Question

part b
because we cannot represent subtraction of vectors on a plane using vectors u and v, find the additive inverse of v, which is -v. use scalar multiplication to find the vector -1v.

Explanation:

Step1: Recall Scalar Multiplication Rule

For a vector \(\mathbf{v}\) with components (or in geometric terms, direction and magnitude), scalar multiplication by a scalar \(k\) is defined as \(k\mathbf{v}\), where each component of \(\mathbf{v}\) is multiplied by \(k\). For \(k = - 1\), we use the rule \(-1\times\mathbf{v}=-\mathbf{v}\).

Step2: Apply Scalar Multiplication to \(\mathbf{v}\)

Let \(\mathbf{v}=\langle v_1,v_2
angle\) (in component form) or \(\mathbf{v}\) be a vector with initial point \(P\) and terminal point \(Q\). When we multiply \(\mathbf{v}\) by \(- 1\), the magnitude of \(-\mathbf{v}\) is the same as the magnitude of \(\mathbf{v}\), but the direction is opposite. If \(\mathbf{v}\) has components \(v_1\) and \(v_2\), then \(-1\mathbf{v}=\langle - v_1,-v_2
angle\). Geometrically, if \(\mathbf{v}\) goes from point \(A\) to point \(B\), then \(-\mathbf{v}\) goes from point \(B\) to point \(A\).

(Note: Since the specific components or geometric representation of \(\mathbf{v}\) is not given, the general process to find \(-\mathbf{v}\) is as described. If we had the vector \(\mathbf{v}=\langle a,b
angle\), then \(-\mathbf{v}=\langle - a,-b
angle\); if \(\mathbf{v}\) was a vector with magnitude \(m\) and direction \(\theta\), then \(-\mathbf{v}\) has magnitude \(m\) and direction \(\theta + 180^{\circ}\) (or \(\theta+\pi\) radians).)

Answer:

To find \(-\mathbf{v}\) (the additive inverse of \(\mathbf{v}\)) using scalar multiplication, we multiply each component of \(\mathbf{v}\) by \(- 1\). If \(\mathbf{v}=\langle v_1,v_2
angle\) (component - form), then \(-1\mathbf{v}=\langle - v_1,-v_2
angle\). Geometrically, \(-\mathbf{v}\) has the same magnitude as \(\mathbf{v}\) but opposite direction.