Vector perpendicular to a plane defined by two vectors

[Mr Davis 97]

Say that I have two vectors that define a plane. How do I show that a third vector is perpendicular to this plane? Do I use the cross product somehow?

 

[jedishrfu]

What about using the dot product?

 

[Mr Davis 97]

What about using the dot product?

I don't see how the dot product would be used... To me, it would make sense to use it when we only need to show that one vector is perpendicular to another. But here we need to show that one vector is perpendicular to a plane (that two other vectors create).

 

[Mark44]

A vector perpendicular to the plane would be perpendicular to each of the two vectors that define the plane, so the dot product could be used, taking the dot product of each of the two vectors with the third. Also, the cross product of the two vectors would produce a normal to the plane that would be a scalar multiple of the third vector.

In short, you could use either the dot product or the cross product.

 

[Mr Davis 97]

Okay, so I am specifically asking this in relation to a specific problem:

Given the vector $\vec{A} = 3\hat{i} + 4\hat{j} -4\hat{k}$,

(1) find a unit vector $\hat{B}$ that lies in the x-y plane and is perpendicular to \vec{A}.

First, we note that $\hat{B} = B_x \hat{i} + B_y \hat{j}$, and that $B_x^2 + B_Y^2 = 1$ since $\hat{B}$ is a unit vector. Solving this system simultaneously comes up with the solution $\displaystyle B_x = \frac{4}{5}, B_y = -\frac{3}{5}$

(2) find a unit vector $\hat{C}$ that is perpendicular to both $\vec{A}$ and $\hat{B}$.

the unit vector that is perpendicular to both $\vec{A}$ and $\hat{B}$ is $\displaystyle \hat{C} = \pm \frac{\vec{A} \times \hat{B}}{| \vec{A} \times \hat{B} |}$. After lots of calculations, we find that $\displaystyle \hat{c} = \frac{-12 \hat{i} - 16 \hat{j} + 7 \hat{k}}{\sqrt{449}}$

(3) show that $\vec{A}$ is perpendicular to the plane defined by $\hat{B}$ and $\hat{C}$

So would showing that $\vec{A} \cdot \hat{B} = 0$ and $\vec{A} \cdot \hat{C} = 0$ be sufficient to show that $\vec{A}$ is perpendicular to the plane?

 

[Mark44]

Okay, so I am specifically asking this in relation to a specific problem:

Given the vector $\vec{A} = 3\hat{i} + 4\hat{j} -4\hat{k}$,

(1) find a unit vector $\hat{B}$ that lies in the x-y plane and is perpendicular to \vec{A}.

First, we note that $\hat{B} = B_x \hat{i} + B_y \hat{j}$, and that $B_x^2 + B_Y^2 = 1$ since $\hat{B}$ is a unit vector. Solving this system simultaneously comes up with the solution $\displaystyle B_x = \frac{4}{5}, B_y = -\frac{3}{5}$

Not how I would do it.
1) Find a vector, B, in the x-y plane that is perp. to A.
2) Normalize B to get the unit vector

Since B lies in the x-y plane, it is of the form <x, y, 0>
$A \cdot B = 0 \Rightarrow <3, 4, -4> \cdot <x, y, 0> = 0 \Rightarrow 3x + 4y = 0$
This leads the vectors <3, -4, 0> and <-3, 4, 0>
The corresponding unit vectors are <3/5, -4/5, 0> and <-3/5, 4/5, 0>

Note that I prefer writing vectors using angle brackets rather than with the i, j, and k unit vectors, as being quicker to write.

(2) find a unit vector $\hat{C}$ that is perpendicular to both $\vec{A}$ and $\hat{B}$.

the unit vector that is perpendicular to both $\vec{A}$ and $\hat{B}$ is $\displaystyle \hat{C} = \pm \frac{\vec{A} \times \hat{B}}{| \vec{A} \times \hat{B} |}$. After lots of calculations, we find that $\displaystyle \hat{c} = \frac{-12 \hat{i} - 16 \hat{j} + 7 \hat{k}}{\sqrt{449}}$

(3) show that $\vec{A}$ is perpendicular to the plane defined by $\hat{B}$ and $\hat{C}$

So would showing that $\vec{A} \cdot \hat{B} = 0$ and $\vec{A} \cdot \hat{C} = 0$ be sufficient to show that $\vec{A}$ is perpendicular to the plane?

Yes it would, and this is something you should do, because you have a mistake, most likely in your cross product.

Here's a tip. When you do the cross product, don't do it using the unit vector you found earlier. Use either <3, -4, 0> or <-3, 4, 0>. The only difference will be the magnitude of the resulting vector. Since you're going to normalize it anyway, all you need is the right direction.