To solve the problem, let's first define the vectors clearly:
- Boat's velocity vector: \( \mathbf{v} = \langle 4, 3 \rangle \)
- Water current vector: \( \mathbf{w} = \langle -2, 1 \rangle \)
Step 1: Finding the Dot Product
The dot product of two vectors \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \) is given by: \[ \mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 \]
For vectors \( \mathbf{v} \) and \( \mathbf{w} \): \[ \mathbf{v} \cdot \mathbf{w} = (4)(-2) + (3)(1) = -8 + 3 = -5 \]
Step 2: Determining the Angle Between the Vectors
To find the angle \( \theta \) between the two vectors, we use the formula involving the dot product and magnitudes: \[ \cos(\theta) = \frac{\mathbf{v} \cdot \mathbf{w}}{|\mathbf{v}| |\mathbf{w}|} \]
First, we calculate the magnitudes of \( \mathbf{v} \) and \( \mathbf{w} \): \[ |\mathbf{v}| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \] \[ |\mathbf{w}| = \sqrt{(-2)^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} \]
Now we can substitute back into the cosine formula: \[ \cos(\theta) = \frac{-5}{5 \cdot \sqrt{5}} = \frac{-1}{\sqrt{5}} \]
To find \( \theta \), we take the arccos: \[ \theta = \cos^{-1}\left(-\frac{1}{\sqrt{5}}\right) \]
Calculating the angle: \[ \theta \approx 116.57^\circ \]
Step 3: Interpretation of the Angle
The angle of approximately \( 116.57^\circ \) indicates that the boat's path and the water current are not aligned. Since the angle is greater than \( 90^\circ \), it suggests that the water current is pushing against the direction of the boat's path. This means the current would hinder the boat's intended trajectory, causing it to drift off-course if no adjustments are made.
Step 4: Adjustment to Counteract the Current
To counteract the effect of the water current while still traveling in the direction of \( \mathbf{v} \), we can calculate the effective velocity needed to compensate for the current \( \mathbf{w} \). We want the resultant vector \( \mathbf{r} = \mathbf{v} + \mathbf{u} \), where \( \mathbf{u} \) is the adjustment vector.
We can find this adjustment by solving: \[ \mathbf{r} = \langle 4 + u_1, 3 + u_2 \rangle \] to ensure that the current is counteracted. This should give an effective path that takes into account \( \mathbf{w} \) as follows: \[ \mathbf{r} = \mathbf{v} - \mathbf{w} = \langle 4 - (-2), 3 - 1 \rangle = \langle 4 + 2, 3 - 1 \rangle = \langle 6, 2 \rangle \]
Thus, the adjustment needed is: \[ \mathbf{u} = \langle 2, -1 \rangle \]
By applying this adjustment, the boat can successfully navigate in the desired direction while counteracting the current's effect.
Summary of Results:
- Dot Product: \( -5 \)
- Angle between vectors: \( \approx 116.57^\circ \)
- Adjustment needed: \( \mathbf{u} = \langle 2, -1 \rangle \) to counteract current.