Point a (1, - 1,2) point B (2,1,1) point C (3, - 2,2) . find the area of this triangle . find the distance from triangle to point d (3,1,1)

Point a (1, - 1,2) point B (2,1,1) point C (3, - 2,2) . find the area of this triangle . find the distance from triangle to point d (3,1,1)

Vector AB = [1,2, - 1] vector AC = [2, - 1,0] cross product of vector AB and vector AC = [2 * 0 - (- 1) * (- 1), 2 * (- 1) - 1 * 0,1 * (- 1) - 2 * 2] = [- 1, - 2, - 5] area of triangle ABC = 0.5 | ab | AC | sin (angle BAC) [| ab | is the module of vector AB, that is, the length of vector AB] = 0.5 *

How to calculate vector
Given that a = (3,1), B = (- 1,0) vector λ a + B is perpendicular to a-2b, what is the value of real number λ?
The key is to teach me the operation of vectors

a=(3,1),b=(-1,0)
λa=(3λ,λ ) 2b=(-2,0)
λ a + B = (3, λ - 1, λ) a-2b = (5, 1). Since the vector λ a + B is perpendicular to a-2b, the multiplication of (λ a + b) by (a-2b) equals zero,
So = (3 λ - 1, λ) times (5,1) = 15 λ - 5 + λ = 16, λ - 5 = 0, λ = 5 / 16

How to calculate the length of the triangle bisector of the angle between two vectors?
The title is as follows: △ ABC three vertices a, (1,4), B (4,8), C (- 7,10)
Find the length of the bisector of A
I don't know where to think. You don't have to give me an answer, just a general idea

You're talking about the equation of the angular line
First calculate the angle a with the vector angle formula, then calculate the half angle Tan, and then apply it to the angle formula
A=90°
The slope of line AB is 4 / 3
The slope of the bisector of a is - 7

How to calculate the length of the row (column) vector of a real symmetric matrix of order 3

The square sum of the three elements in this row or column is re squared, which is the module of the vector