Is there a typo in Calculus:Early Transcedentals?

I just finished doing my homework on Local Linear Approximations in 3-space (Ch.13.4). In one of the problems the answer I got is different from the answer key.
Problem 39. We have a function $f(x,y,z)=xe^{yz}$ and points $P(1,-1,-1)$ and $Q(0.99,-1.01,-0.99)$
Part a) Find the local Linear Approximation at point $P$
Part b) Compare the error in approximation at point $Q$ with the distance between $P$ and $Q$.( By compare they mean $\frac{\text{error in approximation}}{\text{distance between$P$and$Q$}}$).

So the ration I get in part b) is $(-0.00982)$, but in the answer key it is $0.01554$

Attempt: Part a) First I found the local linear approximation at point $P$: $$L(x,y,z)=e+e(x-1)-e(y+1)-e(z+1)$$

Part b)The local linear approximation at $Q$: $$L(0.99,-1.01,-0.99)=e+e(-0.01)-e(-0.01)-e(0.01)=2.691$$

The actual value of $f(x,y,z)$ at $Q$ is $f(0.99,-1.01,-0.99)=2.6908299$.

The error in approximation: $f(0.99,-1.01,-0.99)-L(0.99,-1.01,-0.99)= -0.0001701$

Distance between points $P$ and $Q$: $D=\sqrt{0.01^2+0.01^2+0.01^2}=0.01732$

So now when I compare the error in app and distance I get: $$\frac{\text{error in approximation}}{\text{distance between P and Q}}=-0.00982$$

But the answer key say it should be $0.01554$. Is it a typo or my mistake?

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Don't most modern textbooks have errata pages on the web? – GEdgar May 20 '12 at 23:13
I’d have calculated the error the other way round, making it approximately $0.0001701$, since the estimate is greater than the true value, but apart from that I see nothing wrong. – Brian M. Scott May 20 '12 at 23:13
@GEdgar Lemme check. – Koba May 20 '12 at 23:19
@BrianM.Scott I solved Problems with the same questions and in each one I got the right answer that matched the answer key. But in this problem I get the answer that does not match the key, even though I am using the method that worked in other similar problems. So, it must be a typo. – Koba May 20 '12 at 23:20
@GEdgar Nah. I could not find it. – Koba May 20 '12 at 23:27

The mistake is in the way you're calculating your error. You have kept only three decimal places with $L$ but 5 decimal places with $d(P,Q)$. Notice your answer is correct up to two decimal places. Instead, let's keep 7 with each. $$\left|\frac{2.6908299-2.6910990}{\sqrt{3(.01)^2}}\right| =0.01554$$