a) Write the vector form of the line.
Vector form of a line
The vector form of a line is given by:
\[
\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v}
\]
where \(\mathbf{r}_0\) is a point on the line, \(\mathbf{v}\) is the direction vector, and \(t\) is a parameter.
Substitute given values
Given the point \((2, -1, 5)\) and the direction vector \((-3, 4, -2)\), the vector form of the line is:
\[
\mathbf{r}(t) = (2, -1, 5) + t(-3, 4, -2)
\]
The vector form of the line is \(\boxed{\mathbf{r}(t) = (2, -1, 5) + t(-3, 4, -2)}\).
b) Find the parametric equations for \(x, y\), and \(z\).
Parametric equations from vector form
From the vector form \(\mathbf{r}(t) = (2, -1, 5) + t(-3, 4, -2)\), the parametric equations are:
\[
x = 2 - 3t, \quad y = -1 + 4t, \quad z = 5 - 2t
\]
The parametric equations are:
\[
\boxed{x = 2 - 3t}, \quad \boxed{y = -1 + 4t}, \quad \boxed{z = 5 - 2t}
\]
c) Does the point \((-1, 7, 3)\) lie on this line? Show your reasoning.
Check if the point satisfies the parametric equations
Substitute \((-1, 7, 3)\) into the parametric equations:
\[
-1 = 2 - 3t, \quad 7 = -1 + 4t, \quad 3 = 5 - 2t
\]
Solve for \(t\) in each equation
For \(x\):
\[
-1 = 2 - 3t \implies -3 = -3t \implies t = 1
\]
For \(y\):
\[
7 = -1 + 4t \implies 8 = 4t \implies t = 2
\]
For \(z\):
\[
3 = 5 - 2t \implies -2 = -2t \implies t = 1
\]
Analyze the results
The value of \(t\) is inconsistent across the equations (\(t = 1\) for \(x\) and \(z\), but \(t = 2\) for \(y\)). Therefore, the point \((-1, 7, 3)\) does not lie on the line.
The point \((-1, 7, 3)\) does not lie on the line. \(\boxed{\text{No}}\)
a) The vector form of the line is \(\boxed{\mathbf{r}(t) = (2, -1, 5) + t(-3, 4, -2)}\).
b) The parametric equations are:
\[
\boxed{x = 2 - 3t}, \quad \boxed{y = -1 + 4t}, \quad \boxed{z = 5 - 2t}
\]
c) The point \((-1, 7, 3)\) does not lie on the line. \(\boxed{\text{No}}\)
Answered
