To find the component of a vector perpendicular to another

[brotherbobby]

Homework Statement

Calculate the component of a vector $\vec A = \hat i+\hat j+5\hat k$ perpendicular to $\vec B = 3\hat i + 4\hat j$.

Relevant Equations

1. The vector $\vec B'$ perpendicular to $\vec B$ is such that $\vec B'\cdot \vec B = 0$.

2. The (vector) component of a vector $\vec A$ along a vector $\vec B'$ is $(\vec A)_{B'}=(A\cos\theta_{AB'})\hat B$.

Problem statement : I copy and paste the (slightly different) problem statement as it appeared in the text to the right.

Attempt : By inspection, we find that the vector $\vec B'$ perpendicular to $\vec B = 3\hat i+4\hat j$ is $\boldsymbol{\vec B' = 4\hat i -3\hat j}$, remembering that for perpendicular vectors $\vec B$ and $\vec B'$, $\vec B\cdot \vec B' = 0$.

The vector component of $(\vec A)_{B'}=(A\cos\theta_{AB'})\hat B'=\left(\frac{\vec A\cdot \vec B'}{B'} \right)\hat B'$.
Now $\vec A\cdot \vec B'=\frac{1}{5}$, $B' = B = 5$ and $\hat B' =\frac{4\hat i- 3\hat j}{5}$.
Thus $\boxed{(\vec A)_{B'}= \frac{1}{25}\left( \frac{4}{5}\hat i - \frac{3}{5}\hat j\right)}$.Doubt : My answer is at odds with that of the text, which I copy and paste below.

Question : I agree to the text's method, viz. $\vec A = \vec A_{\parallel}+\vec A_{\perp}$. However, I struggle to see what's wrong with my my method either.

 

[anuttarasammyak]

You may add $$\alpha \mathbf{k}$$ to B' and keep it perpendicular to B.

 

[PeroK]

You got answer C. The book solution has a typo in the last line, with a extra factor of 5 in the denominator.

PS I got the solutions mixed up. It's yours that is wrong, with the extra factor of 5.

 

[brotherbobby]

You got answer C. The book solution has a typo in the last line, with a extra factor of 5 in the denominator.

My answer differs from not having a $5\hat k$ also. So my vector component is confined in the $x-y$ plane. I am wondering if this problem has a unique solution.

 

[PeroK]

My answer differs from not having a $5\hat k$ also. So my vector component is confined in the $x-y$ plane. I am wondering if this problem has a unique solution.

It's a given that you include the $5\hat k$. The solution is unique, as long as you sufficiently define what you mean by component. If we express$$\vec A = \alpha \vec B + \vec C$$Where $\vec B \cdot \vec C = 0$, then $\vec C$ is uniquely defined. The proof is an exercise if you want.

 

[PeroK]

PS this is an example of what I call a "specific techique". This is where a specific technique is used to solve a specific problem, even where a completely general solution exists. Sometimes, a specific technique may be a significant shortcut. But, at other times, it's no simpler than the general approach.

The most common example is the quadratic equation. There are lots of videos on YouTube showing various techniques for solving specific quadratic solutions. And, yet, a general solution (completing the square) exists. Not to mention the general quadratic formula - which is so ubiquitous that it seems worth memorising!

In this case, the proof of the above (which I'll include here) represents a general technique for solving all such problems. This is certainly the way I look at mathematics. We are looking for $\alpha$ and $\vec C$ (the component) such that:$$\vec A = \alpha \vec B + \vec C$$Where $\vec B \cdot \vec C = 0$.

First, we see that:$$\vec A \cdot \vec B = \alpha B^2$$Where $B = |\vec B|$. Hence:
$$\alpha = \frac{ \vec A \cdot \vec B}{B^2} \ \ \text{and} \ \ \vec C = \vec A - \frac{ \vec A \cdot \vec B}{B^2}\vec B$$Let's check that out in this case:
$$\vec A \cdot \vec B = 7, \ B = 5$$$$\vec C = (1, 1, 5) - \frac 7 {25}(3, 4, 0) = (\frac 4 {25}, -\frac 3 {25}, 5)$$That's a general approach to vectors that will not only solve elementary problems, but lay a foundation for more advanced mathematics.

 

[brotherbobby]

Can you check the answer? I am getting $\vec C = \left( \dfrac{4}{25},-\dfrac{3}{25}, 5 \right)$

 

[PeroK]

Can you check the answer? I am getting $\vec C = \left( \dfrac{4}{25},-\dfrac{3}{25}, 5 \right)$

Yes, I had a typo in my final answer! Fixed now.

 

[WWGD]

I'm not even clear in what the " Component of a vector" is, unless one refers to the coefficient I ent of either i,j or k.

 

[Mark44]

I'm not even clear in what the " Component of a vector" is.

It's a fairly simple idea that uses the idea that a vector can be decomposed into a sum of other vectors. Of course, you have to specify the direction of the vector component. A trivial example is that the vector $\vec u = 2i + 3j + 5k$ has a component in the direction of the positive x-axis of 2i.

 

[WWGD]

It's a fairly simple idea that uses the idea that a vector can be decomposed into a sum of other vectors. Of course, you have to specify the direction of the vector component. A trivial example is that the vector $\vec u = 2i + 3j + 5k$ has a component in the direction of the positive x-axis of 2i.

Yes, that's what I thought and wrote in the remainder of my post.

 

[FactChecker]

Attempt : By inspection, we find that the vector $\vec B'$ perpendicular to $\vec B = 3\hat i+4\hat j$ is $\boldsymbol{\vec B' = 4\hat i -3\hat j}$, remembering that for perpendicular vectors $\vec B$ and $\vec B'$, $\vec B\cdot \vec B' = 0$.

Let's do a simple sanity check here. There is an entire plane perpendicular to $\vec B$. How did you pick that vector, $\vec B'$, out of an entire plane with no reference to the vector $\vec A$? The answer should be unique. That should make you sceptical of your method.

 

[WWGD]

Let's do a simple sanity check here. There is an entire plane perpendicular to $\vec B$. How did you pick that vector, $\vec B'$, out of an entire plane with no reference to the vector $\vec A$? The answer should be unique. That should make you sceptical of your method.

Or, Brother Bobby, consider, if you've seen it, the Orthogonal Complement of the vector.

 

[FactChecker]

Or, Brother Bobby, consider, if you've seen it, the Orthogonal Complement of the vector.

In fact, with bad luck, $\vec B'$ might have been completely perpendicular to $\vec A$.

 

[brotherbobby]

Let's do a simple sanity check here. There is an entire plane perpendicular to $\vec B$. How did you pick that vector, $\vec B'$, out of an entire plane with no reference to the vector $\vec A$? The answer should be unique. That should make you sceptical of your method.

Fair enough. But when we talk of a vector $\vec B'$ that is perpendicular to $\vec B$, aren't we free to choose any vector? Let's ignore the contents of this particular problem for a moment.

 

[brotherbobby]

Or, Brother Bobby, consider, if you've seen it, the Orthogonal Complement of the vector.

Can you explain? Does a vector have a complement orthogonal to it?

 

[PeroK]

This problem was done and dusted by post #8, IMO.

 

[brotherbobby]

This problem was done and dusted by post #8, IMO.

Yes, actually in post #6, by @PeroK . However, I like a little post-mortem to clear things that others bring into the discussion.

 

[Mark44]

Fair enough. But when we talk of a vector $\vec B'$ that is perpendicular to $\vec B$, aren't we free to choose any vector?

Yes. However, all of those vectors ($\vec B'$) determine the plane that is perpendicular to $\vec B$.

 

[FactChecker]

Fair enough. But when we talk of a vector $\vec B'$ that is perpendicular to $\vec B$, aren't we free to choose any vector?

No. You better choose the right one if you want it to have the correct component of $\vec A$. You might pick one with anything from a multiple of the desired component of $\vec A$ to one that is perpendicular to $\vec A$. How would you know? This approach gets you nowhere. It is much better to start with $\vec A$ and subtract out the part of $\vec A$ that is parallel to $\vec B$.

 

[Mark44]

No. You better choose the right one if you want it to have the correct component of $\vec A$.

Just to be clear, my answer of "yes" was strictly in response to the question about getting a perpendicular vector to a given one, with nothing to do with components.