Question

Вершинами трикутника є точки M (1; 0; 1), N (-5; 4; 3) і F (0; 3; -1). Знайдіть кут M трикутника.

Answer 1

Відповідь:     ∠M = 60° .  

Пояснення:

   Вершини ΔMNF :     M (1; 0; 1), N (-5; 4; 3) і F (0; 3; -1) . ∠M - ?

 Cкористаємось формулою :  cosM = ( MN * MF )/( |MN| * |MF| ) .

 Знайдемо вектори :   MN(- 6 ;4 ; 2 ) ;   MF(- 1 ; 3 ;- 2 ) .

MN * MF = - 6*(- 1 ) + 4*3 + 2*(- 2 ) = 14 ; |MN| =√((- 6 )² + 4² + 2² ) = 2√14 ;

               |MF| = √((- 1 )² + 3² + (- 2 )²) = √14 . Підставимо значення :

    cosM = 14/( 2√14 * √14 ) = 14/( 2 * 14 ) = 1/2 ; cosM = 1/2 ;  ∠M = 60° .      

   

Answer 2

To find the angle at vertex M of the triangle, you can use the cosine formula for angles in 3D space. The formula is given by:

$\cos(\theta) = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|}$

where $\mathbf{a}$ and $\mathbf{b}$ are vectors representing two sides of the angle, $\cdot$ denotes the dot product, and $\|\mathbf{a}\|$ and $\|\mathbf{b}\|$ are the magnitudes of the vectors.

Let's denote $\mathbf{MN}$ as vector $\mathbf{a}$ and $\mathbf{MF}$ as vector $\mathbf{b}$.

$\mathbf{MN} = \langle -5 - 1, 4 - 0, 3 - 1 \rangle = \langle -6, 4, 2 \rangle$ $\mathbf{MF} = \langle 0 - 1, 3 - 0, (-1) - 1 \rangle = \langle -1, 3, -2 \rangle$

Now, you can calculate the dot product and magnitudes:

$\|\mathbf{MN}\| = \sqrt{(-6)^2 + 4^2 + 2^2}$ $\|\mathbf{MF}\| = \sqrt{(-1)^2 + 3^2 + (-2)^2}$ $\mathbf{MN} \cdot \mathbf{MF} = (-6)(-1) + (4)(3) + (2)(-2)$

Finally, plug these values into the cosine formula to find the angle:

$\cos(\theta) = \frac{\mathbf{MN} \cdot \mathbf{MF}}{\|\mathbf{MN}\| \|\mathbf{MF}\|}$

And then, you can find the angle $\theta$ using the inverse cosine function.

I can help you with the calculations if you want!