Question

what is the area of a parallelogram formed by ( v_1 = langle 2, 6
angle ) and ( v_2 = langle 3, 1
angle )?
(\bigcirc) 2 square units
(\bigcirc) 6 square units
(\bigcirc) 16 square units
(\bigcirc) 20 square units

Explanation:

Step1: Recall the formula for the area of a parallelogram formed by two vectors \(\vec{v_1} = \langle a, b

angle\) and \(\vec{v_2} = \langle c, d
angle\). The area is the absolute value of the determinant of the matrix formed by these vectors as columns (or rows), which is \(|ad - bc|\).
Given \(\vec{v_1} = \langle 2, 6
angle\) and \(\vec{v_2} = \langle 3, 1
angle\), here \(a = 2\), \(b = 6\), \(c = 3\), \(d = 1\).

Step2: Calculate the determinant \(ad - bc\).

Substitute the values: \(ad - bc=(2\times1)-(6\times3)\)
\(= 2 - 18\)
\(= - 16\)

Step3: Take the absolute value of the determinant to get the area.

The area is \(|-16| = 16\) square units.

Answer:

16 square units (corresponding to the option "16 square units")