Proving Perpendicular Vectors with Direction Angles

[vertciel]

Hello everyone,

Thank you in advance for your help!

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Homework Statement

10. A vector $\vec{u}$ with direction angles A1, B1, and Y1, is perpendicular to a vector $\vec{v}$ with direction angles A2, B2, and Y2. Prove that:
$\cos A1 \cos B2 + \cos B1 \cos B2 + \cos Y1 \cos Y2 = 0$.

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The Attempt at a Solution

I let $\vec{u} = [a, b, c], \vec{v} = [x, y, z]$.

Since these are perpendicular, therefore:

$\vec{u} \bullet \vec{v} = ax + by + cz = 0$.

Also, $a, b, c, x, y, z$ would all correspond to their direction cosines.

However, I do not understand how I can prove the above statement with these facts. For example, would $\cos A1 \cos A2 = 0$ simply because they are the components of two vectors which are parallel to each other?

 

[lanedance]

what is a direction angle? i assume if the vector is length r then a = r.cosA1 etc...?

then just put them in your dot product & you're pretty much there

 

[HallsofIvy]

Hello everyone,

Thank you in advance for your help!

---

Homework Statement

10. A vector $\vec{u}$ with direction angles A1, B1, and Y1, is perpendicular to a vector $\vec{v}$ with direction angles A2, B2, and Y2. Prove that:
$\cos A1 \cos B2 + \cos B1 \cos B2 + \cos Y1 \cos Y2 = 0$.

---

The Attempt at a Solution

I let $\vec{u} = [a, b, c], \vec{v} = [x, y, z]$.

Since these are perpendicular, therefore:

$\vec{u} \bullet \vec{v} = ax + by + cz = 0$.

Also, $a, b, c, x, y, z$ would all correspond to their direction cosines.

However, I do not understand how I can prove the above statement with these facts. For example, would $\cos A1 \cos A2 = 0$ simply because they are the components of two vectors which are parallel to each other?

The whole point of "direction cosines" is that if a vector $\vec{v}$ has direction cosines A1, B1, Y1, (I imagine that in your book those are $\Alpha$, $\Beta$, and $\Gamma$ and that you are told that they are the cosines of the angles the vector makes with the x, y, and z axes, respectively) Then $\vec{v}= A1\vec{i}+ B1\vec{j}+ Y1\vec{k}$. That makes this problem simple.

 

[vertciel]

Thank you for your response.

Would you mind elaborating on the proof?

I thought that the direction cosines themselves were the unit vectors, so how would $\cos A1 \cos A2 = 0$? Shouldn't the dot product of these direction cosines = 0?

 

[vertciel]

Could anyone please offer an explanation for how to prove the above?

Thank you!

 

[Dick]

The direction cosines aren't unit vectors. They are the coefficients of the unit vectors. In terms of the direction angles, u=|u|*(cos(A1)*i+cos(B1)*j+cos(Y1)*k) and v=|v|*(cos(A2)*i+cos(B2)*j+cos(Y2)*k). And, yes, u.v=0. Substitute the expressions for the vectors into the dot product.

 

[vertciel]

Thanks for your reply, Dick.

I have:

$$\vec{u} = |\vec{u}| \cos A1 \hat{i} + |\vec{u}| \cos B1 \hat{j} + |\vec{u}| \cos Y1 \hat{k}$$

$$\vec{v} = |\vec{v}| \cos A2 \hat{i} + |\vec{v}| \cos B2 \hat{j} + |\vec{v}| \cos Y2 \hat{k}$$

If I dot these two expressions on the RS, I do not see how I would get 0 as the final expression.

Could you please explain more?

 

[Dick]

Your are GIVEN that the two vectors are perpendicular. If you dot them you will get an expression involving the cosines and |u| and |v|. You can set that equal to zero because you are GIVEN u.v=0.