Known vector a = (1,2), B = (COS) α, sin α), Let m = a + TB (t is a real number) if α= π / 4, find the absolute value of M, and when the minimum value is taken, the value of T

Known vector a = (1,2), B = (COS) α, sin α), Let m = a + TB (t is a real number) if α= π / 4, find the absolute value of M, and when the minimum value is taken, the value of T

b=√2/2（1,1）,m=（1+√2/2t,2+√2/2t）,
|m|=√（2+√2/2t） ²+ （1+√2/2t） ²
=√(4+t ²/ 2+2√2t+1+√2t+t ²/ 2)
=√(t ²+ 3√2t+5)=√[2（t+3√2/2） ²+ 1/2]
When t = - 3 √ 2 / 2, the above formula has the minimum value, which is √ 2 / 2,
So the answer is t = - 3 √ 2 / 2

It is known that a and B are two non-zero known vectors. When the modulus of a + TB (t belongs to R) takes the minimum value, find the value of T and prove that B is perpendicular to a + TB (t belongs to R) Please be specific

When | a + TB | takes the minimum value, that is, | a + TB | 2 takes the minimum value | a + TB | 2 = (a + TB) ^ 2 = a ^ 2 + 2tab + T ^ 2 B ^ 2 = B ^ 2 T ^ 2 + 2abt + A ^ 2 will be regarded as a quadratic function about t. because B ^ 2 > 0, when t = - 2Ab / (2B ^ 2) = AB / b ^ 2, | a + TB | takes the minimum value (note that a and B are vectors and cannot be approximately

Hurry! Given that a vector = (2,1) and B vector = (1,2), to minimize | a vector + TB vector | the value of real number T is? Given that a vector = (2,1) and B vector = (1,2), to minimize | a vector + TB vector | the value of real number T is? The answer is - 4 / 5

To minimize the | a vector + TB vector |
That is, find the minimum root sign {(2 + T) ^ 2 + (1 + 2t) ^ 2}
By simplifying the above formula, the root sign {5T ^ 2 + 8t + 5}:
This is an equation with an opening upward and has a minimum value: because B ＾ 2-4ac = 64-100 ＜ 0
So there is no intersection with the x-axis, that is, the equation cannot be less than 0
So just find his minimum
The minimum value is:
y=(4ac-b^2)/4a=(4*5*5-8^2) / (4*5) =(100-64)/20=1.8
That is, find 5T ^ 2 + 8t + 5 = 1.8
Simplified: 5T ^ 2 + 8t + 3.2 = 0
The solution is: T1 = T2 = - 4 / 5
So the answer is - 4 / 5
g

Known vector a=（ 3, 1), vector b=（sin α- m，cos α），α ∈ R, and a∥ b. Then the minimum value of real number m is __

∵
a∥
b. So sin α- m=
3cos α， I.e. M = sin α −
3cos α= 2sin（ α- π
3) Because α ∈ R, so the minimum value of M is: - 2
So the answer is: -2

The value of the real number x that minimizes the root sign (x ^ 2 + 4) + root sign [(8-x) ^ 2 + 16] is_

Method 1 derivation. I don't know if you have learned the method 2 geometric method. X ^ 2 + 4 = (x-0) ^ 2 + (0-2) ^ 2 (8-x) ^ 2 + 16 = (X-8) ^ 2 + (0-4) ^ 2. Therefore, the root sign x ^ 2 + 4 can be regarded as the distance from point P (x, 0) to point a (0,2). The root sign (8-x) ^ 2 + 16 can be regarded as the distance from point P (x, 0) to point B (8,4). The connecting line between three points should be the shortest

The product of zero vector and non-zero real number vector is Others say vector * vector = real number What about this question The product of a zero vector and any vector is still a zero vector. This sentence is also seen from the Internet

0
Remember the following rules
Vector * vector = real
Vector * real = real * vector = vector
therefore
Product of real number 0 and non-zero vector a = vector 0
Product of vector 0 and non-zero real number a = vector 0
Product of vector 0 and non-zero vector a = real 0