Question

The line integral of the vector field \( \vec{F} = x\hat{i} - 2y\hat{j} + z\hat{k} \) along the straight line path from the point \((-1,2,3)\) to \((2,3,5)\), is ..........

• A. \( -\frac{1}{2} \)
• B. \( \frac{9}{2} \)
• C. \( \frac{13}{2} \)
• D. \( -\frac{7}{2} \)

Hint:

To evaluate a line integral over a straight path, parametrize the path linearly using a parameter \( t \), compute \( \vec{F}(t) \cdot \frac{d\vec{r}}{dt} \), and integrate over \( t \in [0,1] \).

Answer 1

The Correct Option is B

Given vector field: \[ \vec{F} = x\hat{i} - 2y\hat{j} + z\hat{k} \] Let the path be parametrized from point \( A(-1,2,3) \) to point \( B(2,3,5) \).
Let position vector \( \vec{r}(t) = (1 - t)(-1,2,3) + t(2,3,5),\ t \in [0,1] \) So, \[ \vec{r}(t) = (-1 + 3t,\ 2 + t,\ 3 + 2t) \] Then, \[ \frac{d\vec{r}}{dt} = \langle 3,\ 1,\ 2 \rangle \] Now, express \( \vec{F} \) in terms of \( t \): \[ \vec{F}(t) = (x(t),\ -2y(t),\ z(t)) = (-1 + 3t,\ -2(2 + t),\ 3 + 2t) \] Now compute the dot product: \[ \vec{F}(t) \cdot \frac{d\vec{r}}{dt} = ( -1 + 3t )(3) + ( -2(2 + t) )(1) + (3 + 2t)(2) \] \[ = 3(-1 + 3t) + (-4 - 2t) + 2(3 + 2t) \] \[ = -3 + 9t - 4 - 2t + 6 + 4t = (9t - 2t + 4t) + (-3 - 4 + 6) \] \[ = 11t -1 \] Now integrate over \( t \in [0,1] \): \[ \int_0^1 (11t - 1)\ dt = \left[ \frac{11}{2}t^2 - t \right]_0^1 = \frac{11}{2} - 1 = \frac{9}{2} \]