91Ó°ÊÓ

In geometry, parametric equations are used to describe a line, plane, or surface in terms of one or more parameters. Here, the given line equation is \(\frac{x+2}{3}= \frac{y-1}{0}= \frac{z}{4}\). This implies a unique situation where \(y\) is constant (\(y = 1\)), and for the line, we can express \(x\) and \(z\) in terms of a parameter \(t\):

• \(x = 3t - 2\)
• \(y = 1\)
• \(z = 4t\)

This representation helps in finding specific points on the line simply by changing the value of \(t\). For this exercise, points \(B\) and \(C\) are defined using these parametric equations to find their coordinates.
Understanding parametric equations is crucial as they allow us to analyze motion along a path defined via a parameter and simplify complex geometric problems dramatically.

In geometry, the distance formula is fundamental to determine the length between two points in a 3D space. The distance formula is an extension of the Pythagorean Theorem and is represented as:\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \] By applying this formula, we can calculate the length of segment \(\mathrm{BC}\) for the points \(B(3t_1 - 2, 1, 4t_1)\) and \(C(3t_2 - 2, 1, 4t_2)\). Here, the distance \(\mathrm{BC}\) is simplified to equate to 5 units, leading to:

• Distance \(\sqrt{9(t_2 - t_1)^2 + 16(t_2 - t_1)^2} = 5\)
• Simplifying the condition \(t_2 - t_1 = \pm 1\)

This formula is powerful as it not only calculates distances but also informs decisions involving other aspects like triangle properties or spatial relationships.

The cross product is used in vector algebra to find a vector perpendicular to two given vectors in 3D space. It is particularly useful for calculating the area of a triangle formed by three points when vectors are known. For vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) in our example, the cross product \(\overrightarrow{AB} \times \overrightarrow{AC}\) provides a vector whose magnitude is twice the area of triangle \(\mathrm{ABC}\).

• For \(\overrightarrow{AB} = ((3t_1-3),2,(4t_1-2))\)
• For \(\overrightarrow{AC} = ((3t_2 - 3),2,(4t_2-2))\)

The magnitude of this cross product gives the area of the triangle as:\[|\overrightarrow{AB} \times \overrightarrow{AC}| = |17|\]Thus, the area is ensured to be \(2\sqrt{34}\) units squared for option (b).
The utility of the cross product in this context confirms the correctness of the area of the triangle and is a crucial tool in 3D geometry to find perpendicular relationships and areas effectively.