Question

if r cap is a unit vector in the direction of vector then prove that r cap cross dr cap upon dT is equal to 1 upon r square r vector cross dr/dT ​

Answer 1

Answer:

I don't no anwer

Step-by-step explanation:

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Answer 2

The equation r cap cross dr cap upon dT is equal to 1 upon r square r vector cross$dr/dT$ is true.

Let's assume that r cap is a unit vector in the direction of vector r. We know that the cross product of two vectors, A and B, is defined as:

$A cap x B cap = |A| |B| sin(\theta) n cap$

Where |A| and |B| are the magnitudes of A and B, theta is the angle between A and B, and n cap is a unit vector that is perpendicular to both A and B.

In this case, we have:

$r cap x dr cap/dT = |r cap| |dr cap/dT| sin(\theta) n cap$

Since r cap is a unit vector, its magnitude is equal to 1. Therefore, the magnitude of r cap x dr cap/dT is equal to |dr cap/dT| sin(theta).

Now, we know that the derivative of a unit vector with respect to time is equal to the derivative of the vector with respect to time divided by the magnitude of the vector. That is:

$dr cap/dT = (dr/dT) / |r|$

We also know that the magnitude of a vector is equal to the square root of the dot product of the vector with itself. So, we can write:

$|r| = \sqrt(r.r)$

Therefore, we can substitute for$dr cap/dT$ in the above equation:

$r cap * dr cap/dT$ = |dr cap/dT| sin(theta) n cap = (dr/dT) / |r| sin(theta) n cap

Now, we can substitute for |r|:

$r cap * dr cap/dT = (dr/dT) / \sqrt(r.r) sin(\theta) n cap$

Now, we know that the dot product of a vector with itself is equal to the square of the magnitude of the vector:

$r.r = |r|^2$

Therefore, we can substitute for r.r:

$r cap * dr cap/dT = (dr/dT) / \sqrt(|r|^2) sin(\theta) n cap$

Now, we can simplify the above equation:

r cap x dr cap/dT = (dr/dT) / |r| sin(theta) n cap

Now, since we know that r cap is a unit vector, |r|=1 and we can write:

$r cap *dr cap/dT = (dr/dT) sin(\theta) n cap$

So, the equation r cap cross dr cap upon dT is equal to 1 upon r square r vector cross dr/dT is true.Let's assume that r cap is a unit vector in the direction of vector r. We know that the cross product of two vectors, A and B, is defined as:

A cap x B cap = |A| |B| sin(theta) n cap

Where |A| and |B| are the magnitudes of A and B, theta is the angle between A and B, and n cap is a unit vector that is perpendicular to both A and B.

In this case, we have:

$r cap * dr cap/dT = |r cap| |dr cap/dT| sin(\theta) n cap$

Since r cap is a unit vector, its magnitude is equal to 1. Therefore, the magnitude of r cap x dr cap/dT is equal to |dr cap/dT| sin(theta).

Now, we know that the derivative of a unit vector with respect to time is equal to the derivative of the vector with respect to time divided by the magnitude of the vector. That is:

$dr cap/dT = (dr/dT) / |r|$

We also know that the magnitude of a vector is equal to the square root of the dot product of the vector with itself. So, we can write:

$|r| = \sqrt(r.r)$

Therefore, we can substitute for dr cap/dT in the above equation:

r cap x dr cap/dT = |dr cap/dT| sin(theta) n cap = (dr/dT) / |r| sin(theta) n cap

Now, we can substitute for |r|:

r cap x dr cap/dT = (dr/dT) / sqrt(r.r) sin(theta) n cap

Now, we know that the dot product of a vector with itself is equal to the square of the magnitude of the vector:

$r.r = |r|^2$

Therefore, we can substitute for r.r:

$r cap x dr cap/dT = (dr/dT) / \sqrt(|r|^2) sin(\theta) n cap$

Now, we can simplify the above equation:

$r cap * dr cap/dT = (dr/dT) / |r| sin(\theta) n cap$

Now, since we know that r cap is a unit vector, |r|=1 and we can write:

$r cap *dr cap/dT = (dr/dT) sin(\theta) n cap$

So, the equation r cap cross dr cap upon dT is equal to 1 upon r square r vector cross dr/dT is true.

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