Cho 2 vecto a → ( 2 ; 4 ) \overrightarrow{a}\left(2;4\right) a <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> ( 2 ; 4 ) và b → ( − 3 ; 7 ) \overrightarrow{b}\left(-3;7\right) b <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> ( − 3 ; 7 ) . Tính tích vô hướng a → . b → \overrightarrow{a}.\overrightarrow{b} a <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> . b <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> là?

a → . b → = 22 \overrightarrow{a}.\overrightarrow{b}=22 a <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> . b <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> = 2 2

a → . b → = − 12 \overrightarrow{a}.\overrightarrow{b}=-12 a <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> . b <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> = − 1 2

a → . b → = − 34 \overrightarrow{a}.\overrightarrow{b}=-34 a <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> . b <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> = − 3 4

a → . b → = 15 \overrightarrow{a}.\overrightarrow{b}=15 a <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> . b <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> = 1 5

Cho vecto a → ( 2 ; 4 ) \overrightarrow{a}\left(2;4\right) a <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> ( 2 ; 4 ) . Độ dài ∣ a → ∣ \left|\overrightarrow{a}\right| ∣ ∣ ∣ a <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> ∣ ∣ ∣ là?

∣ a → ∣ = 2 5 \left|\overrightarrow{a}\right|=2\sqrt{5} ∣ ∣ ∣ a <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> ∣ ∣ ∣ = 2 5 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

∣ a → ∣ = 20 \left|\overrightarrow{a}\right|=20 ∣ ∣ ∣ a <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> ∣ ∣ ∣ = 2 0

∣ a → ∣ = 6 \left|\overrightarrow{a}\right|=6 ∣ ∣ ∣ a <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> ∣ ∣ ∣ = 6

∣ a → ∣ = 4 5 \left|\overrightarrow{a}\right|=4\sqrt{5} ∣ ∣ ∣ a <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> ∣ ∣ ∣ = 4 5 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

Trong mặt phẳng tọa độ Oxy, cho hai vecto a ⃗ = 4 i ⃗ + 6 j ⃗ \vec{a}=4\vec{i}+6\vec{j} a <path d="M377 20c0-5.333 1.833-10 5.5-14S391 0 397 0c4.667 0 8.667 1.667 12 5 3.333 2.667 6.667 9 10 19 6.667 24.667 20.333 43.667 41 57 7.333 4.667 11 10.667 11 18 0 6-1 10-3 12s-6.667 5-14 9c-28.667 14.667-53.667 35.667-75 63 -1.333 1.333-3.167 3.5-5.5 6.5s-4 4.833-5 5.5c-1 .667-2.5 1.333-4.5 2s-4.333 1 -7 1c-4.667 0-9.167-1.833-13.5-5.5S337 184 337 178c0-12.667 15.667-32.333 47-59 H213l-171-1c-8.667-6-13-12.333-13-19 0-4.667 4.333-11.333 13-20h359 c-16-25.333-24-45-24-59z"> = 4 i <path d="M377 20c0-5.333 1.833-10 5.5-14S391 0 397 0c4.667 0 8.667 1.667 12 5 3.333 2.667 6.667 9 10 19 6.667 24.667 20.333 43.667 41 57 7.333 4.667 11 10.667 11 18 0 6-1 10-3 12s-6.667 5-14 9c-28.667 14.667-53.667 35.667-75 63 -1.333 1.333-3.167 3.5-5.5 6.5s-4 4.833-5 5.5c-1 .667-2.5 1.333-4.5 2s-4.333 1 -7 1c-4.667 0-9.167-1.833-13.5-5.5S337 184 337 178c0-12.667 15.667-32.333 47-59 H213l-171-1c-8.667-6-13-12.333-13-19 0-4.667 4.333-11.333 13-20h359 c-16-25.333-24-45-24-59z"> + 6 j <path d="M377 20c0-5.333 1.833-10 5.5-14S391 0 397 0c4.667 0 8.667 1.667 12 5 3.333 2.667 6.667 9 10 19 6.667 24.667 20.333 43.667 41 57 7.333 4.667 11 10.667 11 18 0 6-1 10-3 12s-6.667 5-14 9c-28.667 14.667-53.667 35.667-75 63 -1.333 1.333-3.167 3.5-5.5 6.5s-4 4.833-5 5.5c-1 .667-2.5 1.333-4.5 2s-4.333 1 -7 1c-4.667 0-9.167-1.833-13.5-5.5S337 184 337 178c0-12.667 15.667-32.333 47-59 H213l-171-1c-8.667-6-13-12.333-13-19 0-4.667 4.333-11.333 13-20h359 c-16-25.333-24-45-24-59z"> và b ⃗ = 3 i ⃗ − 7 j ⃗ \vec{b}=3\vec{i}-7\vec{j} b <path d="M377 20c0-5.333 1.833-10 5.5-14S391 0 397 0c4.667 0 8.667 1.667 12 5 3.333 2.667 6.667 9 10 19 6.667 24.667 20.333 43.667 41 57 7.333 4.667 11 10.667 11 18 0 6-1 10-3 12s-6.667 5-14 9c-28.667 14.667-53.667 35.667-75 63 -1.333 1.333-3.167 3.5-5.5 6.5s-4 4.833-5 5.5c-1 .667-2.5 1.333-4.5 2s-4.333 1 -7 1c-4.667 0-9.167-1.833-13.5-5.5S337 184 337 178c0-12.667 15.667-32.333 47-59 H213l-171-1c-8.667-6-13-12.333-13-19 0-4.667 4.333-11.333 13-20h359 c-16-25.333-24-45-24-59z"> = 3 i <path d="M377 20c0-5.333 1.833-10 5.5-14S391 0 397 0c4.667 0 8.667 1.667 12 5 3.333 2.667 6.667 9 10 19 6.667 24.667 20.333 43.667 41 57 7.333 4.667 11 10.667 11 18 0 6-1 10-3 12s-6.667 5-14 9c-28.667 14.667-53.667 35.667-75 63 -1.333 1.333-3.167 3.5-5.5 6.5s-4 4.833-5 5.5c-1 .667-2.5 1.333-4.5 2s-4.333 1 -7 1c-4.667 0-9.167-1.833-13.5-5.5S337 184 337 178c0-12.667 15.667-32.333 47-59 H213l-171-1c-8.667-6-13-12.333-13-19 0-4.667 4.333-11.333 13-20h359 c-16-25.333-24-45-24-59z"> − 7 j <path d="M377 20c0-5.333 1.833-10 5.5-14S391 0 397 0c4.667 0 8.667 1.667 12 5 3.333 2.667 6.667 9 10 19 6.667 24.667 20.333 43.667 41 57 7.333 4.667 11 10.667 11 18 0 6-1 10-3 12s-6.667 5-14 9c-28.667 14.667-53.667 35.667-75 63 -1.333 1.333-3.167 3.5-5.5 6.5s-4 4.833-5 5.5c-1 .667-2.5 1.333-4.5 2s-4.333 1 -7 1c-4.667 0-9.167-1.833-13.5-5.5S337 184 337 178c0-12.667 15.667-32.333 47-59 H213l-171-1c-8.667-6-13-12.333-13-19 0-4.667 4.333-11.333 13-20h359 c-16-25.333-24-45-24-59z"> . Tính tích vô hướng a ⃗ . b ⃗ \vec{a}.\vec{b} a <path d="M377 20c0-5.333 1.833-10 5.5-14S391 0 397 0c4.667 0 8.667 1.667 12 5 3.333 2.667 6.667 9 10 19 6.667 24.667 20.333 43.667 41 57 7.333 4.667 11 10.667 11 18 0 6-1 10-3 12s-6.667 5-14 9c-28.667 14.667-53.667 35.667-75 63 -1.333 1.333-3.167 3.5-5.5 6.5s-4 4.833-5 5.5c-1 .667-2.5 1.333-4.5 2s-4.333 1 -7 1c-4.667 0-9.167-1.833-13.5-5.5S337 184 337 178c0-12.667 15.667-32.333 47-59 H213l-171-1c-8.667-6-13-12.333-13-19 0-4.667 4.333-11.333 13-20h359 c-16-25.333-24-45-24-59z"> . b <path d="M377 20c0-5.333 1.833-10 5.5-14S391 0 397 0c4.667 0 8.667 1.667 12 5 3.333 2.667 6.667 9 10 19 6.667 24.667 20.333 43.667 41 57 7.333 4.667 11 10.667 11 18 0 6-1 10-3 12s-6.667 5-14 9c-28.667 14.667-53.667 35.667-75 63 -1.333 1.333-3.167 3.5-5.5 6.5s-4 4.833-5 5.5c-1 .667-2.5 1.333-4.5 2s-4.333 1 -7 1c-4.667 0-9.167-1.833-13.5-5.5S337 184 337 178c0-12.667 15.667-32.333 47-59 H213l-171-1c-8.667-6-13-12.333-13-19 0-4.667 4.333-11.333 13-20h359 c-16-25.333-24-45-24-59z"> .

a ⃗ ⋅ b ⃗ = − 30. \vec{a}\cdot\vec{b}=-30. a <path d="M377 20c0-5.333 1.833-10 5.5-14S391 0 397 0c4.667 0 8.667 1.667 12 5 3.333 2.667 6.667 9 10 19 6.667 24.667 20.333 43.667 41 57 7.333 4.667 11 10.667 11 18 0 6-1 10-3 12s-6.667 5-14 9c-28.667 14.667-53.667 35.667-75 63 -1.333 1.333-3.167 3.5-5.5 6.5s-4 4.833-5 5.5c-1 .667-2.5 1.333-4.5 2s-4.333 1 -7 1c-4.667 0-9.167-1.833-13.5-5.5S337 184 337 178c0-12.667 15.667-32.333 47-59 H213l-171-1c-8.667-6-13-12.333-13-19 0-4.667 4.333-11.333 13-20h359 c-16-25.333-24-45-24-59z"> ⋅ b <path d="M377 20c0-5.333 1.833-10 5.5-14S391 0 397 0c4.667 0 8.667 1.667 12 5 3.333 2.667 6.667 9 10 19 6.667 24.667 20.333 43.667 41 57 7.333 4.667 11 10.667 11 18 0 6-1 10-3 12s-6.667 5-14 9c-28.667 14.667-53.667 35.667-75 63 -1.333 1.333-3.167 3.5-5.5 6.5s-4 4.833-5 5.5c-1 .667-2.5 1.333-4.5 2s-4.333 1 -7 1c-4.667 0-9.167-1.833-13.5-5.5S337 184 337 178c0-12.667 15.667-32.333 47-59 H213l-171-1c-8.667-6-13-12.333-13-19 0-4.667 4.333-11.333 13-20h359 c-16-25.333-24-45-24-59z"> = − 3 0 .

a ⃗ ⋅ b ⃗ = 43. \vec{a}\cdot\vec{b}=43. a <path d="M377 20c0-5.333 1.833-10 5.5-14S391 0 397 0c4.667 0 8.667 1.667 12 5 3.333 2.667 6.667 9 10 19 6.667 24.667 20.333 43.667 41 57 7.333 4.667 11 10.667 11 18 0 6-1 10-3 12s-6.667 5-14 9c-28.667 14.667-53.667 35.667-75 63 -1.333 1.333-3.167 3.5-5.5 6.5s-4 4.833-5 5.5c-1 .667-2.5 1.333-4.5 2s-4.333 1 -7 1c-4.667 0-9.167-1.833-13.5-5.5S337 184 337 178c0-12.667 15.667-32.333 47-59 H213l-171-1c-8.667-6-13-12.333-13-19 0-4.667 4.333-11.333 13-20h359 c-16-25.333-24-45-24-59z"> ⋅ b <path d="M377 20c0-5.333 1.833-10 5.5-14S391 0 397 0c4.667 0 8.667 1.667 12 5 3.333 2.667 6.667 9 10 19 6.667 24.667 20.333 43.667 41 57 7.333 4.667 11 10.667 11 18 0 6-1 10-3 12s-6.667 5-14 9c-28.667 14.667-53.667 35.667-75 63 -1.333 1.333-3.167 3.5-5.5 6.5s-4 4.833-5 5.5c-1 .667-2.5 1.333-4.5 2s-4.333 1 -7 1c-4.667 0-9.167-1.833-13.5-5.5S337 184 337 178c0-12.667 15.667-32.333 47-59 H213l-171-1c-8.667-6-13-12.333-13-19 0-4.667 4.333-11.333 13-20h359 c-16-25.333-24-45-24-59z"> = 4 3 .

a ⃗ ⋅ b ⃗ = 3. \vec{a}\cdot\vec{b}=3. a <path d="M377 20c0-5.333 1.833-10 5.5-14S391 0 397 0c4.667 0 8.667 1.667 12 5 3.333 2.667 6.667 9 10 19 6.667 24.667 20.333 43.667 41 57 7.333 4.667 11 10.667 11 18 0 6-1 10-3 12s-6.667 5-14 9c-28.667 14.667-53.667 35.667-75 63 -1.333 1.333-3.167 3.5-5.5 6.5s-4 4.833-5 5.5c-1 .667-2.5 1.333-4.5 2s-4.333 1 -7 1c-4.667 0-9.167-1.833-13.5-5.5S337 184 337 178c0-12.667 15.667-32.333 47-59 H213l-171-1c-8.667-6-13-12.333-13-19 0-4.667 4.333-11.333 13-20h359 c-16-25.333-24-45-24-59z"> ⋅ b <path d="M377 20c0-5.333 1.833-10 5.5-14S391 0 397 0c4.667 0 8.667 1.667 12 5 3.333 2.667 6.667 9 10 19 6.667 24.667 20.333 43.667 41 57 7.333 4.667 11 10.667 11 18 0 6-1 10-3 12s-6.667 5-14 9c-28.667 14.667-53.667 35.667-75 63 -1.333 1.333-3.167 3.5-5.5 6.5s-4 4.833-5 5.5c-1 .667-2.5 1.333-4.5 2s-4.333 1 -7 1c-4.667 0-9.167-1.833-13.5-5.5S337 184 337 178c0-12.667 15.667-32.333 47-59 H213l-171-1c-8.667-6-13-12.333-13-19 0-4.667 4.333-11.333 13-20h359 c-16-25.333-24-45-24-59z"> = 3 .

a ⃗ ⋅ b ⃗ = 30. \vec{a}\cdot\vec{b}=30. a <path d="M377 20c0-5.333 1.833-10 5.5-14S391 0 397 0c4.667 0 8.667 1.667 12 5 3.333 2.667 6.667 9 10 19 6.667 24.667 20.333 43.667 41 57 7.333 4.667 11 10.667 11 18 0 6-1 10-3 12s-6.667 5-14 9c-28.667 14.667-53.667 35.667-75 63 -1.333 1.333-3.167 3.5-5.5 6.5s-4 4.833-5 5.5c-1 .667-2.5 1.333-4.5 2s-4.333 1 -7 1c-4.667 0-9.167-1.833-13.5-5.5S337 184 337 178c0-12.667 15.667-32.333 47-59 H213l-171-1c-8.667-6-13-12.333-13-19 0-4.667 4.333-11.333 13-20h359 c-16-25.333-24-45-24-59z"> ⋅ b <path d="M377 20c0-5.333 1.833-10 5.5-14S391 0 397 0c4.667 0 8.667 1.667 12 5 3.333 2.667 6.667 9 10 19 6.667 24.667 20.333 43.667 41 57 7.333 4.667 11 10.667 11 18 0 6-1 10-3 12s-6.667 5-14 9c-28.667 14.667-53.667 35.667-75 63 -1.333 1.333-3.167 3.5-5.5 6.5s-4 4.833-5 5.5c-1 .667-2.5 1.333-4.5 2s-4.333 1 -7 1c-4.667 0-9.167-1.833-13.5-5.5S337 184 337 178c0-12.667 15.667-32.333 47-59 H213l-171-1c-8.667-6-13-12.333-13-19 0-4.667 4.333-11.333 13-20h359 c-16-25.333-24-45-24-59z"> = 3 0 .

Trong mặt phẳng tọa độ Oxy, tính khoảng cách giữa hai điểm M ( 1 ; − 2 ) M(1;-2) M ( 1 ; − 2 ) và N ( − 3 ; 4 ) N(-3;4) N ( − 3 ; 4 ) .

M N = 2 13 . MN=2\sqrt{13}. M N = 2 1 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> .

M N = 3 6 . MN=3\sqrt{6}. M N = 3 6 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> .

tích vô hướng

Authored by Ngọc Linh Phạm Thị

Mathematics

10th Grade

Used 1+ times

AI Actions

Add similar questions

Adjust reading levels

Convert to real-world scenario

Translate activity

More...

Content View
Student View

40 questions

Show all answers

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Cho hình vuông ABCD, cạnh a. Tính $\overrightarrow{AB}.\overrightarrow{CD}$ bằng

$a^2$

$-a^2$

$2a$

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Cho tam giác ABC đều, cạnh bằng 1. Tính $\overrightarrow{BA}.\overrightarrow{BC}$

$\frac{\sqrt[]{3}}{2}$

$\frac{1}{2}$

MULTIPLE CHOICE QUESTION

1 min • 1 pt

$150^0$

$45^0$

$60^0$

$120^0$

10.

MULTIPLE CHOICE QUESTION

1 min • 1 pt

Cho tam giác ABC vuông cân tại A. Góc giữa hai vector $\overrightarrow{BA}$ và $\overrightarrow{BC}$ là

$90^0$

$45^0$

$135^0$

$60^0$

Access all questions and much more by creating a free account

Create resources

Host any resource

Get auto-graded reports

Continue with Google

Continue with Email

Continue with Microsoft

or continue with

Facebook

Apple

Others

By signing up, you agree to our Terms of Service & Privacy Policy

Already have an account?

Similar Resources on Wayground

35 questions

Sohcahtoa Finding Missing Sides

Quiz

•

9th - 12th Grade

35 questions

Effects of Mean Median and Mode

Quiz

•

9th - 12th Grade

36 questions

Year 9 End of Term Quiz Game

Quiz

•

9th - 10th Grade

35 questions

Mean Median Mode Range Iqr Box And Whisker Plot

Quiz

•

9th - 12th Grade

40 questions

Hy's Class Period 7/8

Quiz

•

8th Grade - University

41 questions

Right Triangle

Quiz

•

8th - 10th Grade

43 questions

Equations Solutions

Quiz

•

8th - 10th Grade

38 questions

Quadratic Solving Multiple Choice

Quiz

•

9th - 10th Grade

Popular Resources on Wayground

10 questions

How much do you know about our Portrait of an Eagle?

Quiz

•

10th Grade

10 questions

Fast Food Slogans

Quiz

•

6th - 8th Grade

21 questions

Continents and Oceans

Quiz

•

6th Grade

20 questions

Parts of Speech

Quiz

•

5th Grade

16 questions

Subject & Predicate

Quiz

•

5th Grade

25 questions

Multiplication Facts

Quiz

•

5th Grade

12 questions

Map Skills

Quiz

•

3rd Grade

22 questions

Continents and Oceans

Quiz

•

5th Grade

Discover more resources for Mathematics

10 questions

Identifying equations

Quiz

•

KG - University

12 questions

Identify Rigid Transformations

Quiz

•

8th - 12th Grade

9 questions

Proving lines are parallel using angles

Quiz

•

9th - 11th Grade

10 questions

Midpoint Formula

Quiz

•

10th Grade

12 questions

Parallel Lines Cut by a Transversal

Quiz

•

10th Grade

15 questions

Triangle Sum Theorem and Exterior Angle Theorem

Quiz

•

10th Grade

15 questions

2.3 (b) Domain & Range Lesson Check

Quiz

•

9th - 12th Grade

20 questions

Classifying Real Numbers

Quiz

•

6th - 12th Grade

tích vô hướng

Cho hình vuông ABCD, cạnh a. Tính AB→.CD→\overrightarrow{AB}.\overrightarrow{CD}AB.CD bằng

Cho tam giác ABC đều, cạnh bằng 1. Tính BA→.BC→\overrightarrow{BA}.\overrightarrow{BC}BA.BC

Cho hình thoi ABCD cạnh a. Tính AC→.BD→\overrightarrow{AC}.\overrightarrow{BD}AC.BD

Cho tam giác ABC đều. Tính cos⁡(AB→,BC→)\cos\left(\overrightarrow{AB},\overrightarrow{BC}\right)cos(AB,BC)

Cho hình vuông ABCD cạnh a. Tính AB→.BD→\overrightarrow{AB}.\overrightarrow{BD}AB.BD

Cho hai vector a→\overrightarrow{a}a và b→\overrightarrow{b}b . Khẳng định nào sau đây đúng?

Cho hai vector a→,b→\overrightarrow{a},\overrightarrow{b}a,b ngược hướng và đều khác 0→\overrightarrow{0}0 . Khẳng định nào sau đây đúng?

Cho tam giác ABC có AB = 3, AC = 4, BC = 5. Tính AB→.AC→\overrightarrow{AB}.\overrightarrow{AC}AB.AC

Cho ∣a→∣=3,∣b→∣=2,a→.b→=−3\left|\overrightarrow{a}\right|=3,\left|\overrightarrow{b}\right|=2,\overrightarrow{a}.\overrightarrow{b}=-3​a​=3,​b​=2,a.b=−3 . Tính góc giữa hai vector a→\overrightarrow{a}a và b→\overrightarrow{b}b

Cho tam giác ABC vuông cân tại A. Góc giữa hai vector BA→\overrightarrow{BA}BA và BC→\overrightarrow{BC}BC là

Access all questions and much more by creating a free account

Similar Resources on Wayground

Popular Resources on Wayground

Discover more resources for Mathematics

Cho hình vuông ABCD, cạnh a. Tính $\overrightarrow{AB}.\overrightarrow{CD}$ bằng

Cho tam giác ABC đều, cạnh bằng 1. Tính $\overrightarrow{BA}.\overrightarrow{BC}$

Cho hình thoi ABCD cạnh a. Tính $\overrightarrow{AC}.\overrightarrow{BD}$

Cho tam giác ABC đều. Tính $\cos\left(\overrightarrow{AB},\overrightarrow{BC}\right)$

Cho hình vuông ABCD cạnh a. Tính $\overrightarrow{AB}.\overrightarrow{BD}$

Cho hai vector $\overrightarrow{a}$ và $\overrightarrow{b}$ . Khẳng định nào sau đây đúng?

Cho hai vector $\overrightarrow{a},\overrightarrow{b}$ ngược hướng và đều khác $\overrightarrow{0}$ . Khẳng định nào sau đây đúng?

Cho tam giác ABC có AB = 3, AC = 4, BC = 5. Tính $\overrightarrow{AB}.\overrightarrow{AC}$

Cho $\left|\overrightarrow{a}\right|=3,\left|\overrightarrow{b}\right|=2,\overrightarrow{a}.\overrightarrow{b}=-3$ . Tính góc giữa hai vector $\overrightarrow{a}$ và $\overrightarrow{b}$

Cho tam giác ABC vuông cân tại A. Góc giữa hai vector $\overrightarrow{BA}$ và $\overrightarrow{BC}$ là