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В правильной шестиугольной пирамиде
SABCDEF, стороны основания которой равны 1 , а боковые ребра равны 2 , найдите косинус угла между плоскостями
SAF и
SBC

В правильной шестиугольной пирамиде $S A B C D E F$ , стороны основания которой равны $1$ , а боковые ребра равны $2$ , найдите косинус угла между плоскостями $S A F$ и $S B C$

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Full solution

Q. В правильной шестиугольной пирамиде

S A B C D E F

, стороны основания которой равны

1

, а боковые ребра равны

2

, найдите косинус угла между плоскостями

S A F

и

S B C

Find Pyramid Height: Let's find the height of the pyramid using the Pythagorean theorem for triangle $SMO$ , where $M$ is the center of the hexagon and $O$ is the foot of the height from $S$ . The distance $MO$ is the radius of the circumscribed circle around the base, which is $\sqrt{3}/2$ * side of the hexagon. So, $MO = (\sqrt{3}/2) * 1 = \sqrt{3}/2$ . Using $SO$ as the height $h$ , we have $SM = \sqrt{3}/2$ and $M$ $0$ , and the hypotenuse is the lateral edge, which is $M$ $1$ . The Pythagorean theorem gives us $M$ $2$ .
Calculate Height: Now we calculate the height $h$ . $h^2 + \frac{3}{4} = 4$ . So, $h^2 = 4 - \frac{3}{4} = \frac{16}{4} - \frac{3}{4} = \frac{13}{4}$ . Therefore, $h = \sqrt{\frac{13}{4}} = \frac{\sqrt{13}}{2}$ .
Find Triangle Areas: Next, we need to find the area of triangle $SAF$ . Since it's a right triangle at $A$ , the area is $(1/2) \times \text{base} \times \text{height} = (1/2) \times 1 \times \sqrt{13}/2 = \sqrt{13}/4$ .
Find Cosine of Angle: We do the same for triangle $SBC$ . Since it's also a right triangle at $B$ , the area is $(1/2) \times \text{base} \times \text{height} = (1/2) \times 1 \times \sqrt{13}/2 = \sqrt{13}/4$ .
Calculate Normal Vectors: Now, let's find the cosine of the angle between the planes using the dot product of their normal vectors. The normal vector for plane SAF can be found using cross product of vectors SA and SF. Let's assume $SA = (0, 0, \sqrt{13}/2)$ and $SF = (1, 0, \sqrt{13}/2)$ . The cross product gives us a normal vector $n_1$ .
Calculate Normal Vectors: Now, let's find the cosine of the angle between the planes using the dot product of their normal vectors. The normal vector for plane SAF can be found using cross product of vectors SA and SF. Let's assume $SA = (0, 0, \sqrt{13}/2)$ and $SF = (1, 0, \sqrt{13}/2)$ . The cross product gives us a normal vector $n_1$ .Calculating the cross product, $n_1 = SA \times SF = \left| \begin{array}{ccc} i & j & k \ 0 & 0 & \sqrt{13}/2 \ 1 & 0 & \sqrt{13}/2 \end{array} \right| = (0i - (\sqrt{13}/2)j) - ((0 - 0)k) = (0, -\sqrt{13}/2, 0)$ .
Calculate Normal Vectors: Now, let's find the cosine of the angle between the planes using the dot product of their normal vectors. The normal vector for plane SAF can be found using cross product of vectors SA and SF. Let's assume $SA = (0, 0, \sqrt{13}/2)$ and $SF = (1, 0, \sqrt{13}/2)$ . The cross product gives us a normal vector $n_1$ .Calculating the cross product, $n_1 = SA \times SF = \left| \begin{array}{ccc} i & j & k \ 0 & 0 & \sqrt{13}/2 \ 1 & 0 & \sqrt{13}/2 \end{array} \right| = (0i - (\sqrt{13}/2)j) - ((0 - 0)k) = (0, -\sqrt{13}/2, 0)$ .We find the normal vector for plane SBC in a similar way. Assume $SB = (0, 0, \sqrt{13}/2)$ and $SC = (\cos(60), \sin(60), \sqrt{13}/2)$ . The cross product gives us a normal vector $n_2$ .
Calculate Normal Vectors: Now, let's find the cosine of the angle between the planes using the dot product of their normal vectors. The normal vector for plane SAF can be found using cross product of vectors SA and SF. Let's assume $SA = (0, 0, \sqrt{13}/2)$ and $SF = (1, 0, \sqrt{13}/2)$ . The cross product gives us a normal vector $n_1$ .Calculating the cross product, $n_1 = SA \times SF = \left| \begin{array}{ccc} i & j & k \ 0 & 0 & \sqrt{13}/2 \ 1 & 0 & \sqrt{13}/2 \end{array} \right| = (0i - (\sqrt{13}/2)j) - ((0 - 0)k) = (0, -\sqrt{13}/2, 0)$ .We find the normal vector for plane SBC in a similar way. Assume $SB = (0, 0, \sqrt{13}/2)$ and $SC = (\cos(60), \sin(60), \sqrt{13}/2)$ . The cross product gives us a normal vector $n_2$ .Calculating the cross product, $n_2 = SB \times SC = \left| \begin{array}{ccc} i & j & k \ 0 & 0 & \sqrt{13}/2 \ 1/2 & \sqrt{3}/2 & \sqrt{13}/2 \end{array} \right| = (0i - (\sqrt{13}/2)j) - ((0 - 0)k) = (0, -\sqrt{13}/2, 0)$ .

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