Calculating the Magnitude of Vector CD Using the Cosine Rule

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To find the magnitude of vector CD, the cosine rule is applied using the known magnitudes of vectors c and d, and their dot product. The dot product indicates the cosine of the angle between the vectors, allowing for the calculation of the angle indirectly. Instead of determining the angle θ, using the relationship cos(θ) = 4/35 simplifies the process. This method leads to an exact value for the magnitude of vector CD, which is √66. The discussion emphasizes the efficiency of using the cosine rule directly with the dot product for precise calculations.
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C, D are points defined by position vectors c and d. Magnitude of c is 5, mag of d is 7, c dotproduct d is 4 ie c.d = 4, find the magnitude of vector CD.

So i started this way

c.d = magc*magdcos@
= 35cos@, @ = 83.4 degrees

But still no idea how do get magnitude of vector cd. Thank you!
 
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Are you sure it isn't the magnitude of vector CD? (i.e. the vector from point C to point D)? If so, use the cosine rule on the triangle, since you know two vectors and an angle.
 
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oops yes you are right, ok i use the cos rule and do get an answer(which is correct), but I am after it as an exact value (root66). How would i get that? Thanks!
 
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?? You say you got an answer but you want an exact value? Did you use a calculator to get cos(\theta). Since you want to use cos(\theta) in the cosine rule, why not just use cos(\theta)= \frac{4}{35} rather than finding \theta itself?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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