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What's the difference between dot multiplication and cross multiplication of vectors
There are two kinds of vector multiplication, namely inner product and outer product
Inner product is also called scalar product, because the result is a number (scalar), and the inner product of vector a and B is | a | B | cos & lt; a, B
(where & lt; a and B denote the angle between a and b)
The outer product of a vector is also called cross product. The result is a vector whose direction is perpendicular to the plane where a and B lie according to the right-handed system | a | B | Sin & lt; a, B
The difference between dot multiplication and cross multiplication is not in vector
It is in the elementary operation of general real numbers and letters (including unknowns and letters representing known numbers)
Point multiplication and cross multiplication (i.e. · and ×) are essentially the same in the multiplication operation of general real numbers and letters. They both represent the product relationship of numbers and numbers, but some writing methods are stipulated, such as: only cross multiplication (2 × 3) can be used between numbers, not dot multiplication (avoid being regarded as decimal point). Generally, dot multiplication (or save) is used between letters
How to determine the direction of the outer product of a vector?
The direction is determined according to the right hand rule, that is, the palm is standing on the vector a of the plane a and B, and the palm is facing B. then the direction of thumb is perpendicular to the plane, which is defined as the direction of outer product
Let a be a matrix of order 3, and the value of determinant A is 1 / 2. What is the determinant value of (2a) ∧ - 1-5a *?
Let a be a fifth order matrix, and the value of determinant A is 1 / 2. What is the determinant value of (2a) ∧ - 1-5a *?
Because a * = |a ^ - 1 = (1 / 2) a ^ - 1
therefore
|(2A)^-1-5A*|
= |(1/2)A^-1-(5/2)A^-1|
= |(-2)A^-1|
=(- 2) ^ 3 | a ^ - 1 | - pay attention here. In the supplementary question, you changed a to be a 5-order square matrix, so you need to change 3 to 5
= -8 |A|^-1
= -16.
What is 15 / 2 divided by (1.1-4 / 3) + 5 / 1 times 5 / 3
15 / 2 divided by (1.1-4 / 3) + 5 / 1 multiplied by 5 / 3
=2/15÷(1.1-0.75)+3/25
=2/15×20/7+3/25
=8/21+3/25
=263/525
Given the proposition p "∀ x ∈ [1,2], x2-a ≥ 0", proposition q "∃ x0 ∈ R, X02 + 2ax0 + 2-A = 0", if the proposition "P and Q" is true, the value range of real number a is obtained
If "P and Q" is a true proposition, then p is a true proposition, and Q is also a true proposition. If P is a true proposition, a ≤ X2 is always true, ∵ x ∈ [1,2], a ≤ 1; if q is a true proposition, that is, X2 + 2aX + 2-A = 0 has a real root, △ = 4a2-4 (2-A) ≥ 0, that is, a ≥ 1 or a ≤ - 2, then {a | a ≤ - 2 or a = 1} can be obtained by finding the intersection of ① and ②. In conclusion, the value range of the real number a is a ≤ - 2 or a = 1
Simple calculation of 7.6 - 1.8 + 2.4 - 8.2
7.6 - 1.8 + 2.4 - 8.2
= 7.6 + 2.4 - 1.8 - 8.2
= (7.6 + 2.4) - (1.8 +8.2 )
=10-10
=0
Through a point a (1,0) in the circle o x ^ 2 + y ^ 2 = 4 as the chord BC of the circle O, find the trajectory equation of the midpoint m of the chord BC
With the direct method, can explain the best, please!
There was a little mistake just now
The correction is as follows:
M（a,b）O（0,0）
Because am is perpendicular to OM
We get B / (A-1) * (B / a) = - 1
We get (A-1 / 2) ^ 2 + B ^ 2 = 1 / 4
The trajectory equation is: (x-1 / 2) ^ 2 + B ^ 2 = 1 / 4
It is a circle with (1 / 2,0) as the center and 1 / 2 as the radius
I hope it can help you~
2100 divided by 75
（25*84）/(25*3)
84/3
Given that the line L passes through P (2, l) and its inclination angle is half of that of the known line M: 3x-4y-17 = 0, the equation of L is obtained
Let the inclination angle of M be a
tana=3/4 cosa=4/5 sina=3/5
tana/2=sina/(1+cosa)=1/3
Let L: y = 1 / 3x + B
Substituting (2,1)
L：y=1/3x+1/3
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