# 1 vectors and 1 limits problem

Hi guys,

Starting revision for my exams and I can't seem to solve these problems. ):

1. Given that three non zero vectors A, B and C are perpendicular to each other and that the vector R = pA + qB + rC is a zero vector, show that p = q = r = 0

2. In another question, given that f(x) is differenciatable at x = 0 and $\frac{Lim}{x->0}$ $\frac{f(x) - f(kx)}{x}$ = a, wjere a and k are constants.

b. Show that f'(0) = a/1-k

What I did was take the defination of differenciation:
$\frac{Lim}{h->0}$ $\frac{f(x+h) - f(x)}{h}$

and I let x = kx, and h = (1-k)x,
$\frac{Lim}{(1-k)x->0}$ $\frac{f(x) - f(kx)}{(1-k)x}$

and I did this (assuming that I can write $\frac{Lim}{(1-k)x->0}$ as $\frac{Lim}{x->0}$

1/(1-k)$\frac{Lim}{x->0}$ $\frac{f(x) - f(kx)}{x}$ = a/1-k

Is this correct?

*Edit* Just realized that I misread the forum title and posted in the wrong one. ):

Last edited:

micromass
Staff Emeritus
Homework Helper
Hi guys,

Starting revision for my exams and I can't seem to solve these problems. ):

1. Given that three non zero vectors A, B and C are perpendicular to each other and that the vector R = pA + qB + rC is a zero vector, show that p = q = r = 0

What can you tell us about the inner products R.A, R.B and R.C??

2. In another question, given that f(x) is differenciatable at x = 0 and $\frac{Lim}{x->0}$ $\frac{f(x) - f(kx)}{x}$ = a, wjere a and k are constants.

b. Show that f'(0) = a/1-k

What I did was take the defination of differenciation:
$\frac{Lim}{h->0}$ $\frac{f(x+h) - f(x)}{h}$

and I let x = kx, and h = (1-k)x,
$\frac{Lim}{(1-k)x->0}$ $\frac{f(x) - f(kx)}{(1-k)x}$

and I did this (assuming that I can write $\frac{Lim}{(1-k)x->0}$ as $\frac{Lim}{x->0}$

1/(1-k)$\frac{Lim}{x->0}$ $\frac{f(x) - f(kx)}{x}$ = a/1-k

Is this correct?

*Edit* Just realized that I misread the forum title and posted in the wrong one. ):
[/QUOTE]

No, you can't do that. You took

$\frac{Lim}{h->0}$ $\frac{f(x+h) - f(x)}{h}$

but the x in there is a constant!! But later

$\frac{Lim}{(1-k)x->0}$ $\frac{f(x) - f(kx)}{(1-k)x}$

your x is suddenly a limit variable.

I'll illustrate it by taking x=2, then you write

$\frac{Lim}{h->0}$ $\frac{f(2+h) - f(2)}{h}$

this is perfectly valid and it equals f'(2). But then you write

$\frac{Lim}{(1-k)2->0}$ $\frac{f(2) - f(k2)}{(1-k)2}$

which obviously makes no sense. (1-k)2 -> 0 makes no sense!!

I'll give you a hint on how to solve it:

\begin{align}
f^\prime(0)
&= \lim_{h\rightarrow 0}{\frac{f(h)-f(0)}{h}}\\
&= \lim_{h\rightarrow 0}{ \frac{\frac{f(h)-f(kh)}{h}}{ \frac{f(h)-f(kh)}{f(h)-f(0)}}}\\
&= \lim_{h\rightarrow 0}{ \frac{\frac{f(h)-f(kh)}{h}}{ \frac{(f(h)-f(0)) - (f(kh)-f(0))}{f(h)-f(0)}}}\\
\end{align}

What can you tell us about the inner products R.A, R.B and R.C??

Oh!

R.R = R.A + R.B + R.C
Since R.R = 0,
R= p = q = r?

I'll give you a hint on how to solve it:

\begin{align}
f^\prime(0)
&= \lim_{h\rightarrow 0}{\frac{f(h)-f(0)}{h}}\\
&= \lim_{h\rightarrow 0}{ \frac{\frac{f(h)-f(kh)}{h}}{ \frac{f(h)-f(kh)}{f(h)-f(0)}}}\\
&= \lim_{h\rightarrow 0}{ \frac{\frac{f(h)-f(kh)}{h}}{ \frac{(f(h)-f(0)) - (f(kh)-f(0))}{f(h)-f(0)}}}\\
\end{align}

How did you get from the first to the second step? )(Actually I'm lost from that step onwards also)

Anyway thanks for your help so far micromass!

mathman
R.A = pA.A Since R=0 then R.A=0, however A.A > 0, therefore p=0. Similarly for q and r.

HallsofIvy
Homework Helper
Oh!

R.R = R.A + R.B + R.C
Since R.R = 0,
R= p = q = r?
Why do you say "R.R= 0"?? And "R= p= q= r" makes no sense. R is a vector while p, q, and r are numbers.

mathman