The vector equation for the line passing through the points (-1, -7, -4) and (-8, 7, 2) is given by \(\mathbf{r}(t) = \langle -7t - 1, 14t - 7, 6t - 4 \rangle\), where \(t\) is the parameter representing points on the line.
To find the vector equation for the line through the points \((-1, -7, -4)\) and \((-8, 7, 2)\), we can use the point-slope form of the equation for a line in three-dimensional space:
\[ \mathbf{r}(t) = \mathbf{v}t + \mathbf{r}_0 \]
where:
- \(\mathbf{r}(t)\) is the position vector for any point on the line,
- \(\mathbf{v}\) is the direction vector of the line, and
- \(\mathbf{r}_0\) is the position vector of a point on the line.
The direction vector \(\mathbf{v}\) can be found by subtracting the initial point \((-1, -7, -4)\) from the terminal point \((-8, 7, 2)\):
\[ \mathbf{v} = \langle -8 - (-1), 7 - (-7), 2 - (-4) \rangle = \langle -7, 14, 6 \rangle \]
Now, choose one of the points, say \((-1, -7, -4)\), as \(\mathbf{r}_0\). The vector equation becomes:
\[ \mathbf{r}(t) = \langle -7, 14, 6 \rangle t + \langle -1, -7, -4 \rangle \]
Therefore, the vector equation for the line through the points \((-1, -7, -4)\) and \((-8, 7, 2)\) is:
\[ \mathbf{r}(t) = \langle -7t - 1, 14t - 7, 6t - 4 \rangle \]
The probable question may be:
Find a vector equation with parameter t for the line through the points (-1, -7,-4) and (-8,7,2). x(t) = Remember to use WebWork vector notation and the parameter t in your answer.