Question

Угол между векторами AB и AC если a(0;2;-1) b(0;2;2) c(0;-3;-2) d(5;1;3)

Answer 1

Ответ:

Пошаговое объяснение:

Найдем вектора AB и AC, вычитая координаты начальной точки вектора из конечной:

AB = (0 - 0; 2 - 2; 2 - (-1)) = (0; 0; 3)

AC = (0 - 0; -3 - 2; -2 - (-1)) = (0; -5; -1)

Из формулы скалярного произведения векторов косинус угла между этими векторами равен:

$cos\alpha=\frac{AB*AC}{|AB|*|AC|}=\frac{0*0+(-5)*0+(-1) *3}{\sqrt{0^{2}+0^{2}+3^{2}}*\sqrt{0^{2}+(-5)^{2}+(-1)^{2}}}=\frac{-3}{\sqrt{9}*\sqrt{26}}=-\frac{1}{\sqrt{26}}\\\alpha=arccos(-\frac{1}{\sqrt{26}})$

Answer 2

To find the angle between two vectors, you can use the dot product formula and then apply the definition of the dot product:

The dot product of two vectors A and B is given by: A · B = |A| * |B| * cos(θ),

where A · B is the dot product of A and B, |A| is the magnitude (length) of vector A, |B| is the magnitude (length) of vector B, and θ is the angle between the two vectors.

Let's calculate the vectors AB and AC:

1. Vector AB: A = (0, 2, -1) B = (0, 2, 2) AB = B - A = (0 - 0, 2 - 2, 2 - (-1)) = (0, 0, 3)

2. Vector AC: A = (0, 2, -1) C = (0, -3, -2) AC = C - A = (0 - 0, -3 - 2, -2 - (-1)) = (0, -5, -1)

Now, we'll calculate the dot product of vectors AB and AC:

AB · AC = (0 * 0) + (0 * -5) + (3 * -1) = 0 - 15 - 3 = -18

Next, we need to calculate the magnitudes of vectors AB and AC:

|AB| = √(0^2 + 0^2 + 3^2) = √9 = 3 |AC| = √(0^2 + (-5)^2 + (-1)^2) = √26

Now, we can find the cosine of the angle θ using the dot product formula:

cos(θ) = (AB · AC) / (|AB| * |AC|) cos(θ) = -18 / (3 * √26)

Finally, we can find the angle θ using the arccosine (cos^-1) function:

θ = arccos(cos(θ)) θ = arccos(-18 / (3 * √26))

Using a calculator, we find that θ ≈ 104.47 degrees.

So, the angle between vectors AB and AC is approximately 104.47 degrees.