Determine a unit vector perpendicular to the given planes

[chwala]

Homework Statement

see attached.

Relevant Equations

vectors

I looked at this question and i wanted to ask if we could also use; $C$ =$c_2$($-\dfrac {3}{2}i$ +$j - 3k)$ ... cheers

This problem can also be solved by using the approach of cross product $A×B$...

 

[Mark44]

Homework Statement:: see attached.
Relevant Equations:: vectors

View attachment 297125

View attachment 297126

I looked at this question and i wanted to ask if we could also use; $C$ =$c_2$($-\dfrac {3}{2}i$ +$j - 3k)$ ... cheers

This problem can also be solved by using the approach of cross product $A×B$...

They're asking for a unit vector, so you will need to solve for $c_2$. Doing so will give you exactly the same two vectors as shown in the solution.

 

[chwala]

I understand that...I just wanted to be certain that we could work with any $c_1, c_2$ or $c_3$...thanks...working to the given solution is straightforward.

 

[chwala]

They're asking for a unit vector, so you will need to solve for $c_2$. Doing so will give you exactly the same two vectors as shown in the solution.

By the way Mark what do you mean by solve for $c_2?$...in the text they worked with $c_3$ to solution...I am assuming that you mean the same will apply for $c_2$.

They're asking for a unit vector, so you will need to solve for $c_2$. Doing so will give you exactly the same two vectors as shown in the solution.

 

[Mark44]

By the way Mark what do you mean by solve for $c_2$?

Because you wrote "$C = c_2 (-\dfrac {3}{2}i + j - 3k)$ and it needs to be a unit vector. There are only two values of $c_2$ that make this a unit vector: one positive and one negative.

 

[chwala]

What i meant;
The unit vector in the direction of c can also be given by;
$±\dfrac{{-\frac {3}{2}}i+j-3k}{\sqrt \frac{49}{4}}...$

using either $c_1, c_2$ or $c_3$ would work as indicated in my post $3$.

 

[Mark44]

What i meant;
The unit vector in the direction of c can also be given by;
$±\dfrac{{-\frac {3}{2}}i+j-3k}{\sqrt \frac{49}{4}}$

Yes, but why would you want to write it this way without simplifying it? Dividing by $\frac{49} 4$ is equivalent to multiplying by 2/7. If you do this, you get the same result as shown in the solution.

 

[chwala]

Yes, but why would you want to write it this way without simplifying it? Dividing by $\frac{49} 4$ is equivalent to multiplying by 2/7. If you do this, you get the same result as shown in the solution.

Yes of course I can simplify that! ...I was just indicating to you on what I meant...we shall end up with same solution...
Cheers Mark. Thanks.

 

[Dr Transport]

Why don't you check the solution by taking the cross-product of \vec{A} and \vec{B} and normalizing it

 

[chwala]

I did check and confirmed that the problem can also be solved by using cross- product...where $A×B=15i-10j+30k$ ...

My only query was specific on $c_1,c_2$ and $c_3$...Cheers

 

[WWGD]

As a side-comment, it seems you're using a book where exercise solutions use round numbers. Notice the scaling was by a nice nimber like 2/7 and not, a.g., $\sqrt 13 /2$.

 

[chwala]

As a side-comment, it seems you're using a book where exercise solutions use round numbers. Notice the scaling was by a nice nimber like 2/7 and not, a.g., $\sqrt 13 /2$.

That's true...am actually using Schaum's textbook on Vector Analysis.