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According to the Pythagorean Theorem, plot the points on the number axis as root 3 and root 5, respectively According to the Pythagorean Theorem, draw points on the number axis which are represented as root 3 and root 5, respectively
Draw a number axis, take the original point as P, and make a vertical line Y passing through the P point, so that the points 1,2,3 and 4 are ABCD respectively. Use a compass to intercept PA'= PA on the vertical line Y, connect AA', and then intercept PB'= AA'= root number on the vertical line Y.2. Connect AB', and AB'= root number 3. Similarly, intercept PC'= AB' on the vertical line Y, and then connect BC', where BC'= root number, and then draw root number 3 and root number 5 on the number axis.
I hope you can keep up with my train of thought.
Given the vector a=(number 3,-1), b=(1/2, root number 3/2), if there is a nonzero real number k, t such that x=a+(t-square-3) b, y=-ka+tb, and x is vertical y. Try to Find the Minimum Value of k+t Square/t Given vector a=(root number 3,-1), b=(1/2, root number 3/2), if there is a nonzero real number k, t such that x=a+(t-square-3) b, y=-ka+tb, and x is vertical y. Try to Find the Minimum Value of k+t Square/t
A=(√3,-1), b=(1/2.√3/2), x=a+(t^2-3) b, y=-ka+tb, x⊥y, then the vector x·y=0,(a+bt^2-3b)·(-ka+tb)=0,-ka^2-kabt^2-3abk+tab+t^3b^2-3b^2=0, where a^2=√(3+1)=2, b^2=1, a·b=-√3/2+√3/2=0,(a+bt^2-3b)·(-ka+tb)=-2k+t^3...
Given that the angle between vector a and b is 30°, IaI = root number 3, IbI =2. Then Ia+bI =? Please write out the intermediate procedure! Given that the angle between vector a and vector b is 30°, IaI = root number 3, IbI =2, then Ia+bI =? Please write down the intermediate procedure!
|A+b|2=|a|2+2a*b+|b|2=3+2×(√3)×2×(√3/2)+4=13
|A+b|=√13.
Given IaI=1, IbI=root number 3, a+b=(root number 3,1), find Ia-bI, the angle between vector a+b and vector a-b
Let (a+b),(a-b) included angle = x|a+b|^2=4=|a 2 b|^2+2a.b4=1+3+2a.ba.b=0|a-b|=|a+b|=2(a-b).(a+b)=|a+b||a-b||cosx||a|^2-||b|^2=|a+b||a-b||cosx1-3=4cosxcosx =-1/2 x=120 degrees
Given vector a=(cosa, sina), vector b=(cosb, sinb), module of Ia-bI=(2 roots 5)/5,(1) the value of cos (a-b) is? (2)-π/2 Is the first question 3/5?
A-b=(cosa-cosb, sina-sinb)
|A-b|^2=(cosa-cosb)^2+(sina-sinb)^2=2-2(cosacosb+sinasinb)=4/5
Cosacosb+sinasinb =3/5
Cos (a-b)= cosacosb+sinasinb=3/5
-π/2
Given vector = a =(1, root number 3), b =(-2,0), then Ia+bI = what So I read the wrong question, and the answer was 2, and I thought it was 5, and I could figure it out.
A + b =(-1, root 3)
So Ia+bI=2
RELATED INFORMATIONS
- 1. Given that the angle between vector a and b is 30°, and |a|= root number 3,|b|=1, find the cosine of the angle between two vectors a+b, a-b Given that the angle between vector a and b is 30°, and |a|= root number 3,|b|=1, find the cosine value of the angle between two vectors a+b, a-b
- 2. Given the plane vector a =(3,-1 under the root), b =(1/2,3/2 under the root). (1) If there are real numbers k and t, x=ta+(t^2-5t+1) b, y=-ka+ab, and x is perpendicular to y, find the relation k=f (t) (2) According to the conclusion of (1), try to find the minimum value of the function k=f (t) on t (0,4). Given the plane vector a =(3,-1 at root), b =(1/2,3/2 at root). (1) If there are real numbers k and t, x=ta+(t^2-5t+1) b, y=-ka+ab, and x is perpendicular to y, find the relation k=f (t) (2) According to the conclusion of (1), try to find the minimum value of the function k=f (t) on t belonging to (0,4).
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