Question

Write a vector in the direction of the vector i-2j+2k that has magnitude 9 units

Answer 1

magnitude of given vector =under root (1^2 +2^2+2^2)
=3
therefore vector that has magnitude 9 = 3(i-2j+2k)

Answer 2

A vector in the direction of the vector i-2j+2k with magnitude 9 is 3i-6j+6k

• Let the given vector be a and required vector be b.
• Given that both vectors are in same direction.
• b = $\alpha$a      
• a = i-2j+2k
• b = α(i-2j+2k)               ------------------------------------(1)
• |b| = $\sqrt{\alpha^{2}(1^{2}+(-2)^{2} + 2^{2}) }$    
• 9 = $\sqrt{\alpha^{2}(1+4+4\\}) }$                                        [since magnitude of vector is 9]
• 9 = $\sqrt{\alpha^{2}(9}) }$  
• 9 = 3$\alpha$
• $\alpha$ = 3
• Substitute $\alpha$ in (1)
• b = 3(i-2j+2k)
• b = 3i-6j+6k