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Given |a|=2,|b|=1, the angle between a and b is π/3, and the angle between vector 2a+3b and 3a-b is obtained Title Direction sign of omitted vector (→) Given |a|=2,|b|=1, the angle between a and b is π/3, and the angle between vector 2a+3b and 3a-b is obtained Topic Direction sign of omitted vector (→)
I changed, this time should be correct cosθ=[(2a+3b)*(3a-b)]/[|2a+3b 3a-b|]=(6a^2+7ab-3b^2)/[√(16+9+12*2*1*1/2)(6^2+1-2*6*1*cosπ/3)]=(21+7*2*1*cosπ/3)/(√37 31)=28/(√37 31)=28/√1147=28√1147/1147≈0.8...
Let vector a.b satisfy |a|=|b|=1,|3a-2b|=3, find |3a+b| Let a.b satisfy |a|=|b|=1,|3a-2b|=3, find |3a+b|
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Given vector a=(2,4) b=(1,1), find:(1)2a-3b;(2)|a+2b| Processed
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If vector a=(3,-4) vector b=(2,1) try to find (a-2b)*(2a+3b) If vector a=(3,-4) vector b=(2,1), try to find (a-2b)*(2a+3b)
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Vector a=(3,-1), b=(-1,2), then |-3a-2b|= what? Come on, guys give me some steps. Thank you!
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What is the value of |a+2b| if the vector |a|=|b|=1 and a.b=-1\2?
Below the square, the answer is 3.
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