Math 221: LINEAR ALGEBRA 4-3. More on the Cross Product Le Chen 1 - PowerPoint PPT Presentation

Math 221: LINEAR ALGEBRA §4-3. More on the Cross Product Le Chen 1 Emory University, 2020 Fall (last updated on 08/27/2020) Creative Commons License (CC BY-NC-SA) 1 Slides are adapted from those by Karen Seyffarth from University of Calgary.

. det Theorem       x 0 x 1 x 2 If � u = y 0  , � v = y 1  , and � w = y 2  . Then    z 0 z 1 z 2   x 0 x 1 x 2  . u · ( � v × � w ) = det y 0 y 1 y 2 �  z 0 z 1 z 2

. Theorem       x 0 x 1 x 2 If � u = y 0  , � v = y 1  , and � w = y 2  . Then    z 0 z 1 z 2   x 0 x 1 x 2  . u · ( � v × � w ) = det y 0 y 1 y 2 �  z 0 z 1 z 2 � � Shorthand: � u · ( � v × � w ) = det u � � v w �

Then Proof.       x 0 x 1 x 2 u = v = w = Let � y 0  , � y 1  , and � y 2     . z 0 z 1 z 2 �

Proof.       x 0 x 1 x 2 u = v = w = Let � y 0  , � y 1  , and � y 2     . Then z 0 z 1 z 2     x 0 y 1 z 2 − z 1 y 2  · � u · ( � v × � w ) = y 0 − ( x 1 z 2 − z 1 x 2 )    z 0 x 1 y 2 − y 1 x 2 = x 0 ( y 1 z 2 − z 1 y 2 ) − y 0 ( x 1 z 2 − z 1 x 2 ) + z 0 ( x 1 y 2 − y 1 x 2 ) � � � � � � y 1 y 2 x 1 x 2 y 1 y 2 � � � � � � = x 0 � − y 0 � + z 0 � � � � � � z 1 z 2 z 1 z 2 z 1 z 2 � � � � � � x 0 x 1 x 2 � � � � = y 0 y 1 y 2 . � � � � z 0 z 1 z 2 � � �

Properties of the Cross Product Theorem w be in R 3 . Let � u ,� v and �

Properties of the Cross Product Theorem w be in R 3 . Let � u ,� v and � 1. � u × � v is a vector.

Properties of the Cross Product Theorem w be in R 3 . Let � u ,� v and � 1. � u × � v is a vector. 2. � u × � v is orthogonal to both � u and � v.

Properties of the Cross Product Theorem w be in R 3 . Let � u ,� v and � 1. � u × � v is a vector. 2. � u × � v is orthogonal to both � u and � v. u × � 0 = � 0 and � u = � 3. � 0 × � 0 .

Properties of the Cross Product Theorem w be in R 3 . Let � u ,� v and � 1. � u × � v is a vector. 2. � u × � v is orthogonal to both � u and � v. u × � 0 = � 0 and � u = � 3. � 0 × � 0 . u = � 4. � u × � 0 .

Properties of the Cross Product Theorem w be in R 3 . Let � u ,� v and � 1. � u × � v is a vector. 2. � u × � v is orthogonal to both � u and � v. u × � 0 = � 0 and � u = � 3. � 0 × � 0 . u = � 4. � u × � 0 . 5. � u × � v = − ( � v × � u ) .

Properties of the Cross Product Theorem w be in R 3 . Let � u ,� v and � 1. � u × � v is a vector. 2. � u × � v is orthogonal to both � u and � v. u × � 0 = � 0 and � u = � 3. � 0 × � 0 . u = � 4. � u × � 0 . 5. � u × � v = − ( � v × � u ) . 6. ( k � u ) × � v = k ( � u × � v ) = � u × ( k � v ) for any scalar k.

Properties of the Cross Product Theorem w be in R 3 . Let � u ,� v and � 1. � u × � v is a vector. 2. � u × � v is orthogonal to both � u and � v. u × � 0 = � 0 and � u = � 3. � 0 × � 0 . u = � 4. � u × � 0 . 5. � u × � v = − ( � v × � u ) . 6. ( k � u ) × � v = k ( � u × � v ) = � u × ( k � v ) for any scalar k. u × ( � v + � w ) = � u × � v + � u × � 7. � w.

Properties of the Cross Product Theorem w be in R 3 . Let � u ,� v and � 1. � u × � v is a vector. 2. � u × � v is orthogonal to both � u and � v. u × � 0 = � 0 and � u = � 3. � 0 × � 0 . u = � 4. � u × � 0 . 5. � u × � v = − ( � v × � u ) . 6. ( k � u ) × � v = k ( � u × � v ) = � u × ( k � v ) for any scalar k. u × ( � v + � w ) = � u × � v + � u × � 7. � w. 8. ( � v + � w ) × � u = � v × � u + � w × � u.

Write and , and work out all the terms. The Lagrange Identity Theorem v ∈ R 3 , then If � u ,� v || 2 = || � v || 2 − ( � u || 2 || � v ) 2 . || � u × � u · �

The Lagrange Identity Theorem v ∈ R 3 , then If � u ,� v || 2 = || � v || 2 − ( � u || 2 || � v ) 2 . || � u × � u · � Proof.     x 1 x 2 Write � u = y 1  and � v = y 2 �    , and work out all the terms. z 1 z 2

As a consequence of the Lagrange Identity and the fact that Taking square roots on both sides yields, are parallel, then and observation that if . This is consistent with our earlier , and , then sin or If . , sin Note that since sin . we have � u · � v = || � u || || � v || cos θ, v || 2 − ( � v || 2 u || 2 || � v ) 2 || � u × � = || � u · � v || 2 − || � v || 2 cos 2 θ u || 2 || � u || 2 || � = || � v || 2 (1 − cos 2 θ ) u || 2 || � = || � v || 2 sin 2 θ. u || 2 || � = || �

As a consequence of the Lagrange Identity and the fact that Taking square roots on both sides yields, are parallel, then and observation that if . This is consistent with our earlier , and , then sin or If . , sin Note that since . we have u · � � v = || � u || || � v || cos θ, v || 2 − ( � v || 2 u || 2 || � v ) 2 || � u × � = || � u · � v || 2 − || � v || 2 cos 2 θ u || 2 || � u || 2 || � = || � v || 2 (1 − cos 2 θ ) u || 2 || � = || � v || 2 sin 2 θ. u || 2 || � = || � || � u × � v || = || � u || || � v || sin θ.

As a consequence of the Lagrange Identity and the fact that Taking square roots on both sides yields, are parallel, then and observation that if . This is consistent with our earlier , and , then sin or If . we have � u · � v = || � u || || � v || cos θ, v || 2 − ( � v || 2 u || 2 || � v ) 2 || � u × � = || � u · � v || 2 − || � v || 2 cos 2 θ u || 2 || � u || 2 || � = || � v || 2 (1 − cos 2 θ ) u || 2 || � = || � v || 2 sin 2 θ. u || 2 || � = || � || � u × � v || = || � u || || � v || sin θ. Note that since 0 ≤ θ ≤ π , sin θ ≥ 0 .

As a consequence of the Lagrange Identity and the fact that Taking square roots on both sides yields, we have u · � � v = || � u || || � v || cos θ, v || 2 − ( � v || 2 u || 2 || � v ) 2 || � u × � = || � u · � v || 2 − || � v || 2 cos 2 θ u || 2 || � u || 2 || � = || � v || 2 (1 − cos 2 θ ) u || 2 || � = || � v || 2 sin 2 θ. u || 2 || � = || � || � u × � v || = || � u || || � v || sin θ. Note that since 0 ≤ θ ≤ π , sin θ ≥ 0 . If θ = 0 or θ = π , then sin θ = 0 , and || � u × � v || = 0 . This is consistent with our earlier v = � observation that if � u and � v are parallel, then � u × � 0 .

sin The area of the parallelogram defjned by sin . Therefore, the area is , we see that Since sin parallelogram. is the height of the , where is and Theorem v be nonzero vectors in R 3 , and let θ denote the angle between � Let � u and � u and � v. 1. || � u × � v || = || � u || || � v || sin θ , and is the area of the parallelogram defined by � u and � v. v = � u × � 0 . 2. � u and � v are parallel if and only if �

sin parallelogram. sin . Therefore, the area is , we see that Since sin Theorem v be nonzero vectors in R 3 , and let θ denote the angle between � Let � u and � u and � v. 1. || � u × � v || = || � u || || � v || sin θ , and is the area of the parallelogram defined by � u and � v. v = � u × � 0 . 2. � u and � v are parallel if and only if � Proof of area of parallelogram. The area of the parallelogram defjned by � u and � v is || � u || h , where h is the height of the � v h θ � u

sin parallelogram. Therefore, the area is Theorem v be nonzero vectors in R 3 , and let θ denote the angle between � Let � u and � u and � v. 1. || � u × � v || = || � u || || � v || sin θ , and is the area of the parallelogram defined by � u and � v. v = � u × � 0 . 2. � u and � v are parallel if and only if � Proof of area of parallelogram. The area of the parallelogram defjned by � u and � v is || � u || h , where h is the height of the � v h θ � u h Since sin θ = v || , we see that h = || � v || sin θ . || �

parallelogram. Theorem v be nonzero vectors in R 3 , and let θ denote the angle between � Let � u and � u and � v. 1. || � u × � v || = || � u || || � v || sin θ , and is the area of the parallelogram defined by � u and � v. v = � u × � 0 . 2. � u and � v are parallel if and only if � Proof of area of parallelogram. The area of the parallelogram defjned by � u and � v is || � u || h , where h is the height of the v � h θ u � h Since sin θ = v || , we see that h = || � v || sin θ . Therefore, the area is || � || � u || || � v || sin θ. �

Theorem The volume of the parallelepiped determined by the three vectors � u, � v, and w in R 3 is � | � w · ( � u × � v ) | .

Problem Find the area of the triangle having vertices A (3 , − 1 , 2) , B (1 , 1 , 0) and C (1 , 2 , − 1) .