I need help w/ "Given a=i+2j-2k, find y, z so b=yj+zk is perp. unit vector."

iremos

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I need help w/ "Given a=i+2j-2k, find y, z so b=yj+zk is perp. unit vector."

Hi everyone!

I have an admission exam coming up and they gave us a sample exam paper with no answer.

Below is the question that I'm trying to figure out.

Q: Given the vector a=i+2j-2k, find a possible set of value for y and z such that b=yj+zk is a unit vector that is perpendicular to a.

This is what i get so far,

since a is perpendicular to b, their dot product would be 0, therefore y and z = 0.

Would really appreciate your help! Thank you!
 
Hi everyone!

I have an admission exam coming up and they gave us a sample exam paper with no answer.

Below is the question that I'm trying to figure out.

Q: Given the vector a=i+2j-2k, find a possible set of value for y and z such that b=yj+zk is a unit vector that is perpendicular to a.

This is what i get so far,

since a is perpendicular to b, their dot product would be 0, therefore y and z = 0.

Would really appreciate your help! Thank you!
Unfortunately you did not get too far. What is a\(\displaystyle \cdot\)b? Now set that equal to 0 and solve.

Of course, if b=0, then a\(\displaystyle \cdot\)b=0. Do you think that is the only case for a dot product to be 0?
 
Hi Jomo, thanks for your reply!

i figured out that a.b = 2y-2z, therefor y=z, am i right?

and b = 1, since it is a unit vector.

i'm stuck here and i'm not sure how to solve it after trying for 30 mins :(
 
Hi Jomo, thanks for your reply!

i figured out that a.b = 2y-2z, therefor y=z, am i right?

and b = 1, since it is a unit vector.

i'm stuck here and i'm not sure how to solve it after trying for 30 mins :(
Why is b=1? Why do you think that you are not done?

I will give you a few vectors for b and I would like you to compute a.b for each of them.

b = <0,5,5>, b = <0,3,3>, b = <0,-8,-8> and b = <0,2/3,2/3>
 
Hi Jomo! Thank you for your reply.

b= 1 because it is an unit vector hence it has a magnitude of 1?

The answer to all the a.b you stated are 0! but the magnitude of those vectors will not be 1.

I think i have figured it out.
since y=z
b.b = y^2+ y^2
1= 2y^2
y = sqrt(1/2)

therefore y and z will be sqrt(1/2) or -sqrt(1/2)

Please correct me if I'm wrong.
 
Well, in the future, if you're ever unsure of your answer, you can always check it yourself by working the problem "in reverse" to see if you end up with the given information. Your proposed solutions are y=sqrt(1/2) and y=-sqrt(1/2), with z always being equal to y. The vector b is defined as <0, y, z> so let's plug them in and see what happens:

\(\displaystyle b=<0,\sqrt{\frac{1}{2}},\sqrt{\frac{1}{2}}>\) and \(\displaystyle ||b||=\sqrt{0^2+\left(\sqrt{\frac{1}{2}}\right)^2+\left(\sqrt{\frac{1}{2}}\right)^2}=1\)

\(\displaystyle b=<0,-\sqrt{\frac{1}{2}},-\sqrt{\frac{1}{2}}>\) and \(\displaystyle ||b||=\sqrt{0^2+\left(-\sqrt{\frac{1}{2}}\right)^2+\left(-\sqrt{\frac{1}{2}}\right)^2}=1\)

Both of those fit with the given information that b is a unit vector (magnitude 1). Now, let's make sure they're orthogonal to a.

\(\displaystyle a=<1,2,-2>\) so \(\displaystyle a\cdot b=<1,2,-2>\cdot <0,\sqrt{\frac{1}{2}},\sqrt{\frac{1}{2}}>=0\)

\(\displaystyle a\cdot b=<1,2,-2>\cdot <0,-\sqrt{\frac{1}{2}},-\sqrt{\frac{1}{2}}>=0\)

A dot product of 0 means the vectors are orthogonal, so it all checks out. Your answers are correct.
 
Hi Jomo! Thank you for your reply.

b= 1 because it is an unit vector hence it has a magnitude of 1?

The answer to all the a.b you stated are 0! but the magnitude of those vectors will not be 1.

I think i have figured it out.
since y=z
b.b = y^2+ y^2
1= 2y^2
y = sqrt(1/2)

therefore y and z will be sqrt(1/2) or -sqrt(1/2)

Please correct me if I'm wrong.
Ah, you did say it was a unit vector. Sorry!
 
Well, in the future, if you're ever unsure of your answer, you can always check it yourself by working the problem "in reverse" to see if you end up with the given information. Your proposed solutions are y=sqrt(1/2) and y=-sqrt(1/2), with z always being equal to y. The vector b is defined as <0, y, z> so let's plug them in and see what happens:

\(\displaystyle b=<0,\sqrt{\frac{1}{2}},\sqrt{\frac{1}{2}}>\) and \(\displaystyle ||b||=\sqrt{0^2+\left(\sqrt{\frac{1}{2}}\right)^2+\left(\sqrt{\frac{1}{2}}\right)^2}=1\)

\(\displaystyle b=<0,-\sqrt{\frac{1}{2}},-\sqrt{\frac{1}{2}}>\) and \(\displaystyle ||b||=\sqrt{0^2+\left(-\sqrt{\frac{1}{2}}\right)^2+\left(-\sqrt{\frac{1}{2}}\right)^2}=1\)

Both of those fit with the given information that b is a unit vector (magnitude 1). Now, let's make sure they're orthogonal to a.

\(\displaystyle a=<1,2,-2>\) so \(\displaystyle a\cdot b=<1,2,-2>\cdot <0,\sqrt{\frac{1}{2}},\sqrt{\frac{1}{2}}>=0\)

\(\displaystyle a\cdot b=<1,2,-2>\cdot <0,-\sqrt{\frac{1}{2}},-\sqrt{\frac{1}{2}}>=0\)

A dot product of 0 means the vectors are orthogonal, so it all checks out. Your answers are correct.

Hello Ksdhart2, thank you for your info, appreciate it very much! :)
 
Q: Given the vector a=i+2j-2k, find a possible set of value for y and z such that b=yj+zk is a unit vector that is perpendicular to a.
Why, oh why oh make this so difficult?
Look: \(\displaystyle y^2+z^2=1\) to be a unit vector.
For \(\displaystyle \bf{a}\bot \bf{b}\) we must have \(\displaystyle 2y-2z=0\) or \(\displaystyle y=z\).
That means \(\displaystyle y=z=\pm\frac{\sqrt 2}{2}\).
 
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