Question
(1/3sqrt(2) )*<1,1,4> is a unit vector pointing in a direction opposite to the vector <-3,-3,-12>.
The answer is TRUE
Can you show me the working step of this question? Thank you very much.
There are a couple of ways. I think the simplest would be to find the unit vector in the direction of <-3, -3, -12> and show that it adds to $\displaystyle \frac{1}{3\sqrt{2}}<1, 1, 4>$ to make 0.
So the magnitude of the vector <-3,-3,-12> is $\displaystyle \sqrt{162} = 9\sqrt{2}$. Thus the unit vector in the direction of <-3, -3, -12> is
$\displaystyle \frac{<-3, -3, -12>}{9\sqrt{2}} = -\frac{1}{3\sqrt{2}}<1, 1, 4>$.
You can easily see if you add this vector to the unit vector $\displaystyle \frac{1}{3\sqrt{2}}<1, 1, 4>$ that the sum is 0.
-Dan