Question #d7dc2

1 Answer
A08
Jan 23, 2017

We know with reference to the given figure that by Law of parallelogram of vectors

|vecR|=sqrt(|vecP|^2+|vecQ|^2+2|vecP|cdot|vecQ|cdot cos theta) .....(1)
and alpha=tan^-1((|vecQ|sin theta)/(|vecP|+|vecQ|cos theta)) .....(2)

Inserting given values in equation (1) we get

|vecR|=sqrt(|5|^2+|-3|^2+2|5|cdot|-3|cdot cos 60)
=>|vecR|=sqrt(23+9+2xx5xx3xx 0.5)
=>|vecR|=sqrt47
=>|vecR|=6.86, rounded to two decimal places

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