MHB Component of a vector perpendicular to another

[Drain Brain]

Given $\overline{A}=-4a_{x}+2a_{y}+3a_{z}$ and $\overline{B}=3a_{x}+4a_{y}-a_{x}$.
1.Find the vector component of A parallel to B
2.Find the vector component of A perpendicular to B

my solution for 1.

$\overline{A}\cdot\overline{b}=-1.372$

$(\overline{A}\cdot\overline{b})(\overline{b})=-0.806a_{x}-1.075a_{y}+0.268a_{z}$

I'm not sure if my method is correct in 1, but it matched the answer at the back of my book. For 2 I have no idea how to attack it. please help. TIA.

 

[Ackbach]

Given $\overline{A}=-4a_{x}+2a_{y}+3a_{z}$ and $\overline{B}=3a_{x}+4a_{y}-a_{x}$.

I'm assuming you meant $\overline{B}=3a_x+4a_y-a_{z}$.

1.Find the vector component of A parallel to B
2.Find the vector component of A perpendicular to B

my solution for 1.

$\overline{A}\cdot\overline{b}=-1.372$

$(\overline{A}\cdot\overline{b})(\overline{b})=-0.806a_{x}-1.075a_{y}+0.268a_{z}$

Hmm; I get that $\overline{A}\cdot\overline{B}=(-4)(3)+(2)(4)+(3)(-1)=-12+8-3=-7.$ This is assuming that $a_x, a_y,$ and $a_z$ are orthonormal. Then $(\overline{A}\cdot\overline{B})\overline{B}=-21a_x-28a_y+7a_z$. But if you're trying to find the component of $\overline{A}$ parallel to $\overline{B}$, you would need to compute
$$\text{proj}_{\overline{B}}(\overline{A})=\frac{\overline{A}\cdot\overline{B}}{\overline{B}\cdot \overline{B}} \, \overline{B},$$
or
$$\frac{1}{9+16+1}(-21a_x-28a_y+7a_z)=-0.807a_x -1.077a_y+0.269a_z,$$
pretty much what you got.

I'm not sure if my method is correct in 1, but it matched the answer at the back of my book. For 2 I have no idea how to attack it. please help. TIA.

Why not subtract the result of your previous calculation from $\overline{A}$? If you take $\overline{A}$, and subtract off from it however much of it is in the direction of $\overline{B}$, you should be left with how much of it is perpendicular to $\overline{B}$, right?

This, by the way, is the beginning of the Gram-Schmidt Orthonormalization Procedure, of great importance in linear algebra and quantum mechanics.

 

[Drain Brain]

I'm assuming you meant $\overline{B}=3a_x+4a_y-a_{z}$.
Hmm; I get that $\overline{A}\cdot\overline{B}=(-4)(3)+(2)(4)+(3)(-1)=-12+8-3=-7.$ This is assuming that $a_x, a_y,$ and $a_z$ are orthonormal. Then $(\overline{A}\cdot\overline{B})\overline{B}=-21a_x-28a_y+7a_z$. But if you're trying to find the component of $\overline{A}$ parallel to $\overline{B}$, you would need to compute
$$\text{proj}_{\overline{B}}(\overline{A})=\frac{\overline{A}\cdot\overline{B}}{\overline{B}\cdot \overline{B}} \, \overline{B},$$
or
$$\frac{1}{9+16+1}(-21a_x-28a_y+7a_z)=-0.807a_x -1.077a_y+0.269a_z,$$
pretty much what you got.
Why not subtract the result of your previous calculation from $\overline{A}$? If you take $\overline{A}$, and subtract off from it however much of it is in the direction of $\overline{B}$, you should be left with how much of it is perpendicular to $\overline{B}$, right?

This, by the way, is the beginning of the Gram-Schmidt Orthonormalization Procedure, of great importance in linear algebra and quantum mechanics.

By the way, $\overline{b}$ is the unit vector in the direction of $\overline{B}$

 

[Drain Brain]

By the way, $\overline{b}$ is the unit vector in the direction of $\overline{B}$

Is there a general formula for prob. 2 like what you suggested in prob 1? Thnaks!

 

[Ackbach]

Is there a general formula for prob. 2 like what you suggested in prob 1? Thnaks!

I'd recommend you take a closer look at the tail end of Post #2.

 

[requiredCaptcha]

1) The component of vector parallel to another vector is found by the formula

u . v/ l v l
u refers to first vector, . refers to dot product, v is second vector and l v l is magnitude of second vector.

2) The component of vector perpendicular to another vector is found by the formula

P - ( P . Q^) Q^

P refers to first vector, - refers to subtraction, . refers to dot product, Q^ refers to the unit vector in the direction of second vector.

You can ask me for more doubts,

I am a student studying in 8th grade.