find the area of parallelogram whose adjacent sides are +3+3 and -3-2+.
To find the area of a parallelogram, you need the lengths of its adjacent sides. In this case, we have two adjacent sides as vectors: +3+3 and -3-2+.
To find the area, we can calculate the magnitude (length) of the cross product of these vectors. The cross product of two vectors, A and B, is given by:
|A x B| = |A| |B| sin(theta),
where |A x B| is the magnitude of the cross product, |A| and |B| are the magnitudes of vectors A and B, and theta is the angle between them.
First, let's calculate the magnitudes of the vectors +3+3 and -3-2+:
|+3+3| = sqrt((3)^2 + (3)^2) = sqrt(18) = 3sqrt(2),
|-3-2+| = sqrt((-3)^2 + (-2)^2 + 0^2) = sqrt(13).
Next, we need to calculate the angle between the two vectors. Let's call this angle theta.
The dot product of two vectors, A and B, is given by:
A · B = |A| |B| cos(theta).
Since the dot product is also equal to the sum of the component-wise products, we have:
(+3+3) · (-3-2+) = 3sqrt(2) * sqrt(13) * cos(theta).
Expanding this equation:
(-3)(3) + (-2)(3) + (0)(1) = 3sqrt(2) * sqrt(13) * cos(theta),
-9 - 6 = 3sqrt(2) * sqrt(13) * cos(theta),
-15 = 3sqrt(26) * cos(theta).
Dividing both sides by 3sqrt(26):
-15 / (3sqrt(26)) = cos(theta).
Simplifying the left side:
-5 / sqrt(26) = cos(theta).
Finally, we can use the inverse cosine function to find the value of theta:
theta = arccos(-5 / sqrt(26)).
Now we have all the necessary values to calculate the area:
Area = |+3+3 x -3-2+| = |+3+3| |(-3-2+)| sin(theta)
= (3sqrt(2)) * (sqrt(13)) * sin(theta)
= (3sqrt(2)) * (sqrt(13)) * sin(arccos(-5 / sqrt(26))).
Using a calculator, you can find the value of sin(arccos(-5 / sqrt(26))) and simplify to get the area of the parallelogram.