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Vector Refresher Part 3 • Vector Dot Product Definitions • Some Properties • The Angle Between 2 Vectors • Scalar Projections • Vector Projections Dot Product • One form of vector multiplication • Yields a SCALAR quantity • Can be used to find the angle between 2 vectors • Can also be used to find the projection of a vector in a given direction Symbolism • The dot product is symbolized with a dot between 2 vectors Symbolism • The dot product is symbolized with a dot between 2 vectors • The following means “Vector A dotted with vector B” A· B One Definition The dot product is defined as the sum of the product of similar components of a vector One Definition The dot product is defined as the sum of the product of similar components of a vector If we have the following 2 vectors: U = aiˆ + bjˆ + ckˆ V = diˆ + ejˆ + fkˆ One Definition The dot product is defined as the sum of the product of similar components of a vector If we have the following 2 vectors: U = aiˆ + bjˆ + ckˆ V = diˆ + ejˆ + fkˆ U ·V = (ad)+ (be)+ (cf ) One Definition The dot product is defined as the sum of the product of similar components of a vector If we have the following 2 vectors: U = aiˆ + bjˆ + ckˆ V = diˆ + ejˆ + fkˆ U ·V = (ad)+ (be)+ (cf ) NOTE: This is a SCALAR term whose units are the product of the units of the 2 vectors Another Definition The dot product is also related to the angle produced by arranging 2 vectors tail to tail. Another Definition The dot product is also related to the angle produced by arranging 2 vectors tail to tail. If we have the following 2 vectors: U = aiˆ + bjˆ + ckˆ V = diˆ + ejˆ + fkˆ U ·V = U V cos(q ) U θ V Properties of the Dot Product Commutative: U ·V = V ·U Properties of the Dot Product Commutative: U ·V = V ·U Associative: k(U ·V ) = (kU)·V = (kV)·U Properties of the Dot Product Commutative: U ·V = V ·U Associative: k(U ·V ) = (kU)·V = (kV)·U Distributive: (U +V)·W = (U ·W )+ (V ·W ) The Angle Between 2 Vectors The dot product is a useful tool in determining the angle between 2 vectors U θ V The Angle Between 2 Vectors The dot product is a useful tool in determining the angle between 2 vectors U ·V = U V cos(q ) U θ V The Angle Between 2 Vectors The dot product is a useful tool in determining the angle between 2 vectors U ·V = U V cos(q ) U ·V = cos(q ) U V U θ V The Angle Between 2 Vectors The dot product is a useful tool in determining the angle between 2 vectors U ·V = U V cos(q ) U ·V = cos(q ) U V æ U ·V ö ÷ =q cos-1 çç ÷ U V è ø U θ V The Angle Between 2 Vectors The dot product is a useful tool in determining the angle between 2 vectors U ·V = U V cos(q ) U ·V = cos(q ) U V æ U ·V ö ÷ =q cos-1 çç ÷ U V è ø U θ V If 2 vectors are orthogonal, their dot product is 0 Scalar Projection The dot product is also used to determine how much of a vector is acting in a particular direction. Scalar Projection The dot product is also used to determine how much of a vector is acting in a particular direction. U θ V Scalar Projection The dot product is also used to determine how much of a vector is acting in a particular direction. If we want to find how much of U acts in the direction of V , (length of the green line) we can use the dot product U θ V Scalar Projection The dot product is also used to determine how much of a vector is acting in a particular direction. If we want to find how much of U acts in the direction of V , (length of the green line) we can use the dot product U θ V V U · uˆV = U · V Scalar Projection The dot product is also used to determine how much of a vector is acting in a particular direction. If we want to find how much of U acts in the direction of V , (length of the green line) we can use the dot product U V θ U // V U // = U · uˆV = U · V Note that this result is a SCALAR quantity, meaning that it has no direction associated. Scalar Projection The dot product is also used to determine how much of a vector is acting in a particular direction. If we want to find how much of U acts in the direction of V , (length of the green line) we can use the dot product U V θ U // V U // = U · uˆV = U · V Note that this result is a SCALAR quantity, meaning that it has no direction associated. Thus, this calculation is the scalar projection Vector Projection The scalar projection can be used to determine a vector projection We can transform the scalar projection, in this case U · uˆ , into a V vector by multiplying the scalar projection and the unit vector that described the direction of interest, in this case uˆV U V θ U // æ V ç U // = (U · uˆV )uˆV = çU · V è öV ÷ ÷V ø This is a VECTOR quantity that describes the vector shown by the green arrow Applications of the Vector Projection We can use the vector projection to determine the vector parallel and perpendicular to a given direction U V θ U // Applications of the Vector Projection We can use the vector projection to determine the vector parallel and perpendicular to a given direction A vector can be described as its vector component parallel to a direction plus its component perpendicular to a direction U =U// +U^ U V θ U // Applications of the Vector Projection We can use the vector projection to determine the vector parallel and perpendicular to a given direction A vector can be described as its vector component parallel to a direction plus its component perpendicular to a direction U^ =U -U// U V θ U // U =U// +U^ Example Problem If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2 vectors, the projection of U onto V , and the component of U that is perpendicular to V . Example Problem If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2 vectors, the projection of U onto V , and the component of U that is perpendicular to V . Looking at this formula, we need to U ·V = cos(q ) U V determine the magnitude of each vector and evaluate the dot product Example Problem If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2 vectors, the projection of U onto V , and the component of U that is perpendicular to V . We can start by finding the magnitude of U ·V = cos(q ) U V vector U U = (2)2 + (2)2 + (-1)2 Example Problem If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2 vectors, the projection of U onto V , and the component of U that is perpendicular to V . We can start by finding the magnitude of U ·V = cos(q ) U V vector U U = (2)2 + (2)2 + (-1)2 U = 4 + 4 +1 = 9 Example Problem If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2 vectors, the projection of U onto V , and the component of U that is perpendicular to V . We can start by finding the magnitude of U ·V = cos(q ) U V vector U U = (2)2 + (2)2 + (-1)2 U = 4 + 4 +1 = 9 U =3 Example Problem If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2 vectors, the projection of U onto V , and the component of U that is perpendicular to V . Now, we can do the same for vector V U ·V = cos(q ) (3) V V = (-2)2 + (3)2 + (6)2 Example Problem If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2 vectors, the projection of U onto V , and the component of U that is perpendicular to V . Now, we can do the same for vector V U ·V = cos(q ) (3) V V = (-2)2 + (3)2 + (6)2 V = 4 + 9 + 36 = 49 Example Problem If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2 vectors, the projection of U onto V , and the component of U that is perpendicular to V . Now, we can do the same for vector V U ·V = cos(q ) (3)(7) V = (-2)2 + (3)2 + (6)2 V = 4 + 9 + 36 = 49 V =7 Example Problem If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2 vectors, the projection of U onto V , and the component of U that is perpendicular to V . Next, we’ll take the dot product to -4 = cos(q ) (3)(7) complete the formula. U ·V = (2)(-2)+ (2)(3)+ (-1)(6) U ·V = -4 + 6 - 6 = -4 Example Problem If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2 vectors, the projection of U onto V , and the component of U that is perpendicular to V . Now, we can use the inverse cosine function to find the angle -4 = cos(q ) (3)(7) Example Problem If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2 vectors, the projection of U onto V , and the component of U that is perpendicular to V . Now, we can use the inverse cosine function to find the angle -4 = cos(q ) (3)(7) -1 æ -4 ö cos ç ÷ = q è 21 ø 101° = q Example Problem If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2 vectors, the projection of U onto V , and the component of U that is perpendicular to V . To find the projection of U onto V, we 101° = q U// = U · uˆV need to use the formula to the left, which means we need the unit vector that describes the direction of V Example Problem If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2 vectors, the projection of U onto V , and the component of U that is perpendicular to V . We already calculated the magnitude of 101° = q U// = U · uˆV V. We’ll use that to find the unit vector V uˆV = V V =7 Example Problem If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2 vectors, the projection of U onto V , and the component of U that is perpendicular to V . We already calculated the magnitude of 101° = q U// = U · uˆV V. We’ll use that to find the unit vector V uˆV = V V =7 2ˆ 3 ˆ 6 ˆ uˆV = - i + j + k 7 7 7 Example Problem If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2 vectors, the projection of U onto V , and the component of U that is perpendicular to V . Now, we can take the dot product to find 101° = q U// = U · uˆV the scalar projection. V uˆV = V V =7 2ˆ 3 ˆ 6 ˆ uˆV = - i + j + k 7 7 7 æ -2 ö æ 3ö æ6ö ˆ U · uV = (2)ç ÷ + (2)ç ÷ + (-1)ç ÷ è 7ø è 7ø è 7ø Example Problem If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2 vectors, the projection of U onto V , and the component of U that is perpendicular to V . Now, we can take the dot product to find 101° = q U// = U · uˆV the scalar projection. V uˆV = V V =7 2ˆ 3 ˆ 6 ˆ uˆV = - i + j + k 7 7 7 æ -2 ö æ 3ö æ6ö ˆ U · uV = (2)ç ÷ + (2)ç ÷ + (-1)ç ÷ è 7ø è 7ø è 7ø U · uˆV = -4 7 Example Problem If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2 vectors, the projection of U onto V , and the component of U that is perpendicular to V . To find the vector projection, we’ll apply 101° = q -4 U // = 7 the scalar projection to the unit vector that describes the direction of V. U// = U// uˆV Example Problem If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2 vectors, the projection of U onto V , and the component of U that is perpendicular to V . To find the vector projection, we’ll apply 101° = q -4 U // = 7 the scalar projection to the unit vector that describes the direction of V. U// = U// uˆV æ -4 öæ 2 ˆ 3 ˆ 6 ˆ ö U// = ç ÷ç - i + j + k ÷ è 7 øè 7 7 7 ø Example Problem If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2 vectors, the projection of U onto V , and the component of U that is perpendicular to V . To find the vector projection, we’ll apply 101° = q -4 U // = 7 the scalar projection to the unit vector that describes the direction of V. U// = U// uˆV æ -4 öæ 2 ˆ 3 ˆ 6 ˆ ö U// = ç ÷ç - i + j + k ÷ è 7 øè 7 7 7 ø 8 ˆ 12 ˆ 24 ˆ U// = i j- k 49 49 49 Example Problem If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2 vectors, the projection of U onto V , and the component of U that is perpendicular to V . Finally, we can subtract the component 101° = q -4 U // = 7 8 ˆ 12 ˆ 24 ˆ U// = i j- k 49 49 49 of U parallel to V from U to get the part of U that is perpendicular to V. U^ =U -U// Example Problem If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2 vectors, the projection of U onto V , and the component of U that is perpendicular to V . Finally, we can subtract the component 101° = q -4 U // = 7 of U parallel to V from U to get the part of U that is perpendicular to V. U^ =U -U// æ æ -24 öö ˆ 8 ö ˆ æ æ -12 öö ˆ æ U^ = ç 2 - ÷ i + ç 2 - ç ÷÷ j + ç -1- ç ÷÷ k è 49 ø è è 49 øø è è 49 øø 90 ˆ 110 ˆ 25 ˆ U^ = i + j- k 49 49 49 8 ˆ 12 ˆ 24 ˆ U// = i j- k 49 49 49 Example Problem If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2 vectors, the projection of U onto V , and the component of U that is perpendicular to V . We can check our work with the 101° = q U // = -4 7 90 110 ˆ 25 ˆ U^ = iˆ + j- k 49 49 49 U// = 8 ˆ 12 ˆ 24 ˆ ij- k 49 49 49 following formula 2 U = U// + U^ 2 because the parallel and perpendicular components of U form a right triangle, with U as the hypotenuse. Example Problem If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2 vectors, the projection of U onto V , and the component of U that is perpendicular to V . We can check our work with the 101° = q U // = -4 7 U// = following formula 8 ˆ 12 ˆ 24 ˆ ij- k 49 49 49 2 U = U// + U^ 90 110 ˆ 25 ˆ U^ = iˆ + j- k 49 49 49 æ 90 ö æ 110 ö æ -25 ö U^ = ç ÷ + ç ÷ +ç ÷ è 49 ø è 49 ø è 49 ø 2 2 2 2 because the parallel and perpendicular components of U form a right triangle, with U as the hypotenuse. Example Problem If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2 vectors, the projection of U onto V , and the component of U that is perpendicular to V . We can check our work with the 101° = q U // = -4 7 U// = following formula 8 ˆ 12 ˆ 24 ˆ ij- k 49 49 49 2 U = U// + U^ 90 110 ˆ 25 ˆ U^ = iˆ + j- k 49 49 49 æ 90 ö æ 110 ö æ -25 ö U^ = ç ÷ + ç ÷ +ç ÷ è 49 ø è 49 ø è 49 ø 2 2 2 2 because the parallel and perpendicular components of U form a right triangle, with U as the hypotenuse. æ 8100 ö æ 12100 ö æ 625 ö U^ = ç ÷+ç ÷+ç ÷ è 2401 ø è 2401 ø è 2401 ø æ 20825 ö U^ = ç ÷ è 2401 ø Example Problem If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2 vectors, the projection of U onto V , and the component of U that is perpendicular to V . We can check our work with the 101° = q U // = -4 7 U// = 8 ˆ 12 ˆ 24 ˆ ij- k 49 49 49 90 110 ˆ 25 ˆ U^ = iˆ + j- k 49 49 49 æ 20825 ö U^ = ç ÷ è 2401 ø -4 U // = 7 following formula 2 U = U// + U^ 2 because the parallel and perpendicular components of U form a right triangle, with U as the hypotenuse. Example Problem If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2 vectors, the projection of U onto V , and the component of U that is perpendicular to V . We can check our work with the 101° = q U // = -4 7 U// = 8 ˆ 12 ˆ 24 ˆ ij- k 49 49 49 90 110 ˆ 25 ˆ U^ = iˆ + j- k 49 49 49 -4 U // = 7 æ 20825 ö U^ = ç ÷ è 2401 ø 2 U = æ 20825 ö æ -4 ö ç ÷ +ç ÷ è 2401 ø è 7 ø 2 following formula 2 U = U// + U^ 2 because the parallel and perpendicular components of U form a right triangle, with U as the hypotenuse. Example Problem If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2 vectors, the projection of U onto V , and the component of U that is perpendicular to V . We can check our work with the 101° = q U // = -4 7 U// = 8 ˆ 12 ˆ 24 ˆ ij- k 49 49 49 90 110 ˆ 25 ˆ U^ = iˆ + j- k 49 49 49 -4 U // = 7 æ 20825 ö U^ = ç ÷ è 2401 ø 2 U = æ 20825 ö æ -4 ö ç ÷ +ç ÷ è 2401 ø è 7 ø 2 following formula 2 U = U// + U^ 2 because the parallel and perpendicular components of U form a right triangle, with U as the hypotenuse. 20825 16 U = + 2401 49 Example Problem If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2 vectors, the projection of U onto V , and the component of U that is perpendicular to V . We can check our work with the 101° = q U // = -4 7 U// = 8 ˆ 12 ˆ 24 ˆ ij- k 49 49 49 90 110 ˆ 25 ˆ U^ = iˆ + j- k 49 49 49 -4 U // = 7 æ 20825 ö U^ = ç ÷ è 2401 ø 2 U = æ 20825 ö æ -4 ö ç ÷ +ç ÷ è 2401 ø è 7 ø 2 following formula 2 U = U// + U^ 2 because the parallel and perpendicular components of U form a right triangle, with U as the hypotenuse. 20825 16 U = + 2401 49 U= 9 Example Problem If U = 2iˆ + 2 jˆ - kˆ and V = -2iˆ + 3 jˆ + 6kˆ, find the angle between the 2 vectors, the projection of U onto V , and the component of U that is perpendicular to V . We can check our work with the 101° = q U // = -4 7 U// = 8 ˆ 12 ˆ 24 ˆ ij- k 49 49 49 90 110 ˆ 25 ˆ U^ = iˆ + j- k 49 49 49 -4 U // = 7 æ 20825 ö U^ = ç ÷ è 2401 ø 2 U = æ 20825 ö æ -4 ö ç ÷ +ç ÷ è 2401 ø è 7 ø 2 following formula 2 U = U// + U^ 2 because the parallel and perpendicular components of U form a right triangle, with U as the hypotenuse. 20825 16 U = + 2401 49 U= 9 U =3