Find a vector equation and parametric equations for the line. (Use the parameter $t$). The line through the point $(5, 2, 3)$ and parallel to the vector $4 \mathbf{i}- \mathbf{j} + 5 \mathbf{k}$.

Find

• $r(t) = \boxed \,$
• $(x(t), y(t), z(t)) = \boxed \,$

Neetesh Kumar | December 15, 2024

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This is the solution to Math 1C
Assignment: 12.5 Question Number 3
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Step-by-step solution:

Given that the line passes through the point $(5, 2, 3)$ and is parallel to the vector $4 \mathbf{i} - \mathbf{j} + 5 \mathbf{k}$, we can find the vector and parametric equations of the line.

Step 1: Vector Equation:

The vector equation of a line is:

$\mathbf{r}(t) = \mathbf{r}_0 + t \mathbf{v}$

Where:

• $\mathbf{r}_0$ is the position vector of a point on the line, which is $(5, 2, 3)$,
• $\mathbf{v}$ is the direction vector, which is $4 \mathbf{i} - \mathbf{j} + 5 \mathbf{k}$,
• $t$ is the parameter.

Thus, the vector equation becomes:

$\mathbf{r}(t) = \langle 5, 2, 3 \rangle + t \langle 4 , -1, 5 \rangle$

Simplifying:

$\mathbf{r}(t) = \langle 5 + 4t \mathbf{i}, 2 - t \mathbf{j}, 3 + 5t \mathbf{k} \rangle$

So, the vector equation is:

$\mathbf{r}(t) = \boxed{\langle 5 + 4t \mathbf{i}, 2 - t \mathbf{j}, 3 + 5t \mathbf{k} \rangle}$

Step 2: Parametric Equations:

The parametric equations are obtained by separating the components of the vector equation:

• $x(t) = 5 + 4t$
• $y(t) = 2 - t$
• $z(t) = 3 + 5t$

So, the parametric equations are:

$(x(t), y(t), z(t)) = \boxed{5 + 4t, 2 - t, 3 + 5t}$

Final Answer:

$\mathbf{r}(t) = \boxed{\langle 5 + 4t \mathbf{i}, 2 - t \mathbf{j}, 3 + 5t \mathbf{k} \rangle}$

$(x(t), y(t), z(t)) = \boxed{5 + 4t, 2 - t, 3 + 5t}$

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