show that |a x b|^2 + |a.b|^2 = |a|^2 |b|^2. the "x" is supposed to mean cross product and the "." is dot product
|axb| = |a| |b| sinθ
a.b = |a| |b| cosθ
now square and add them up
thanks!!
To show that |a x b|^2 + |a . b|^2 = |a|^2 |b|^2, where "x" represents the cross product and "." represents the dot product, we can use the properties of vector algebra and the definitions of the cross product and dot product.
First, let's start with the left side of the equation:
|a x b|^2 + |a . b|^2
The cross product of two vectors, a and b, can be defined as:
|a x b| = |a| |b| sin(theta)
where |a| and |b| are the magnitudes of vectors a and b, and theta is the angle between the two vectors.
The dot product of two vectors, a and b, can be defined as:
|a . b| = |a| |b| cos(theta)
where |a| and |b| are the magnitudes of vectors a and b, and theta is the angle between the two vectors.
Now, let's substitute these values into the left side of the equation:
|a x b|^2 + |a . b|^2
= (|a| |b| sin(theta))^2 + (|a| |b| cos(theta))^2
Expanding and simplifying this expression, we get:
= |a|^2 |b|^2 sin^2(theta) + |a|^2 |b|^2 cos^2(theta)
Using the trigonometric identity sin^2(theta) + cos^2(theta) = 1, we can rewrite the above expression as:
= |a|^2 |b|^2 (sin^2(theta) + cos^2(theta))
Now, sin^2(theta) + cos^2(theta) equals 1, so we have:
= |a|^2 |b|^2 (1)
Since anything multiplied by 1 is equal to itself, we can simplify the expression further:
= |a|^2 |b|^2
This is the same as the right side of the equation.
Therefore, we have shown that |a x b|^2 + |a . b|^2 = |a|^2 |b|^2.