The sine of the angle between □ □ and (applying the Pythagorean theorem in the right triangle □ □ □). Point □ is the middle of the rectangle’s diagonals, The magnitude of □ □ is simply the length 5 4 4 0 | | ⃑ □ ,Īs ⃑ □ is a unit vector perpendicular to the plane of the rectangle. The cross product of □ □and □ □ is From this, we find that □ □ = ( 2 2, 1 6. This second method implies that we have to find the magnitude of both vectors and s i n □. Or we apply □ □ × □ □ = ‖ ‖ □ □ ‖ ‖ ‖ ‖ □ □ ‖ ‖ □ ⃑ □ s i n,īetween □ □ and □ □ and ⃑ □ is a unit vector perpendicular to the plane of the rectangle. Either we find the components of both vectors, for instance in the coordinate To find □ □ × □ □, we have two possibilities. Let us now find the cross product of two vectors whose components are not explicitly given but that are defined with specific points in a rectangle. Substituting this expression for □ into equation (2) gives We have now two linear equations with two unknownsįrom equation (1), we find that □ = 2 □ + 3 . We can now write their cross products with ⃑ □ as Let us extract the components of ⃑ □ and ⃑ □ from their expressions in terms of It follows that ⃑ □ + ⃑ □ × ⃑ □ = ⃑ □ × ⃑ □ + ⃑ □ × ⃑ □ the cross product is therefore distributive.Įxample 3: Finding a Vector Given its Cross Product with Two Known Vectors It can be found out easily when looking more closely at the determinant we used to With the previous example, we can wonder whether the cross product is distributive, We can now calculate ⃑ □ + ⃑ □ × ⃑ □ as We can write the following definition of the cross product of two vectors in theĮxample 2: Calculating the Cross Product of Two 2D Vectors If we combine this with the fact that ⃑ □ × ⃑ □ is a vector parallel to ⃑ □, Since s i n c o s c o s s i n s i n □ □ − □ □ = □ . Let us make a matrix with the components in terms of S i n s i n c o s c o s s i n □ = □ □ − □ □. Using the subtraction trigonometric identity s i n s i n c o s c o s s i n ( □ − □ ) = □ □ − □ □ , we find that
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