5 −1 0 4 | j + | 5 2 0 −1 | k = ( 8 1 ) i ( 20 0 ) j + ( −5 0 ) k = 7 i 20 j 5 k .

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

a × b = ( 7 ) 2 + ( −20 ) 2 + ( −5 ) 2 = 474 .

Thus, 7 474 , −20 474 , −5 474 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 .

−3 194 , −13 194 , 4 194

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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 × 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.

Area of a parallelogram = base × height = u ( v sin θ ) = u × v

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 P Q = 0 1 , 1 0 , 0 0 = −1 , 1 , 0 and P R = 0 1 , 0 0 , 1 0 = −1 , 0 , 1 . The area of the parallelogram with adjacent sides P Q and P R is given by P Q × P R :

P Q × P R = | i j k −1 1 0 −1 0 1 | = ( 1 0 ) i ( −1 0 ) j + ( 0 ( −1 ) ) k = i + j + k P Q × P R = 1 , 1 , 1 = 1 2 + 1 2 + 1 2 = 3 .

The area of Δ P Q R is half the area of the parallelogram, or 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 13

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Questions & Answers

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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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