How to calculate the addition of this vector group? one two -1 four Minus one sixth one -1 four six How to calculate this?

How to calculate the addition of this vector group? one two -1 four Minus one sixth one -1 four six How to calculate this?

Vector (1,2, - 1,4) - 1 / 6 * (1, - 1,4,6)
=(1,2,-1,4)-(1/6,-1/6,2/3,1)
=(1-1/6,2+1/6,-1-2/3,4-1)
=(5/6,13/6,-5/3,3)

The topic of vector operation
Both a and B are vectors. Given that a + 3b is perpendicular to 7a-5b and a-4b is perpendicular to 7a-2b, find the angle between a and B

Solution a + 3b is perpendicular to 7a-5b
(a+3b)(7a-5b)=7a^2-15b^2+16a*b=0(1)
A-4b is perpendicular to 7a-2b
(a-4b)(7a-2b)=7a^2+8b^2-30a*b=0(2)
(1)*15+(2)*8,161a^2-161b^2=0,|a|=|b|
(1)+(2),-7|b|^2-14|a||b|cosθ=0,cosθ=-1/2,θ=120

On the problem of vector cross product
Vector a × vector b=
| i j k|
|a1 b1 c1|
|a2 b2 c2|

Vector cross product can be regarded as opposite to dot product
But they are different
The result of dot multiplication is a constant
The cross product turns out to be a vector
Module of dot multiplication = module of a * module of B * cos angle
Module of cross product = module of a * module of B * sin angle
Have you ever learned determinants
This is the content of analytic geometry
Write two vector coordinates in order (by column), add a column, such as (1,1,1), then add the determinant sign to get the module of the cross product of the two vectors

What is the function of vector cross product in solid geometry
Can you give a few simple examples

The vector has direction and size. The cross product is the third vector perpendicular to the two vectors. Generally, the two vectors are regarded as the normal vectors of the two planes. Their cross product represents the normal vectors of the third plane perpendicular to the two planes. It can be used to find the plane