3D Geometry
Vector Equation of Line
Grade 12
Question:
<p>The vector equation of the line passing through the point \((1, 2, -4)\) and perpendicular to the two lines \(\frac{x - 8}{3} = \frac{y + 19}{-16} = \frac{z - 10}{7}\) and \(\frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{-5}\) is</p>
<p>(a) \(\vec{r} = (\vec{i} + 2\vec{j} - 4\vec{k}) + \lambda(2\vec{i} + 3\vec{j} + 6\vec{k})\)</p>
<p>(b) \(\vec{r} = (2\vec{i} + 3\vec{j} - 6\vec{k}) + \lambda(\vec{i} + 2\vec{j} - 4\vec{k})\)</p>
<p>(c) \(\vec{r} = (\vec{i} + 2\vec{j} - 4\vec{k}) + \lambda(3\vec{i} + 8\vec{j} - 5\vec{k})\)</p>
<p>(d) \(\vec{r} = (\vec{i} + 2\vec{j} - 4\vec{k}) + \lambda(3\vec{i} - 16\vec{j} - 7\vec{k})\)</p>
Step-by-Step Solution
Key Concept: The direction vector of the required line is the cross product of the direction vectors of the two given lines.
The direction vector perpendicular to both given lines is found by taking the cross product of their direction vectors.
Correct Answer: A