Find the vector equation of a line passing through origin and is parallel to a vector formed by
joining point A(2, 4, 6) and B(-3, 2, 5).

Answer

To find the vector equation of a line passing through the origin and parallel to the vector formed by joining points A(2, 4, 6) and B(-3, 2, 5), we can follow these steps:

### Step 1: Find the direction vector AB
To find the direction vector from point A to point B, we use the formula:
\[ \vec{AB} = \vec{B} - \vec{A} \]
Where:
- \(\vec{A} = (2, 4, 6)\)
- \(\vec{B} = (-3, 2, 5)\)

Calculating the components:
\[ \vec{AB} = (-3 - 2, 2 - 4, 5 - 6) = (-5, -2, -1) \]

### Step 2: Write the vector equation of the line
The vector equation of a line can be expressed in the form:
\[ \vec{r} = \vec{a} + \lambda \vec{d} \]
Where:
- \(\vec{a}\) is a point on the line (in this case, the origin, which is \((0, 0, 0)\))
- \(\vec{d}\) is the direction vector (which we found to be \((-5, -2, -1)\))
- \(\lambda\) is a scalar parameter

Since the line passes through the origin, we have:
\[ \vec{a} = (0, 0, 0) \]
Thus, the equation becomes:
\[ \vec{r} = (0, 0, 0) + \lambda (-5, -2, -1) \]
This simplifies to:
\[ \vec{r} = \lambda (-5, -2, -1) \]

### Step 3: Finalize the vector equation
We can express this in component form:
\[ \vec{r} = -5\lambda \hat{i} - 2\lambda \hat{j} - \lambda \hat{k} \]

### Conclusion
The vector equation of the line passing through the origin and parallel to the vector formed by joining points A and B is:
\[ \vec{r} = \lambda (-5, -2, -1) \]