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Given v=-i+6j and w=2i-3j, find: (i)Magnitude of each vector v and w. (ii)The dot product v*w (iii )The angle between the two vectors v and w. Find the angle in degrees and round to one decimal place.

User Sfuqua
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(i) The magnitude of a vector is calculated using the formula ||v|| = sqrt(v1^2 + v2^2). Using this formula for vector v which equals -i + 6j, we square each of the components and sum them, then take the square root of the result. In this case, this results to roughly 6.08, which is the magnitude of vector v.
To find the magnitude of vector w which equals 2i - 3j, we use the same approach. Squaring the components of the vector, summing them and taking the square root yields a value of approximately 3.61.

(ii) The dot product of two vectors is calculated using the formula v*w = v1*w1 + v2*w2. Applying this formula, we have (-1 x 2) + (6 x -3) = -2 - 18 = -20. Therefore, the dot product of v and w is -20.

(iii)The angle between two vectors is calculated using the formula angle = acos((v*w) / (||v||*||w||)). Given the dot product of the two vectors is -20 and their magnitudes are roughly 6.08 and 3.61, this formula gives us an angle in radians. In order to get the angle in degrees, which is often more readable, we convert this radian value into degrees by using the conversion formula (180/pi)*angle in radians.

After converting the radian value to degrees, we found the angle to be approximately 155.8°. We then rounded this value to one decimal place for simplicity, giving the final answer of 155.8°.

So, in summary, the magnitude of vector v is approximately 6.08, the magnitude of vector w is approximately 3.61, the dot product of the vectors is -20, and the angle between them when rounded to one decimal place is 155.8°.

User Giles Smith
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