Find a vector of magnitude 5 units, and parallel to the resultant of vectors  \(\vec{a}\) = 2\(\hat{i}\) +3\(\hat{j}\) -  \(\hat{k}\) and \(\vec{b}\) = \(\hat{i}\) -2\(\hat{j}\) + \(\hat{k}\)

±5. {(3 \(\hat{i}\) + \(\hat{j}\))/ √(10)

We have  \(\vec{a}\) = 2\(\hat{i}\) +3\(\hat{j}\) -  \(\hat{k}\) and \(\vec{b}\) = \(\hat{i}\) -2\(\hat{j}\) + \(\hat{k}\)

let \(\vec{c}\)  be the resultant of \(\vec{a}\)  and \(\vec{b}\)

Then, 

\(\vec{c}\) = \(\vec{a}\)  + \(\vec{b}\) =(2\(\hat{i}\) +3\(\hat{j}\)-  \(\hat{k}\)) + ( \(\hat{i}\) -2\(\hat{j}\)  + \(\hat{k}\)) = 3 \(\hat{i}\) +\(\hat{j}\)  

|\(\vec{c}\)| = √(3)² +(1)²

|\(\vec{c}\)|= √10

so unit vector \(\hat{c}\) = \(\vec{c}\) / |\(\vec{c}\)| = (3 \(\hat{i}\)+\(\hat{j}\) Ì‚)/ √10

Hence vector of magnitude 4 units, and parallel to the resultant of vectors  \(\vec{a}\)  and \(\vec{b}\)  is = ±5. {(3 \(\hat{i}\) + \(\hat{j}\))/ √(10)