Solving a Vector Problem: Length (17^1/2) and Same Direction as v = <7,0,-6>

[Kawrae]

Find the vector that satisfies the stated conditions: Length (17^1/2) and same direction as v = <7,0,-6>.

The book gives the answer as being u = (1/5)<7,0,-6>. Having some trouble getting this answer... here is how I went about attempting to solve it:

1. Finding the vector length of the given vector.
||v|| = (7^2 + 0^2 + -6^2)^(1/2)
= (49+0+36)^(1/2)
= (85)^(1/2)

2. Setting the two vectors equal...
||u|| = (17)^(1/2) = k||v|| Since multiples of the vector will give same direction
(17)^(1/2)/||v|| = k

So k = (17)^(1/2)/(85)^(1/2)

So then my answer would be u = (17)^(1/2)/(85)^(1/2) <7,0,-6>.


Can anyone help point out where I am messing up?? Thank you

 

[radou]

Work with vectors. Your first condition is u = k v. The second one is ||u||=||kv|| = 17^(1/2). This should work just fine.

 

[driscoll79]

I'd recommend first finding the unit vector $$v = \frac{u}{||u||}$$, because you know that vector has a magnitude of 1. Then, you know that $$||kv||=\sqrt{17}$$, and you know that $$||v||=1$$, so $$k=\sqrt{17}$$. So, your new vector $$w=kv=<\frac{7\sqrt{17}}{\sqrt{85}},0,\frac{-6\sqrt{17}}{\sqrt{85}}>=<7\sqrt{\frac{1}{5}},0,-6\sqrt{\frac{1}{5}}>$$ has a magnitude of $$\sqrt{17}$$. So, your answer should be correct... I'd check the answer in your book with your instructor.

 

[NateTG]

Can anyone help point out where I am messing up?? Thank you

Are you sure that the book doesn't have the following?
$$\sqrt{\frac{1}{5}} \left<7,0-6\right>$$

(Which is one of many possible expressions equivalent to the answer you've given.)