Proof that a parametrized curve orthogonal to a certain vector v

**x3bnm** · April 19th 2012, 08:06 AM

Let $\alpha\colon I\rightarrow R^3$ be a parametrized curve and let $\mathbf{v} \in R^3$ be a fixed vector. Assume that $\alpha'(t)$ is orthogonal to $\mathbf{v}$
for all $t \in I$ and that $\alpha(0)$ is also orthogonal to $\mathbf{v}$

Prove that $\alpha(t)$ is orthogonal to $\mathbf{v}$ for all $t \in I$

We are given that $\alpha'(t)$ and $\alpha(0)$ is orthogonal to $\mathbf{v}$

Let $\mathbf{v} = (v_1, v_2, v_3)$ and $\alpha(t) = (x(t), y(t), z(t))$

$\therefore \alpha'(t) = (x'(t), y'(t), z'(t))$

So we have:
$\alpha'(t) \cdot\, \mathbf{v} = (x'(t) * \, v_1 +\,\, y'(t) * v_2 + \,\, z'(t) * v_3) = 0 \text{ [Given]}$

$\alpha(0) \cdot\, \mathbf{v} = (x(0) *\, v_1 +\,\, y(0) * v_2 +\,\, z(0) * v_3) = 0 \text{ [Given]}$

Now how do I use this info and prove that $\alpha(t)$ is orthogonal to $\mathbf{v}$ for all $t \in I$ ?

Is it possible for anyone to kindly give some hints on how to prove this?

**x3bnm** · April 19th 2012, 09:56 AM

I think I solved it. For those who are interested:

Since

$\alpha'(t) \cdot\, \mathbf{v} = (x'(t) *\, v_1)+\,\,( y'(t) * v_2) +\,\, (z'(t) * v_3) = 0 \text{ [Given]}$

Particularly:

$(x'(t) *\, v_1)+\,\, (y'(t) * v_2) +\,\, (z'(t) * v_3) = 0$

Taking integration of both sides:

$(v_1 * x(t)) + C_1 +\,\,( v_2 * (y(t)) + C_2 +\,\, (v_3 * z(t)) + C_3 = C$

By taking all constants to the right side we get:

$(v_1 * x(t)) + (v_2 * y(t)) + (v_3 * z(t)) = C - C_1 - C_2 - C_3 \text{ ................(1)}$

We also know that:

$\alpha(0) \cdot\, \mathbf{v} = (x(0) *\, v_1) +\,\, (y(0) * v_2) +\,\, (z(0) * v_3) = 0 \text{ [Given]}$

So pluging in $t = 0$ into eq(1) we get:

$C - C_1 - C_2 - C_3 = 0$

therefore:

$(v_1 * x(t)) + (v_2 * y(t)) + (v_3 * z(t)) = 0$

But this means: $\alpha(t) \cdot\,\, \mathbf{v} = 0$

So $\alpha(t)$ is orthogonal to $\mathbf{v}$ .[As was to be shown]

**HallsofIvy** · April 19th 2012, 04:15 PM

Looks good but I think I would go the "other way"- that is, differentiate $\alpha$ rather than integrate $\alpha'$ .

Let $u(t)= \alpha(t)\cdot v$ . Then, since v is a constant vector, $u'= \alpha\cdot v= 0$ . Since the derivative of u is 0, u is a constant. And since $u(0)= \alpha(0)\cdot v= 0$ it follows that $u(t)= \alpha(t)\cdot v= 0$ for all t.

**x3bnm** · April 19th 2012, 04:37 PM

Originally Posted by HallsofIvy

Looks good but I think I would go the "other way"- that is, differentiate $\alpha$ rather than integrate $\alpha'$ .

Let $u(t)= \alpha(t)\cdot v$ . Then, since v is a constant vector, $u'= \alpha\cdot v= 0$ . Since the derivative of u is 0, u is a constant. And since $u(0)= \alpha(0)\cdot v= 0$ it follows that $u(t)= \alpha(t)\cdot v= 0$ for all t.

Thank you for showing another way of solving this problem. I believe your way is the correct and better method. Again thanks.

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