Question

Find \((\mathbf{a}) \mathbf{u} \cdot \mathbf{v},(\mathbf{b}) \mathbf{u} \cdot \mathbf{u},(\mathbf{c})\|\mathbf{u}\|^{2},(\mathbf{d})(\mathbf{u} \cdot \mathbf{v}) \mathbf{v}\) and \((e) u \cdot(2 v)\). $$ \mathbf{u}=\langle-4,8\rangle, \quad \mathbf{v}=\langle 6,3\rangle $$

Short answer

So: (a) \(\mathbf{u} \cdot \mathbf{v}\) is 0. (b) \(\mathbf{u} \cdot \mathbf{u}\) is 80. (c)\(\|\mathbf{u}\|^{2}\) is 80. (d)(\(\mathbf{u} \cdot \mathbf{v}) \mathbf{v}=< 0,0>\). (e) \( \mathbf{u} \cdot (2*\mathbf{v})\) is 0.

Step-by-step solution

Calculation of Dot Product

To calculate the dot product between two vectors we multiply corresponding entries and then sum all those products. Let's do this for vectors \(\mathbf{u}\) and \(\mathbf{v}\), for parts (a) and (d). For \(\mathbf{u} \cdot \mathbf{v}\), this results in (-4*6)+(8*3)=-24+24=0. For (d), we need it times vector \(\mathbf{v}\), so it will be 0*\(\mathbf{v}=\langle 0,0\rangle\).

Self Dot Product

For part (b), we calculate \(\mathbf{u} \cdot \mathbf{u}\), also known as self-dot product, in the same way as before: (-4*(-4))+(8*8)=16+64=80.

Norm Squared

For part (c), we calculate \(\|\mathbf{u}\|^{2}\) (the square of the norm of vector \(\mathbf{u}\)). It's the same as the self-dot product, which we have already calculated, so it's just 80.

Dot Product with Scalar Vector

For part (e), first, we multiply vector \(\mathbf{v}\) by 2, \(\mathbf{v}' = 2*\mathbf{v} = \langle 12,6\rangle\), then calculate the dot product with \(\mathbf{u}\) following the same procedure: \(\mathbf{u} \cdot 2*\mathbf{v}=\mathbf{u} \cdot \langle 12,6\rangle =(-4*12)+(8*6)=-48+48=0.

Key concepts

Dot Product Calculation

Understanding the dot product is crucial for anyone studying vectors. Essentially, the dot product is a mathematical operation that takes two equal-length sequences of numbers (typically coordinate vectors) and returns a single number. This operation reveals the extent to which two vectors are pointing in the same direction.

For two-dimensional vectors \(\mathbf{u}\) and \(\mathbf{v}\), given by \(\mathbf{u}=\langle u_1, u_2 \rangle\) and \(\mathbf{v}=\langle v_1, v_2 \rangle\), their dot product \(\mathbf{u} \cdot \mathbf{v}\) is calculated by multiplying corresponding components and summing these products: \(\mathbf{u} \cdot \mathbf{v} = u_1 \times v_1 + u_2 \times v_2\).

In our exercise, we applied this formula to vectors \(\mathbf{u}\) and \(\mathbf{v}\), resulting in a dot product of zero, indicating that they are orthogonal, or at a right angle to each other. This computation is the foundation for many concepts in vector analysis and physics, including the projection of one vector onto another and determining if vectors are perpendicular.

Norm of Vector

The norm of a vector, often denoted as \(\|\mathbf{u}\|\), is a measure of the vector's length or magnitude. In a physical context, you might consider it equivalent to the intensity or strength of the vector quantity.

To find the norm of a vector, you can calculate the square root of the dot product of the vector with itself. Mathematically, for a vector \(\mathbf{u}\), its norm is \(\|\mathbf{u}\| = \sqrt{\mathbf{u} \cdot \mathbf{u}}\). However, for our exercise's purpose in part (c), we're interested in the square of the norm, \(\|\mathbf{u}\|^2\), which simplifies to \(\mathbf{u} \cdot \mathbf{u}\), eliminating the need for taking the square root.

For vector \(\mathbf{u}=\langle-4,8\rangle\), the square of its norm was found to be 80, which is the sum of the squares of its components. This value is crucial in many applications, like normalizing a vector, which means resizing it to a unit vector—having a norm (length) of one—without changing its direction.

Scalar Multiplication in Vectors

Scalar multiplication involves multiplying a vector by a scalar (a single number), altering the magnitude of the vector, but not its direction. When you multiply a vector by a positive scalar, it stretches or shrinks the vector; a negative scalar also flips its direction.

Mathematically, for a scalar \(k\) and a vector \(\mathbf{v}=\langle v_1, v_2 \rangle\), the product is \(k\mathbf{v}=\langle k \cdot v_1, k \cdot v_2 \rangle\). In the context of our exercise, we multiplied vector \(\mathbf{v}\) by 2, effectively doubling its length, to get a new vector \(\langle 12,6 \rangle\). The subsequent dot product of this scalar multiplied vector with vector \(\mathbf{u}\) again resulted in zero, emphasizing that the dot product is unaffected by scalar multiplication when vectors are orthogonal.

This concept is not just a mathematical exercise; it's an essential operation in physics (to scale forces, for example), computer graphics (to resize or reflect images), and economics (to adjust prices in cost functions).