Find the projection of v onto u

Find the projection of v onto u.

The Correct Answer and Explanation is :

The projection of vector v onto vector u is a vector that represents the component of v in the direction of u. This projection is calculated using the formula:

[ \text{proj}_{\mathbf{u}} \mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{u}}{\mathbf{u} \cdot \mathbf{u}} \mathbf{u} ]

Where:

• v · u is the dot product of v and u.
• u · u is the dot product of u with itself, which equals the square of the magnitude of u.

Step-by-Step Calculation:

1. **Compute the Dot Product of *v* and u:**
   [ \mathbf{v} \cdot \mathbf{u} = v_1 \times u_1 + v_2 \times u_2 + v_3 \times u_3 ]
   This operation multiplies corresponding components of v and u and sums the results.
2. **Compute the Dot Product of *u* with Itself:**
   [ \mathbf{u} \cdot \mathbf{u} = u_1^2 + u_2^2 + u_3^2 ]
   This gives the square of the magnitude of u.
3. Calculate the Scalar Projection:
   [ \text{Scalar Projection} = \frac{\mathbf{v} \cdot \mathbf{u}}{\mathbf{u} \cdot \mathbf{u}} ]
   This scalar represents the magnitude of the projection of v onto u.
4. Find the Vector Projection:
   [ \text{proj}_{\mathbf{u}} \mathbf{v} = \text{Scalar Projection} \times \mathbf{u} ]
   This step scales u by the scalar projection to obtain the vector projection.

Geometric Interpretation:

The vector projection of v onto u represents the component of v that points in the same direction as u. Geometrically, it is the shadow or footprint of v when a light source is placed directly above u. This projection is particularly useful in physics and engineering for resolving forces, velocities, or other vector quantities into components along specific directions.

Example:

Consider vectors v = (3, 4, 0) and u = (1, 2, 0).

1. **Dot Product of *v* and u:**
   [ \mathbf{v} \cdot \mathbf{u} = (3 \times 1) + (4 \times 2) + (0 \times 0) = 3 + 8 = 11 ]
2. **Dot Product of *u* with Itself:**
   [ \mathbf{u} \cdot \mathbf{u} = (1^2) + (2^2) + (0^2) = 1 + 4 = 5 ]
3. Scalar Projection:
   [ \text{Scalar Projection} = \frac{11}{5} = 2.2 ]
4. Vector Projection:
   [ \text{proj}_{\mathbf{u}} \mathbf{v} = 2.2 \times (1, 2, 0) = (2.2, 4.4, 0) ]

Thus, the projection of v onto u is the vector (2.2, 4.4, 0).

For a visual explanation and further clarification, you might find the following video helpful:

videoHow to Find the Projection of u Onto v and the Vector Component of u Orthogonal to vturn0search4