Answer :
Answer:
Option 2.
Step-by-step explanation:
It is given that
a = <6,-9,3>
We need to compare the maginitude of vertor a and vector 4a.
If a vector is defined as v = <x,y,z>, then its magnitude is defined as
[tex]|v|=\sqrt{x^2+y^2+z^2}[/tex]
Using the above formula we get
[tex]|a|=\sqrt{6^2+(-9)^2+(3)^2}[/tex]
[tex]|a|=\sqrt{126}[/tex]
[tex]|a|=\sqrt{9\times 14}[/tex]
[tex]|a|=3\sqrt{14}[/tex]
The vector 4a is
[tex]4a=<4(6),4(-9),4(3)>=<24,-36,12>[/tex]
[tex]|4a|=\sqrt{(24)^2+(-36)^2+(12)^2}[/tex]
[tex]|4a|=\sqrt{4^2(6^2+(-9)^2+(3)^2)}[/tex]
[tex]|4a|=4\sqrt{6^2+(-9)^2+(3)^2}[/tex]
[tex]|4a|=4|a|[/tex]
The magnitude of 4a is a factor of 4 greater than the magnitude of a.
Therefore, the correct option is 2.