Question

Начало вектора а задано координатами (2;2) , а конец (5;-1).Найти проекции на оси : ах=?ау=?|вектора а|=?

Answer 1

Запишем вектор в координатах, а затем найдем его модуль:
$\vec a_x=B_x-A_x=5-2=3; \\ \vec a_y=B_y-A_y=-1-2=-3; \\ \vec a(3;-3); \\ |\vec a|= \sqrt{3^2+(-3)^2}= \sqrt{2*3^2}=3 \sqrt{2}$

Answer 2

Решение:
ax=5-2=3  
ay=-1-2=-3 
a=корень((3^2+(-3)^2))=корень(18)

Answer 3

Finding the Projections and Magnitude of Vector a

To find the projections of vector a onto the x and y axes, and the magnitude of vector a, we can use the given coordinates of the start and end points of the vector.

Given: Start point of vector a: (2, 2) End point of vector a: (5, -1)

Projections of Vector a onto the x and y Axes

The projections of vector a onto the x and y axes can be calculated using the differences in the x and y coordinates of the start and end points of the vector.

Projection onto the x-axis (a_x): The projection of vector a onto the x-axis is the difference in the x-coordinates of the end and start points of the vector. a_x = x_end - x_start = 5 - 2 = 3

Projection onto the y-axis (a_y): The projection of vector a onto the y-axis is the difference in the y-coordinates of the end and start points of the vector. a_y = y_end - y_start = -1 - 2 = -3

Magnitude of Vector a

The magnitude (or length) of vector a can be calculated using the distance formula between the start and end points of the vector.

Magnitude of vector a (|a|): The magnitude of vector a can be calculated using the distance formula: |a| = sqrt((x_end - x_start)^2 + (y_end - y_start)^2) |a| = sqrt((5 - 2)^2 + (-1 - 2)^2) |a| = sqrt(3^2 + (-3)^2) |a| = sqrt(9 + 9) |a| = sqrt(18) |a| = 3√2

So, the projections of vector a onto the x and y axes are: a_x = 3 a_y = -3

And the magnitude of vector a is: |a| = 3√2