The dot product of two vectors $$\langle a, b \rangle$$ and $$\langle c, d \rangle$$ is the scalar quantity $$ac + bd$$. Therefore, $$\vec{v} \cdot \vec{w} = (-3)(2) + (6)(4) = -6 + 24 = 18$$.

The angle $$\theta$$ between two vectors is found using the formula $$\cos(\theta) = \frac{\vec{a} \cdot \vec{b}}{\Vert\vec{a}\Vert \Vert\vec{b}\Vert}$$. The dot product is $$\vec{a} \cdot \vec{b} = (2)(0) + (2)(3) = 6$$. The magnitudes are $$\Vert\vec{a}\Vert = \sqrt{2^2+2^2} = \sqrt{8} = 2\sqrt{2}$$ and $$\Vert\vec{b}\Vert = \sqrt{0^2+3^2} = 3$$. So, $$\cos(\theta) = \frac{6}{(2\sqrt{2})(3)} = \frac{6}{6\sqrt{2}} = \frac{1}{\sqrt{2}}$$. The angle whose cosine is $$\frac{1}{\sqrt{2}}$$ is $$\theta = \frac{\pi}{4}$$ radians.

First, find the component form of the vector: $$\vec{w} = \langle -4 - 1, 3 - 9 \rangle = \langle -5, -6 \rangle$$. A vector $$\langle a, b \rangle$$ can be written as the linear combination $$a\vec{i} + b\vec{j}$$. Therefore, $$\vec{w} = -5\vec{i} - 6\vec{j}$$.

Two vectors are perpendicular if and only if their dot product is zero. The dot product is $$\vec{u} \cdot \vec{v} = (k)(3) + (-4)(6) = 3k - 24$$. Setting the dot product to zero gives the equation $$3k - 24 = 0$$. Solving for $$k$$, we get $$3k = 24$$, so $$k = 8$$.

This question tests AP Precalculus skills in vectors, specifically equilibrium conditions and vector addition. Vectors are quantities having both magnitude and direction, essential in physics for analyzing forces in static equilibrium. In this scenario, three forces act on a crate at rest, requiring the tension vector T to balance the weight and push forces so their sum equals zero. Choice A is correct because for equilibrium, W + P + T = 0, so T = -(W + P) = -(-50j + 20i) = -20i + 50j N, which exactly balances the other forces. Choice C is incorrect because it only considers magnitudes without proper vector addition, failing to account for the vertical component needed to balance the weight. To help students: Emphasize that equilibrium means the vector sum equals zero, not just balancing magnitudes. Draw free body diagrams showing all forces and practice setting up equilibrium equations component by component.

This question tests AP Precalculus skills in vectors, specifically vector addition and magnitude calculation using the Pythagorean theorem. Vectors combine effects acting simultaneously, such as a drone's air velocity and wind velocity producing a ground velocity. In this scenario, the ground velocity g = d + w = (6-2)i + (1+3)j = 4i + 4j represents the combined effect of drone motion and wind. Choice B is correct because the magnitude of g = 4i + 4j is calculated as √(4² + 4²) = √(16 + 16) = √32, using the formula |v| = √(x² + y²) for a 2D vector. Choice C is incorrect because it appears to calculate √(6² + 4²) = √52, possibly using incorrect components or confusing which vector's magnitude to find. To help students: Emphasize performing vector addition first, then calculating magnitude. Practice identifying when to add vectors versus when to find magnitudes, as these operations occur in different orders.

This question tests AP Precalculus skills in vectors, specifically scalar multiplication and its effect on magnitude and direction. Scalar multiplication of a vector changes its magnitude by the absolute value of the scalar and reverses direction if the scalar is negative. In this scenario, multiplying force vector $\vec{F}=\langle 3,-4,0\rangle$ by scalar $k=2$ gives $2\vec{F}=\langle 6,-8,0\rangle$. Choice B is correct because scalar multiplication by 2 doubles the magnitude while keeping the same direction (since 2 is positive). The original magnitude is $|\vec{F}|=\sqrt{9+16}=5$ kN, and after multiplication, $|2\vec{F}|=\sqrt{36+64}=10$ kN, confirming the doubling. Choice A is incorrect because it describes multiplication by -0.5, not 2. To help students: Emphasize that positive scalars preserve direction while changing magnitude proportionally. Practice calculating magnitudes before and after scalar multiplication, and watch for confusion between the effects of positive versus negative scalars.

This question tests AP Precalculus skills in vectors, specifically three-dimensional vector addition. Vectors are quantities having both magnitude and direction, extended to 3D space for robotics and spatial applications. In this scenario, a robot arm's displacement is corrected by adding another vector, requiring component-wise addition in three dimensions. Choice A is correct because vector addition works component-wise in 3D: (2i - 1j + 3k) + (-4i + 5j - 1k) = (2-4)i + (-1+5)j + (3-1)k = -2i + 4j + 2k cm. Choice B is incorrect because it appears to have sign errors in the i-component, possibly forgetting the negative sign on -4i. To help students: Extend 2D vector addition rules to 3D by treating each component independently. Use organized layouts showing i, j, and k components separately to avoid errors.

This question tests AP Precalculus skills in vectors, specifically vector subtraction in structural analysis. Vector subtraction $\vec{T}_2-\vec{T}_1$ finds the difference between two vectors, showing how one vector must change to become the other. In this scenario, $\vec{T}_2-\vec{T}_1 = \langle 3,4,0\rangle - \langle 8,0,0\rangle = \langle 3-8, 4-0, 0-0\rangle = \langle -5,4,0\rangle$. Choice B is correct because it properly subtracts each component: x-component is 3-8=-5, y-component is 4-0=4, z-component is 0-0=0. Choice A is incorrect because it appears to add the vectors instead of subtracting them. To help students: Remember that vector subtraction means subtracting corresponding components, and be careful with the order (which vector comes first matters). Practice rewriting $\vec{a}-\vec{b}$ as $\vec{a}+(-\vec{b})$ to reinforce the concept.

This question tests AP Precalculus skills in vectors, specifically vector addition in projectile motion with gravity effects. Vectors can represent velocities with magnitude (speed) and direction, and velocity changes are found by vector addition. In this scenario, the new velocity after 1 second is $\vec{v}_0+\Delta\vec{v} = \langle 12,16,0\rangle + \langle 0,-9.8,0\rangle = \langle 12+0, 16+(-9.8), 0+0\rangle = \langle 12,6.2,0\rangle$. Choice A is correct because it properly adds the initial velocity and the change due to gravity: the x-component remains 12 (no horizontal acceleration), the y-component becomes 16-9.8=6.2 (gravity reduces upward velocity), and z remains 0. Choice B is incorrect because it appears to add 9.8 instead of subtracting it, ignoring that gravity acts downward. To help students: Emphasize that gravity contributes negative y-velocity in standard coordinates. Practice problems involving multiple time steps, and watch for sign errors with gravity.