Suppose once again that the minute hand of a clock is a vector of magnitude 4 and the hour hand is a vector of magnitude 2. If, at 5 o'clock, one were to take the cross product of the minute hand the hour hand, what would the magnitude of resultant vector be?

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Solution

The cross product of two vectors is given by:
$\to R = \to A \times \to B = A B sin θ^n$
where $A$ magnitude of vector A,
$B$ magnitude of vector B,
$θ =$ smaller angle between $\to A$ and $\to B$
$^n =$ unit vector in the direction of $\to A \times \to B$ .

Given: $A = 4, B = 2$ , at 5O' clock the angle between minute and hour hand will be , $θ = 150^{o}$ ,
therefore, $R = 4 \times 2 sin 150 = 8 \times 1 / 2 = 4$ ,
direction of $\to R$ will be out of he page by right hand rule.

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