Orthogonal Projection Of Vectors U And V

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Let's dive into finding the orthogonal projection of vector u⃗\vec{u} onto vector v⃗\vec{v}. This is a common problem in linear algebra and vector calculus, and understanding the process is super useful for various applications in physics, engineering, and computer graphics. We'll break it down step by step, making sure it's crystal clear.

Understanding Orthogonal Projection

Before we jump into the calculations, let's get a solid grasp of what orthogonal projection actually means. Imagine you have two vectors, u⃗\vec{u} and v⃗\vec{v}. The orthogonal projection of u⃗\vec{u} onto v⃗\vec{v}, often denoted as projv⃗u⃗proj_{\vec{v}} \vec{u}, is essentially the "shadow" of u⃗\vec{u} that falls onto v⃗\vec{v} when you shine a light perpendicularly onto v⃗\vec{v}. This shadow is also a vector, and it lies along the same line as v⃗\vec{v}.

Why is this important, guys? Well, orthogonal projection helps us decompose a vector into components that are parallel and perpendicular to another vector. This is incredibly handy when you need to analyze forces, velocities, or any vector quantity in a specific direction. Think about resolving forces acting on an object on an inclined plane – that's orthogonal projection in action!

The formula for calculating the orthogonal projection of u⃗\vec{u} onto v⃗\vec{v} is given by:

projvβƒ—uβƒ—=uβƒ—β‹…vβƒ—βˆ₯vβƒ—βˆ₯2vβƒ—proj_{\vec{v}} \vec{u} = \frac{\vec{u} \cdot \vec{v}}{\|\vec{v}\|^2} \vec{v}

Where:

  • uβƒ—β‹…vβƒ—\vec{u} \cdot \vec{v} is the dot product of vectors uβƒ—\vec{u} and vβƒ—\vec{v}.
  • βˆ₯vβƒ—βˆ₯2\|\vec{v}\|^2 is the squared magnitude (or squared length) of vector vβƒ—\vec{v}.

Now that we have the formula, let's apply it to the given vectors.

Given Vectors

We are given two vectors:

uβƒ—=(0,2,2)\vec{u} = (0, 2, 2) and vβƒ—=(βˆ’2,0,2)\vec{v} = (-2, 0, 2)

Our mission is to find projv⃗u⃗proj_{\vec{v}} \vec{u}.

Step-by-Step Calculation

1. Calculate the Dot Product u⃗⋅v⃗\vec{u} \cdot \vec{v}

The dot product of two vectors u⃗=(u1,u2,u3)\vec{u} = (u_1, u_2, u_3) and v⃗=(v1,v2,v3)\vec{v} = (v_1, v_2, v_3) is calculated as:

u⃗⋅v⃗=u1v1+u2v2+u3v3\vec{u} \cdot \vec{v} = u_1v_1 + u_2v_2 + u_3v_3

Plugging in the values for our vectors, we get:

uβƒ—β‹…vβƒ—=(0)(βˆ’2)+(2)(0)+(2)(2)=0+0+4=4\vec{u} \cdot \vec{v} = (0)(-2) + (2)(0) + (2)(2) = 0 + 0 + 4 = 4

So, u⃗⋅v⃗=4\vec{u} \cdot \vec{v} = 4.

2. Calculate the Squared Magnitude of vβƒ—\vec{v}, βˆ₯vβƒ—βˆ₯2\|\vec{v}\|^2

The magnitude of a vector v⃗=(v1,v2,v3)\vec{v} = (v_1, v_2, v_3) is given by:

βˆ₯vβƒ—βˆ₯=v12+v22+v32\|\vec{v}\| = \sqrt{v_1^2 + v_2^2 + v_3^2}

Therefore, the squared magnitude is:

βˆ₯vβƒ—βˆ₯2=v12+v22+v32\|\vec{v}\|^2 = v_1^2 + v_2^2 + v_3^2

For our vector vβƒ—=(βˆ’2,0,2)\vec{v} = (-2, 0, 2), we have:

βˆ₯vβƒ—βˆ₯2=(βˆ’2)2+(0)2+(2)2=4+0+4=8\|\vec{v}\|^2 = (-2)^2 + (0)^2 + (2)^2 = 4 + 0 + 4 = 8

So, βˆ₯vβƒ—βˆ₯2=8\|\vec{v}\|^2 = 8.

3. Calculate the Orthogonal Projection projv⃗u⃗proj_{\vec{v}} \vec{u}

Now we have all the pieces we need. We can plug the values we calculated into the formula:

projvβƒ—uβƒ—=uβƒ—β‹…vβƒ—βˆ₯vβƒ—βˆ₯2vβƒ—=48vβƒ—=12vβƒ—proj_{\vec{v}} \vec{u} = \frac{\vec{u} \cdot \vec{v}}{\|\vec{v}\|^2} \vec{v} = \frac{4}{8} \vec{v} = \frac{1}{2} \vec{v}

Since vβƒ—=(βˆ’2,0,2)\vec{v} = (-2, 0, 2), we multiply each component by 12\frac{1}{2}:

projvβƒ—uβƒ—=12(βˆ’2,0,2)=(βˆ’1,0,1)proj_{\vec{v}} \vec{u} = \frac{1}{2}(-2, 0, 2) = (-1, 0, 1)

4. Express the Result in Terms of Unit Vectors

The vector (βˆ’1,0,1)(-1, 0, 1) can be expressed in terms of the unit vectors i^\hat{i}, j^\hat{j}, and k^\hat{k} as:

(βˆ’1,0,1)=βˆ’1i^+0j^+1k^=βˆ’i^+k^(-1, 0, 1) = -1\hat{i} + 0\hat{j} + 1\hat{k} = -\hat{i} + \hat{k}

Final Answer

Therefore, the orthogonal projection of vector u⃗\vec{u} onto vector v⃗\vec{v} is:

projvβƒ—uβƒ—=βˆ’i^+k^proj_{\vec{v}} \vec{u} = -\hat{i} + \hat{k}

So the correct answer is A. βˆ’i^+k^-\hat{i} + \hat{k}.

Key Takeaways

  • Orthogonal Projection: Understand the concept of projecting one vector onto another.
  • Dot Product: Master the dot product calculation.
  • Magnitude: Know how to calculate the magnitude (or length) of a vector.
  • Formula Application: Be comfortable applying the formula for orthogonal projection.

Additional Tips

  • Visualize: Try to visualize the vectors and their projection. This can help you understand the concept better.
  • Practice: Practice with different vectors to solidify your understanding.
  • Check Your Work: Always double-check your calculations to avoid errors.

By following these steps and practicing regularly, you'll become a pro at finding orthogonal projections of vectors. Keep up the great work, guys!

Common Mistakes to Avoid

When calculating orthogonal projections, there are a few common mistakes that students often make. Being aware of these pitfalls can save you from unnecessary errors. Let's take a look at some of them:

  1. Incorrect Dot Product Calculation: The dot product is a fundamental part of the projection formula. A mistake in calculating the dot product will lead to an incorrect projection. Always double-check your multiplication and addition when computing the dot product.

    Example: If u⃗=(1,2,3)\vec{u} = (1, 2, 3) and v⃗=(4,5,6)\vec{v} = (4, 5, 6), the dot product is (1)(4)+(2)(5)+(3)(6)=4+10+18=32(1)(4) + (2)(5) + (3)(6) = 4 + 10 + 18 = 32. Make sure you don't mix up the components or miscalculate the sum.

  2. Forgetting to Square the Magnitude: The formula involves dividing by the squared magnitude of the vector onto which you are projecting. Many students forget to square the magnitude, which leads to an incorrect result. Always remember to square each component, sum them up, and that's your squared magnitude.

    Example: If vβƒ—=(2,βˆ’1,3)\vec{v} = (2, -1, 3), the squared magnitude is (2)2+(βˆ’1)2+(3)2=4+1+9=14(2)^2 + (-1)^2 + (3)^2 = 4 + 1 + 9 = 14. Don't forget to square each term!

  3. Applying the Formula Backwards: It's crucial to understand which vector is being projected onto which. Applying the formula backwards will give you the projection of v⃗\vec{v} onto u⃗\vec{u} instead of u⃗\vec{u} onto v⃗\vec{v}. Pay close attention to the problem statement and ensure you're using the correct vectors in the formula.

    Example: Make sure you are calculating projv⃗u⃗proj_{\vec{v}} \vec{u} and not proju⃗v⃗proj_{\vec{u}} \vec{v} if the problem asks for the projection of u⃗\vec{u} onto v⃗\vec{v}.

  4. Incorrectly Multiplying by the Vector: After calculating the scalar part of the projection (i.e., uβƒ—β‹…vβƒ—βˆ₯vβƒ—βˆ₯2\frac{\vec{u} \cdot \vec{v}}{\|\vec{v}\|^2}), you need to multiply this scalar by the vector vβƒ—\vec{v}. Ensure you multiply each component of vβƒ—\vec{v} by the scalar correctly.

    Example: If the scalar is 12\frac{1}{2} and vβƒ—=(βˆ’2,0,2)\vec{v} = (-2, 0, 2), then the multiplication should be 12Γ—(βˆ’2,0,2)=(βˆ’1,0,1)\frac{1}{2} \times (-2, 0, 2) = (-1, 0, 1).

  5. Not Simplifying the Result: Sometimes, the problem requires you to express the result in a specific form, such as in terms of unit vectors. Failing to simplify or express the result correctly can lead to losing marks, even if the numerical calculation is correct.

    Example: If you get the result (βˆ’1,0,1)(-1, 0, 1), make sure to express it as βˆ’i^+k^-\hat{i} + \hat{k} if the options are given in that format.

  6. Misunderstanding the Question: Always read the question carefully. Sometimes the question might ask for the component of uβƒ—\vec{u} orthogonal to vβƒ—\vec{v}, which is uβƒ—βˆ’projvβƒ—uβƒ—\vec{u} - proj_{\vec{v}} \vec{u}, rather than just the projection itself.

By keeping these common mistakes in mind and double-checking your work, you can significantly improve your accuracy when solving orthogonal projection problems. Happy calculating, guys!