∣ 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

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15 Qs

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Penilaian harian Vektor Kelas XI By Kimshin

Quiz

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Mathematics

•

1st - 3rd Grade

•

Practice Problem

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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 $\overrightarrow{\ a}\cdot\overrightarrow{\ b}=...$

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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 $\overrightarrow{\ a}\cdot\overrightarrow{\ b}=...$