Косинус угла между прямыми

Noticing the setup

I’m working with a standard cube orientation: base square ABCD and top square A1B1C1D1. Point K is on BC, and I need to find the cosine of the angle between AK and B1D. координаты.

Mapping the vectors

I’m thinking through the vectors AK and B1D by choosing origin A at (0,0,0) and marking point K on BC as (1, t, 0).

Figuring out the angle

I’m thinking about the 15° angle at vertex A, formed by lines AB and AK, which translates to the angle between vectors A->B and A->K being 15°.

Calculating angle cosine

I’m working through the cosine of the angle BAK by using the dot product and magnitudes of vectors AB and AK. This simplifies to $\frac{1}{\sqrt{1 + t^2}}$.

Isolating t

I’m pulling together the steps to isolate $t$ by transforming $\cos(15^\circ) = \frac{\sqrt{6} + \sqrt{2}}{4}$ into $\sqrt{1 + t^2} = \sqrt{6} - \sqrt{2}$.

Simplifying the expression

I’m isolating $t$ by rewriting $1 + t^2$ and solving for $t$. Expressing $t$ as a nested radical form helps to clarify the structure.

Confirming the value

I confirmed $t = 2 - \sqrt{3}$ by checking if $(2 - \sqrt{3})^2$ equals 7 - 4√3. This gives the coordinates of point K and direction vector AK.

Calculating direction

I’m piecing together the direction vector $\vec{B1D}$ by subtracting the coordinates of B1 from D, resulting in $\vec{B1D} = (-1,1,-1)$.

Piecing together

I’m working through the cosine between vectors $\vec{AK}$ and $\vec{B1D}$ by calculating their dot product and magnitudes. This helps in finding the cosine of the angle $\alpha$.

Breaking down the expression

I’m simplifying the cosine expression by separating and reducing the components, aiming for a more straightforward form. This involves breaking the fraction into parts and addressing each one.

Rationalizing and simplifying

OK, let me see. I’m rationalizing the denominator $\sqrt{6} - \sqrt{2}$ and simplifying the expression step by step. This involves multiplying and reducing the components to a more concise form.

Breaking down the product

I’m combining and simplifying fractions with square roots, canceling terms, and rationalizing denominators. This results in $\cos(\alpha) = -\frac{\sqrt{2}}{2\sqrt{3}}$.

Verifying the simplification

I’m confirming that $\cos(\alpha) = -\frac{\sqrt{6}}{6}$ is equivalent to $-\frac{1}{\sqrt{6}}$ by cross-multiplying and checking the consistency.

Verifying the equivalence

I’m piecing together the verification of the cosine expression, ensuring $\frac{\sqrt{6}}{6}$ matches $\frac{1}{\sqrt{6}}$. This confirms the angle is obtuse.

Determining the range

I’m figuring out that the angle between the lines is between 90° and 180°, which makes the cosine negative. The smaller angle is typically the one used in these problems.

Understanding angle restrictions

I’m gathering info on how the angle between lines in vector geometry is strictly between 0 and 180 degrees, with the cosine remaining negative for obtuse angles.

Accepting negative cosine

I’m concluding that a negative cosine is valid since the angle is between 90° and 180°, and the problem does not specify the use of the acute angle.

Thought about угол между прямыми for 48 seconds

Решение в координатах

1. Разместим куб в декартовой системе координат так, чтобы:

   • $A = (0,0,0)$,
   • $B = (1,0,0)$,
   • $C = (1,1,0)$,
   • $D = (0,1,0)$,
   • $A_1 = (0,0,1)$,
   • $B_1 = (1,0,1)$,
   • $C_1 = (1,1,1)$,
   • $D_1 = (0,1,1)$.

   Тогда ребро $BC$ лежит на отрезке между точками

   $$B=(1,0,0)\quad \text{и} \quad C=(1,1,0).$$

   Любая точка $K$ на $BC$ имеет вид

   $$K = (1,t,0),\quad t\in [0,1].$$

2. Условие $\angle BAK = 15^\circ$.

   • Вектор $\overrightarrow{AB} = B - A = (1,0,0)$.
   • Вектор $\overrightarrow{AK} = K - A = (1,t,0)$.

   Косинус угла между векторами $\overrightarrow{AB}$ и $\overrightarrow{AK}$ даёт:

   $$\cos(\angle BAK) = \frac{\overrightarrow{AB}\cdot \overrightarrow{AK}} {\|\overrightarrow{AB}\|\;\|\overrightarrow{AK}\|}.$$

   Подставим:

   $$\overrightarrow{AB}\cdot \overrightarrow{AK} = (1,0,0)\cdot(1,t,0) = 1,$$ $$\|\overrightarrow{AB}\| = \sqrt{1^2+0^2+0^2} = 1, \quad \|\overrightarrow{AK}\| = \sqrt{1^2 + t^2 + 0^2} = \sqrt{1 + t^2}.$$

   Следовательно,

   $$\cos(\angle BAK) = \frac{1}{1 \cdot \sqrt{1 + t^2}} = \frac{1}{\sqrt{1 + t^2}}.$$

   По условию $\angle BAK = 15^\circ$, значит

   $$\cos(15^\circ) = \frac{1}{\sqrt{1 + t^2}}.$$

   Известно, что $\cos(15^\circ) = \tfrac{\sqrt{6}+\sqrt{2}}{4}$. Отсюда

   $$\sqrt{1 + t^2} = \frac{1}{\cos(15^\circ)} = \frac{4}{\sqrt{6} + \sqrt{2}}.$$

   Умножая числитель и знаменатель на $\sqrt{6}-\sqrt{2}$, получаем

   $$\sqrt{1 + t^2} = \frac{4(\sqrt{6}-\sqrt{2})}{(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})} = \frac{4(\sqrt{6}-\sqrt{2})}{6 - 2} = \sqrt{6} - \sqrt{2}.$$

   Тогда

   $$1 + t^2 = (\sqrt{6} - \sqrt{2})^2 = 6 + 2 - 2\sqrt{12} = 8 - 4\sqrt{3},$$ $$t^2 = 7 - 4\sqrt{3}.$$

   Заметим, что $7 - 4\sqrt{3} = (2 - \sqrt{3})^2$. Значит,

   $$t = 2 - \sqrt{3} \quad (\text{берём положительное }t\in [0,1]).$$

   Итак,

   $$K = (1,\;2-\sqrt{3},\;0).$$

3. Вектор, задающий направление прямой $B_1D$.

   • $B_1 = (1,0,1)$,
   • $D = (0,1,0)$.

   Возьмём, например, $\overrightarrow{B_1D} = D - B_1$:

   $$\overrightarrow{B_1D} = (0,1,0) - (1,0,1) = (-1,1,-1).$$

4. Ищем угол между прямыми $AK$ и $B_1D$.

   Направляющий вектор прямой $AK$ возьмём $\overrightarrow{AK} = (1,\;2-\sqrt{3},\;0)$.
   Направляющий вектор прямой $B_1D$ — $\overrightarrow{B_1D} = (-1,1,-1)$.

   Пусть $\theta$ — угол между этими двумя векторами. Тогда

   $$\cos(\theta) = \frac{\overrightarrow{AK}\cdot \overrightarrow{B_1D}} {\|\overrightarrow{AK}\|\;\|\overrightarrow{B_1D}\|}.$$

   1. Скалярное произведение:

      $$\overrightarrow{AK} \cdot \overrightarrow{B_1D} = (1,\;2-\sqrt{3},\;0)\cdot(-1,1,-1) = 1\cdot(-1) \;+\; (2-\sqrt{3})\cdot 1 \;+\; 0\cdot(-1) = -1 + (2 - \sqrt{3}) = 1 - \sqrt{3}.$$

   2. Длины векторов:

      $$\|\overrightarrow{AK}\| = \sqrt{1^2 + (2-\sqrt{3})^2 + 0^2} = \sqrt{1 + (2-\sqrt{3})^2} = \sqrt{1 + (7 - 4\sqrt{3})} = \sqrt{8 - 4\sqrt{3}} = \sqrt{6} \;-\; \sqrt{2},$$ $$\|\overrightarrow{B_1D}\| = \sqrt{(-1)^2 + 1^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3}.$$

   3. Косинус искомого угла:

      $$\cos(\theta) = \frac{1 - \sqrt{3}} {\bigl(\sqrt{6} - \sqrt{2}\bigr)\,\sqrt{3}}.$$

      Упростим. Сначала заметим:

      $$\frac{1 - \sqrt{3}}{\sqrt{3}} = \frac{1}{\sqrt{3}} - 1.$$

      А также

      $$\frac{1}{\sqrt{6} - \sqrt{2}} = \frac{\sqrt{6} + \sqrt{2}}{(\sqrt{6}-\sqrt{2})(\sqrt{6}+\sqrt{2})} = \frac{\sqrt{6} + \sqrt{2}}{6 - 2} = \frac{\sqrt{6} + \sqrt{2}}{4}.$$

      Перемножая, получаем (или просто аккуратно раскрывая скобки в числителе):

      $$\cos(\theta) = \frac{(1-\sqrt{3})(\sqrt{6}+\sqrt{2})}{4\,\sqrt{3}} = \frac{-2\sqrt{2}}{4\sqrt{3}} = -\frac{\sqrt{2}}{2\sqrt{3}} = -\frac{\sqrt{6}}{6} = -\frac{1}{\sqrt{6}}.$$

Таким образом,

Это означает, что угол между прямыми $AK$ и $B_1D$ является тупым (ведь косинус отрицательный), и его косинус равен $-\tfrac{\sqrt{6}}{6}$.