find the projection of v onto u and find the angle between v and u.

find the projection of v onto u and find the angle between v and u.

The Correct Answer and Explanation is:

To find the projection of a vector v\mathbf{v}v onto another vector u\mathbf{u}u, we use the formula:projuv=v⋅uu⋅uu\text{proj}_{\mathbf{u}} \mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{u}}{\mathbf{u} \cdot \mathbf{u}} \mathbf{u}proju​v=u⋅uv⋅u​u

Here’s how to break it down:

1. Dot product v⋅u\mathbf{v} \cdot \mathbf{u}v⋅u: This is the sum of the products of corresponding components of vectors v\mathbf{v}v and u\mathbf{u}u. If v=⟨v1,v2⟩\mathbf{v} = \langle v_1, v_2 \ranglev=⟨v1​,v2​⟩ and u=⟨u1,u2⟩\mathbf{u} = \langle u_1, u_2 \rangleu=⟨u1​,u2​⟩, the dot product is: v⋅u=v1u1+v2u2\mathbf{v} \cdot \mathbf{u} = v_1 u_1 + v_2 u_2v⋅u=v1​u1​+v2​u2​
2. Magnitude squared of u\mathbf{u}u: This is the sum of the squares of the components of u\mathbf{u}u: u⋅u=u12+u22\mathbf{u} \cdot \mathbf{u} = u_1^2 + u_2^2u⋅u=u12​+u22​
3. Projection formula: Using the values obtained from the above steps, we can find the projection of v\mathbf{v}v onto u\mathbf{u}u.

---

Example:

Let’s say v=⟨3,4⟩\mathbf{v} = \langle 3, 4 \ranglev=⟨3,4⟩ and u=⟨1,2⟩\mathbf{u} = \langle 1, 2 \rangleu=⟨1,2⟩.

1. Dot product v⋅u\mathbf{v} \cdot \mathbf{u}v⋅u:

v⋅u=3×1+4×2=3+8=11\mathbf{v} \cdot \mathbf{u} = 3 \times 1 + 4 \times 2 = 3 + 8 = 11v⋅u=3×1+4×2=3+8=11

2. Magnitude squared of u\mathbf{u}u:

u⋅u=12+22=1+4=5\mathbf{u} \cdot \mathbf{u} = 1^2 + 2^2 = 1 + 4 = 5u⋅u=12+22=1+4=5

3. Projection of v\mathbf{v}v onto u\mathbf{u}u:

projuv=115⟨1,2⟩=⟨115,225⟩\text{proj}_{\mathbf{u}} \mathbf{v} = \frac{11}{5} \langle 1, 2 \rangle = \langle \frac{11}{5}, \frac{22}{5} \rangleproju​v=511​⟨1,2⟩=⟨511​,522​⟩

So, the projection of v\mathbf{v}v onto u\mathbf{u}u is ⟨115,225⟩\langle \frac{11}{5}, \frac{22}{5} \rangle⟨511​,522​⟩.

---

Angle Between v\mathbf{v}v and u\mathbf{u}u:

The formula for the angle θ\thetaθ between two vectors v\mathbf{v}v and u\mathbf{u}u is:cos⁡θ=v⋅u∣v∣∣u∣\cos \theta = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}| |\mathbf{u}|}cosθ=∣v∣∣u∣v⋅u​

Where:

• ∣v∣|\mathbf{v}|∣v∣ is the magnitude of v\mathbf{v}v, calculated as v12+v22\sqrt{v_1^2 + v_2^2}v12​+v22​​
• ∣u∣|\mathbf{u}|∣u∣ is the magnitude of u\mathbf{u}u, calculated as u12+u22\sqrt{u_1^2 + u_2^2}u12​+u22​​

Using the values from the example:

1. Magnitude of v\mathbf{v}v:

∣v∣=32+42=9+16=25=5|\mathbf{v}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5∣v∣=32+42​=9+16​=25​=5

2. Magnitude of u\mathbf{u}u:

∣u∣=12+22=1+4=5|\mathbf{u}| = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}∣u∣=12+22​=1+4​=5​

3. Cosine of the angle:

cos⁡θ=115×5=115×2.236≈1111.18≈0.983\cos \theta = \frac{11}{5 \times \sqrt{5}} = \frac{11}{5 \times 2.236} \approx \frac{11}{11.18} \approx 0.983cosθ=5×5​11​=5×2.23611​≈11.1811​≈0.983

4. Angle θ\thetaθ:

θ=cos⁡−1(0.983)≈10.3∘\theta = \cos^{-1}(0.983) \approx 10.3^\circθ=cos−1(0.983)≈10.3∘

Thus, the angle between v\mathbf{v}v and u\mathbf{u}u is approximately 10.3∘10.3^\circ10.3∘.