Question

where $m_1$ and $m_2$ are the slopes of $l_1$ and $l_2$, respectively. (assume that $m_1m_2\
eq - 1$.)

1. $l_1:3x - 2y = 5$

$l_2:x + y = 1$

1. $l_1:2x - y = 8$

$l_2:x - 5y = - 4$

1. geometry determine the angle between the diagonal of a cube and the diagonal of its base, as shown in the figure.
2. geometry determine the angle between the diagonal of a cube and its edge, as shown in the figure.

Explanation:

Step1: Label the cube's edge length

Let the edge length of the cube be \(a\).

Step2: Find the length of the base - diagonal

Using the Pythagorean theorem in the base (a square with side - length \(a\)), the length of the base - diagonal \(d_{base}=\sqrt{a^{2}+a^{2}}=\sqrt{2a^{2}}=\sqrt{2}a\).

Step3: Find the length of the space - diagonal

Using the Pythagorean theorem in the right - triangle formed by the base - diagonal, the edge of the cube, and the space - diagonal. The length of the space - diagonal \(d_{space}=\sqrt{(\sqrt{2}a)^{2}+a^{2}}=\sqrt{2a^{2}+a^{2}}=\sqrt{3a^{2}}=\sqrt{3}a\).

Step4: Use the cosine formula

We know that \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\). In the right - triangle formed by the base - diagonal, the space - diagonal, and the edge of the cube (for the angle between the space - diagonal and the base - diagonal), the adjacent side to the angle \(\theta\) is the length of the base - diagonal (\(\sqrt{2}a\)) and the hypotenuse is the length of the space - diagonal (\(\sqrt{3}a\)). So, \(\cos\theta=\frac{\sqrt{2}a}{\sqrt{3}a}=\frac{\sqrt{2}}{\sqrt{3}}\).

Step5: Calculate the angle

\(\theta=\cos^{- 1}(\frac{\sqrt{2}}{\sqrt{3}})\approx35.3^{\circ}\)

Answer:

The angle between the diagonal of a cube and the diagonal of its base is approximately \(35.3^{\circ}\)