∣ a → ∣ = 4 \left|\overrightarrow{\ a}\right|=4 ∣ ∣ ∣ 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 , ∣ b → ∣ = 5 \left|\overrightarrow{\ b}\right|=5 ∣ ∣ ∣ 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"> ∣ ∣ ∣ = 5 dan sudut antara vektor a → \overrightarrow{\ a}\ 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"> dan b → \overrightarrow{\ b}\ 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"> adalah 60 ° 60\degree 6 0 ° . maka a → ⋅ b → = . . . \overrightarrow{\ a}\cdot\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"> = . . .

10 2 10\sqrt{2} 1 0 2 <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">

10 3 10\sqrt{3} 1 0 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">

20 3 20\sqrt{3} 2 0 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">

panjang vektor a → = 2 i − j + 2 k \overrightarrow{\ a}=2i-j+2k 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 i − j + 2 k adalah ...

2 \sqrt{2} 2 <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">

3 \sqrt{3} 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">

u → = 3 i − k \overrightarrow{\ u}=3i-k u <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 i − k diubah ke vektor baris, yaitu ...

( 3 , 0 , − 1 ) \left(3,\ 0,-1\right)\ ( 3 , 0 , − 1 )

( 3 , − 1 ) \left(3,\ -1\right)\ ( 3 , − 1 )

( 3 , − 1 , 0 ) \left(3,\ -1,\ 0\right)\ ( 3 , − 1 , 0 )

( 3 , 1 ) \left(3,\ 1\right)\ ( 3 , 1 )

( 3 , 0 , 1 ) \left(3,\ 0,\ 1\right)\ ( 3 , 0 , 1 )

p → = ( 5 , − 1 ) \overrightarrow{\ p}=\left(5,\ -1\right) p <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"> = ( 5 , − 1 ) maka 2 p → = . . . 2\overrightarrow{\ p}=... 2 p <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"> = . . .

( 10 , − 2 ) \left(10,-2\right)\ ( 1 0 , − 2 )

( 5 , − 1 ) \left(5,-1\right)\ ( 5 , − 1 )

( 10 , − 1 ) \left(10,-1\right)\ ( 1 0 , − 1 )

( 5 , − 2 ) \left(5,-2\right)\ ( 5 , − 2 )

( 10 , 2 ) \left(10,\ 2\right)\ ( 1 0 , 2 )

a → = 9 i − 2 j − 22 k \overrightarrow{\ a}=9i-2j-22k 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"> = 9 i − 2 j − 2 2 k dan b → = − i + 8 j − 2 k \overrightarrow{\ b}=-i+8j-2k 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"> = − i + 8 j − 2 k maka 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"> = . . .

8 i + 6 j − 24 k 8i+6j-24k 8 i + 6 j − 2 4 k

10 i − 10 j − 20 k 10i-10j-20k 1 0 i − 1 0 j − 2 0 k

8 i − 10 j − 24 k 8i-10j-24k 8 i − 1 0 j − 2 4 k

8 i − 6 j + 24 k 8i-6j+24k 8 i − 6 j + 2 4 k

10 i − 6 j + 20 k 10i-6j+20k 1 0 i − 6 j + 2 0 k

a → = 3 i + j + k \overrightarrow{\ a}=3i+j+k 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"> = 3 i + j + k dan b → = 2 i + 5 j + k \overrightarrow{\ b}=2i+5j+k 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 i + 5 j + k . Proyeksi skalar ortogonal a pada b adalah ...

12 30 \frac{12}{\sqrt{30}} 3 0 <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"> 1 2

12 16 \frac{12}{\sqrt{16}} 1 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"> 1 2

12 11 \frac{12}{\sqrt{11}} 1 1 <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"> 1 2

a → = 3 i + j + k \overrightarrow{\ a}=3i+j+k 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"> = 3 i + j + k dan b → = 2 i + 5 j + k \overrightarrow{\ b}=2i+5j+k 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 i + 5 j + k . Proyeksi vektor ortogonal a pada b adalah

12 30 ( 2 i + 5 j + k ) \frac{12}{30}\left(2i+5j+k\right)\ 3 0 1 2 ( 2 i + 5 j + k )

12 16 ( 2 i + 5 j + k ) \frac{12}{16}\left(2i+5j+k\right)\ 1 6 1 2 ( 2 i + 5 j + k )

12 11 ( 2 i + 5 j + k ) \frac{12}{11}\left(2i+5j+k\right)\ 1 1 1 2 ( 2 i + 5 j + k )

12 30 ( 3 i + j + k ) \frac{12}{30}\left(3i+j+k\right)\ 3 0 1 2 ( 3 i + j + k )

12 11 ( 3 i + j + k ) \frac{12}{11}\left(3i+j+k\right)\ 1 1 1 2 ( 3 i + j + k )

c → = ( 0 , − 18 , 10 ) \overrightarrow{\ c}=\left(0,\ -18,\ 10\right) c <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"> = ( 0 , − 1 8 , 1 0 ) ubah ke vektor basis

− 18 j + 10 k -18j+10k − 1 8 j + 1 0 k

− 18 i + 10 j -18i+10j − 1 8 i + 1 0 j

− 18 i + 10 k -18i+10k − 1 8 i + 1 0 k

18 j + 10 k 18j+10k 1 8 j + 1 0 k

18 i + 10 j 18i+10j 1 8 i + 1 0 j

Penilaian harian Vektor Kelas XI By Kimshin

Quiz

•

Mathematics

•

1st - 3rd Grade

•

Practice Problem

•

Medium

Kania Mireshtika

Used 131+ times

FREE Resource

15 questions

Show all answers

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Panjang vektor OA jika O (0, 0, 0) dan A (7, -1, 0)

$5\sqrt{2}$

$5\sqrt{3}$

$3\sqrt{2}$

$2\sqrt{3}$

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

diketahui vektor $\overrightarrow{\ a}\ =i+2j+mk\$ dan $\overrightarrow{\ b}=2i-10j+2k$ . Jika $\overrightarrow{\ a\ }\cdot\overrightarrow{\ b\ }=0$ nilai m adalah ...

-16

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

jika sudut antara vektor $\overrightarrow{\ a}=\left(2,\ 1,\ -3\right)$ dan vektor $\overrightarrow{\ b}=\left(-1,\ 3,-2\right)$ adalah $\alpha$ . besarnya $\alpha=...$

$45\degree$

$60\degree$

$90\degree$

$120\degree$

$150\degree$

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

diketahui vektor $\overrightarrow{\ a}=\left(1,\ 4\right)\$ dan $\overrightarrow{\ b}=\left(2,\ 3\right)\$ . Vektor $2\overrightarrow{\ a}+3\overrightarrow{\ b}$ bila dinyatakan dalam bentuk vektor basis adalah ...

$3i+7j$

$6i+14j$

$9i+12j$

$8i+17j$

$8i+21j$

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

diketahui titik A (4, -2) dan B (3, -1) maka vektor AB adalah ...

$\left(1,-1\right)\$

$\left(3,-5\right)\$

$\left(5,-5\right)\$

$\left(-5,\ 5\right)\$

$\left(-1,\ 1\right)\$

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

$\left|\overrightarrow{\ a}\right|=4$ , $\left|\overrightarrow{\ b}\right|=5$ dan sudut antara vektor $\overrightarrow{\ a}\$ dan $\overrightarrow{\ b}\$ adalah $60\degree$ . maka $\overrightarrow{\ a}\cdot\overrightarrow{\ b}=...$

$10\sqrt{2}$

$10\sqrt{3}$

$20\sqrt{3}$

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

panjang vektor $\overrightarrow{\ a}=2i-j+2k$ adalah ...

$\sqrt{2}$

$\sqrt{3}$

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Penilaian harian Vektor Kelas XI By Kimshin

15 questions

Panjang vektor OA jika O (0, 0, 0) dan A (7, -1, 0)

diketahui vektor $\overrightarrow{\ a}\ =i+2j+mk\$ dan $\overrightarrow{\ b}=2i-10j+2k$ . Jika $\overrightarrow{\ a\ }\cdot\overrightarrow{\ b\ }=0$ nilai m adalah ...

jika sudut antara vektor $\overrightarrow{\ a}=\left(2,\ 1,\ -3\right)$ dan vektor $\overrightarrow{\ b}=\left(-1,\ 3,-2\right)$ adalah $\alpha$ . besarnya $\alpha=...$

diketahui vektor $\overrightarrow{\ a}=\left(1,\ 4\right)\$ dan $\overrightarrow{\ b}=\left(2,\ 3\right)\$ . Vektor $2\overrightarrow{\ a}+3\overrightarrow{\ b}$ bila dinyatakan dalam bentuk vektor basis adalah ...

diketahui titik A (4, -2) dan B (3, -1) maka vektor AB adalah ...

$\left|\overrightarrow{\ a}\right|=4$ , $\left|\overrightarrow{\ b}\right|=5$ dan sudut antara vektor $\overrightarrow{\ a}\$ dan $\overrightarrow{\ b}\$ adalah $60\degree$ . maka $\overrightarrow{\ a}\cdot\overrightarrow{\ b}=...$

panjang vektor $\overrightarrow{\ a}=2i-j+2k$ adalah ...

$\overrightarrow{\ u}=3i-k$ diubah ke vektor baris, yaitu ...

$\overrightarrow{\ p}=\left(5,\ -1\right)$ maka $2\overrightarrow{\ p}=...$

$\overrightarrow{\ a}=2i+5j$ dan $\overrightarrow{\ b}=i-2j$ maka $\overrightarrow{\ a}\cdot\overrightarrow{\ b}=...$

Create a free account and access millions of resources

Similar Resources on Wayground

Popular Resources on Wayground

Discover more resources for Mathematics

Penilaian harian Vektor Kelas XI By Kimshin

15 questions

Panjang vektor OA jika O (0, 0, 0) dan A (7, -1, 0)

diketahui vektor a→ =i+2j+mk \overrightarrow{\ a}\ =i+2j+mk\ a =i+2j+mk dan b→=2i−10j+2k\overrightarrow{\ b}=2i-10j+2k b=2i−10j+2k . Jika a →⋅ b →=0\overrightarrow{\ a\ }\cdot\overrightarrow{\ b\ }=0 a ⋅ b =0 nilai m adalah ...

jika sudut antara vektor a→=(2, 1, −3)\overrightarrow{\ a}=\left(2,\ 1,\ -3\right) a=(2, 1, −3) dan vektor b→=(−1, 3,−2)\overrightarrow{\ b}=\left(-1,\ 3,-2\right) b=(−1, 3,−2) adalah α\alphaα . besarnya α=...\alpha=...α=...

diketahui vektor a→=(1, 4) \overrightarrow{\ a}=\left(1,\ 4\right)\ a=(1, 4) dan b→=(2, 3) \overrightarrow{\ b}=\left(2,\ 3\right)\ b=(2, 3) . Vektor 2 a→+3 b→2\overrightarrow{\ a}+3\overrightarrow{\ b}2 a+3 b bila dinyatakan dalam bentuk vektor basis adalah ...

diketahui titik A (4, -2) dan B (3, -1) maka vektor AB adalah ...

panjang vektor a→=2i−j+2k\overrightarrow{\ a}=2i-j+2k a=2i−j+2k adalah ...

u→=3i−k\overrightarrow{\ u}=3i-k u=3i−k diubah ke vektor baris, yaitu ...

p→=(5, −1)\overrightarrow{\ p}=\left(5,\ -1\right) p​=(5, −1) maka 2 p→=...2\overrightarrow{\ p}=...2 p​=...

a→=2i+5j\overrightarrow{\ a}=2i+5j a=2i+5j dan b→=i−2j\overrightarrow{\ b}=i-2j b=i−2j maka a→⋅ b→=...\overrightarrow{\ a}\cdot\overrightarrow{\ b}=... a⋅ b=...

Create a free account and access millions of resources

Similar Resources on Wayground

Popular Resources on Wayground

Discover more resources for Mathematics

diketahui vektor $\overrightarrow{\ a}\ =i+2j+mk\$ dan $\overrightarrow{\ b}=2i-10j+2k$ . Jika $\overrightarrow{\ a\ }\cdot\overrightarrow{\ b\ }=0$ nilai m adalah ...

jika sudut antara vektor $\overrightarrow{\ a}=\left(2,\ 1,\ -3\right)$ dan vektor $\overrightarrow{\ b}=\left(-1,\ 3,-2\right)$ adalah $\alpha$ . besarnya $\alpha=...$

diketahui vektor $\overrightarrow{\ a}=\left(1,\ 4\right)\$ dan $\overrightarrow{\ b}=\left(2,\ 3\right)\$ . Vektor $2\overrightarrow{\ a}+3\overrightarrow{\ b}$ bila dinyatakan dalam bentuk vektor basis adalah ...

panjang vektor $\overrightarrow{\ a}=2i-j+2k$ adalah ...

$\overrightarrow{\ u}=3i-k$ diubah ke vektor baris, yaitu ...

$\overrightarrow{\ p}=\left(5,\ -1\right)$ maka $2\overrightarrow{\ p}=...$

$\overrightarrow{\ a}=2i+5j$ dan $\overrightarrow{\ b}=i-2j$ maka $\overrightarrow{\ a}\cdot\overrightarrow{\ b}=...$