Question

Даны вектора a{1;4}, b{1;2}, c{7;2}. Найдите координаты вектора d=3a-2b+c

Answer 1

$\vec{d}=3\vec{a}-2\vec{b}+\vec{c}= \{3\cdot1-2\cdot1+7; \ 3\cdot4-2\cdot2+2\}=\\\\{} \ \ =\{3-2+7; \ 12-4+2\}=\{8; 10\}$

Answer 2

d {3-2+7;12-4+2}

d {8;10}

Answer 3

To find the coordinates of vector d, which is defined as d = 3a - 2b + c, we need to perform vector addition and scalar multiplication. Given the vectors a{1;4}, b{1;2}, and c{7;2}, let's calculate d as follows:

1. First, perform the scalar multiplication: 3a = 3 * a{1;4} = {3 * 1; 3 * 4} = {3; 12} -2b = -2 * b{1;2} = {-2 * 1; -2 * 2} = {-2; -4}

2. Now, perform vector addition: d = 3a - 2b + c = {3; 12} + {-2; -4} + {7; 2}

3. Add the corresponding components: d{X; Y} = {3 - 2 + 7; 12 - 4 + 2} = {8; 10}

So, the coordinates of vector d are d{8; 10}.