Viscosity is the measurement of a "thickness" of a liquid. Molasses, for example, is a more viscous fluid than water is.

Vector measurements are defined by a magnitude and a direction. For a liquid to have a measureable "thickness" is logical, but a liquid cannot have a viscosity in a direction. To say that a fluid has a viscosity of \(\displaystyle 1.0\) East makes no sense. Viscosity is a scalar quantity.

Displacement, force, velocity, and acceleration all have associated directions and are classified as vector quantities.

Vector quantities are defined by both the magnitude of the parameter and the direction of action. In contract, scalar quantities are independent of direction and rely only on the magnitude of the parameter.

Mass, distance, time, and brightness are all scalar quantities. This is to say that none of these terms can be applied in a given direction. It would be illogical to have "three grams west" or "eighteen seconds to the left." Distance is the scalar equivalent of the displacement vector.

Force is always a vector quantity, since the direction of the force matters in defining the parameter. "Four Newtons to the right" is quantifiably different from "four Newtons downward" or "four Newtons to the left."

The magnitude of the vector is found using the distance formula:

\(\displaystyle \sqrt{x^2+y^2}=\sqrt{(-2)^2+(3)^2}=\sqrt{13}=3.61\)

To calculate the angle we must first find the inverse tangent of \(\displaystyle \frac{y}{x}\):

\(\displaystyle \arctan(\frac{y}{x})=\arctan(\frac{3}{-2})=-56.3^o\)

This angle value is the principal arctan, but it is in the fourth quadrant while our vector is in the second. We must add the angle 180° to this value to arrive at our final answer.

\(\displaystyle \theta=180^o+(-56.3^o)=124^o\)

For vector \(\displaystyle 2\vec{A}\), the magnitude is doubled, but the direction remains the same.

For our calculation, we use a magnitude of:

\(\displaystyle m=2\times3.61=7.22\)

The x-coordinate is the magnitude times the cosine of the angle, while the y-coordinate is the magnitude times the sine of the angle.

\(\displaystyle x = m \cos \theta = 7.22 \cos (124^o) = -4.00\)

\(\displaystyle y = m \sin \theta = 7.22 \sin (124^o) = 6.00\)

The resultant vector is: \(\displaystyle -4.00 i + 6.00 j\).

When multiplying a vector by a constant (called scalar multiplication), we multiply each component by the constant.

\(\displaystyle 2\vec{A}=2(-1-2j)=-2i-4j\)

The magnitude of this new vector is found with these new components:

\(\displaystyle \sqrt{x^2+y^2}=\sqrt{(-2)^2+(-4)^2}=\sqrt{4+16}=\sqrt{20}=4.47\)

To calculate the angle we must first find the inverse tangent of \(\displaystyle \frac{y}{x}\):

\(\displaystyle \arctan(\frac{y}{x})=\arctan(\frac{-4}{-2})=\arctan(2)=63.4^o\)

This is the principal arctan, but it is in the first quadrant while our vector is in the third. We to add the angle 180° to this value to arrive at our final answer.

\(\displaystyle \theta=180^o+63.4^o=243^o\)

First, construct the two vectors using ruler and protractor:

Place the tail of \(\displaystyle \vec{B}\) at the nose of \(\displaystyle \vec{A}\):

Construct the resultant \(\displaystyle \vec{R}\) from the tail of \(\displaystyle \vec{A}\) to the nose of \(\displaystyle \vec{B}\):

With our ruler and protractor, we find that \(\displaystyle \vec{R}\) is 4.12 at an angle of 14.0° CCW from the x-axis.

Construct \(\displaystyle \vec{A}\) and \(\displaystyle \vec{B}\) from their x- and y-components:

Since we are subtracting, reverse the direction of \(\displaystyle \vec{B}\):

Form \(\displaystyle \vec{A}-\vec{B}=\vec{A}+(-\vec{B})\) by placing the tail of \(\displaystyle -\vec{B}\) at the nose of \(\displaystyle \vec{A}\):

Construct and measure the resultant, \(\displaystyle \vec{R}\), from the tail of \(\displaystyle \vec{A}\)to the nose of \(\displaystyle -\vec{B}\) using a ruler and protractor:

\(\displaystyle m_{\vec{R}}=5.10,\ \theta_{\vec{R}}=101^o\)

The x-coordinate is the magnitude times the cosine of the angle, while the y-coordinate is the magnitude times the sine of the angle.

\(\displaystyle \text{magnitude}=2.24,\ \theta=63.4^o\)

\(\displaystyle x = 2.24 \cos(63.4^o) = 1.00\)

\(\displaystyle y = 2.24 \sin(63.4^o) = 2.00\)

The resultant vector is: \(\displaystyle \text{1.00i + 2.00j}\).

First, construct the two vectors using ruler and protractor:

\(\displaystyle 2\vec{B}\) is twice the length of \(\displaystyle \vec{B}\), but in the same direction:

Since we are subtracting, reverse the direction of \(\displaystyle 2\vec{B}\):

Form \(\displaystyle \vec{A}-2\vec{B}=\vec{A}+(-2\vec{B})\) by placing the tail of \(\displaystyle -2\vec{B}\) at the nose of \(\displaystyle \vec{A}\):

Construct and measure the resultant \(\displaystyle \vec{R}\) from the tail of \(\displaystyle \vec{A}\) to the nose of \(\displaystyle -2\vec{B}\) with a ruler and protractor.

\(\displaystyle m_{\vec{R}}=6.40,\ \theta_{\vec{R}}=321^o\)

First, construct the two vectors using ruler and protractor:

Place the tails of both vectors at the same point:

Construct a parallelogram:

Construct and measure the resultant using ruler and protractor:

\(\displaystyle m_{\vec{R}}=4.12,\ \theta_{\vec{R}}=14.0^o\)