Question

If A(1,2,3) B(3,7,-2) and D(-1,0,-1) are points in a plane, then the vector equation of the line passing through the centroids of △ABD and △ACD is

• A. r = (2i - j) + (j + 4k)
• B. r = (1 + t)i +3j + 3tk
• C. r = (2i +3j + 3k) +t(i + 3j)
• D. r = (i + j + k) + t(2i - j)

Answer 1

The Correct Option is B

We are given four points in 3D space: $A = (1, 2, 3)$, $B = (3, 7, -2)$, $C = (6, 7, 7)$, and $D = (-1, 0, -1)$.

1. Find Centroid of Triangle ABD ($G_1$):
The centroid is calculated as the average of the vertices' coordinates:
$G_1 = \frac{A+B+D}{3} = \left(\frac{1+3-1}{3}, \frac{2+7+0}{3}, \frac{3-2-1}{3}\right) = (1, 3, 0)$

2. Find Centroid of Triangle ACD ($G_2$):
Similarly, for triangle ACD:
$G_2 = \frac{A+C+D}{3} = \left(\frac{1+6-1}{3}, \frac{2+7+0}{3}, \frac{3+7-1}{3}\right) = (2, 3, 3)$

3. Determine Direction Vector:
The direction vector from $G_1$ to $G_2$ is:
$G_2 - G_1 = (2-1, 3-3, 3-0) = (1, 0, 3)$

4. Write Vector Equation of Line:
Using point $G_1$ and the direction vector:
$\vec{r} = G_1 + t(G_2 - G_1) = (1, 3, 0) + t(1, 0, 3)$
This gives the parametric form:
$\vec{r} = (1+t)\vec{i} + 3\vec{j} + 3t\vec{k}$

Alternative Parametric Equations:
The line can also be expressed as:
$x = 1 + t$
$y = 3$
$z = 3t$

Final Answer:

The vector equation of the line is ${\vec r = (1+t)\vec i + 3\vec j + 3t\vec k}$