3D Geometry
Equations of Lines in Space
Grade 12
Question:
<p>The equation of the reflected ray is the line joining <i>Q</i>(6, 5, -2) and <i>B</i>(-10, -15, -14). Express this in standard form.</p>
<p>(a) $$\frac{x}{16} = \frac{y}{20} = \frac{z}{12}$$</p>
<p>(b) $$\frac{x + 10}{4} = \frac{y + 15}{5} = \frac{z + 14}{3}$$</p>
<p>(c) $$\frac{x + 10}{16} = \frac{y + 15}{20} = \frac{z + 14}{12}$$</p>
<p>(d) $$\frac{x - 6}{4} = \frac{y - 5}{5} = \frac{z + 2}{3}$$</p>
Step-by-Step Solution
Key Concept: Find the direction vector between two points and use one point to write the equation of the line in symmetric form.
Solution: The reflected ray passes through Q (6, 5, -2) and B (-10, -15, -14). Direction vector: QB = (-10 - 6, -15 - 5, -14 + 2) = (-16, -20, -12) Simplifying by dividing by -4: (4, 5, 3) Using point B (-10, -15, -14): $\frac{x + 10}{4} = \frac{y + 15}{5} = \frac{z + 14}{3}$
Correct Answer: B