In Exercises 22–24, sketch each vector as a position vector and find its magnitude.
v = -3j

Verified step by step guidance

1. Identify the vector given: \( \mathbf{v} = -3\mathbf{j} \). This means the vector has no \( \mathbf{i} \) (x-direction) component and a \( -3 \) component in the \( \mathbf{j} \) (y-direction).
2. Express the vector in component form as \( \mathbf{v} = (0, -3) \), where 0 is the x-component and -3 is the y-component.
3. To sketch the vector as a position vector, start at the origin \( (0,0) \) and draw an arrow pointing straight down to the point \( (0, -3) \) on the Cartesian plane.
4. Recall that the magnitude (or length) of a vector \( \mathbf{v} = (x, y) \) is given by the formula \( \| \mathbf{v} \| = \sqrt{x^2 + y^2} \).
5. Calculate the magnitude of \( \mathbf{v} \) by substituting the components: \( \| \mathbf{v} \| = \sqrt{0^2 + (-3)^2} = \sqrt{9} \). This gives the magnitude of the vector.

Key Concepts

Position Vector

A position vector represents the location of a point in space relative to the origin. It is typically expressed in component form, such as v = ai + bj, where i and j are unit vectors along the x and y axes. In this question, the vector v = -3j points 3 units in the negative y-direction.

Vector Components and Unit Vectors

Vectors are broken down into components along coordinate axes using unit vectors i (x-axis) and j (y-axis). The given vector v = -3j has no x-component and a y-component of -3, indicating direction and magnitude along the y-axis only.

Magnitude of a Vector

The magnitude of a vector is its length and is found using the Pythagorean theorem: |v| = √(a² + b²). For v = -3j, the magnitude is |v| = √(0² + (-3)²) = 3, representing the distance from the origin to the point defined by the vector.