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$v=(4,0,-5)$ and I am given the hint: Construct two nonparallel vectors orthogonal to $v$ in $\Bbb R^3$.

I've looked at this post Find the equation of the plane passing through a point and a vector orthogonal but I don't see how that relates to the hint.

I am able to solve the question if it is a line in $\Bbb R^2$ that passes through the origin and orthogonal to a vector $v$. An example: if $v=(-2,3)$ then a vector orthogonal to $v$ is $v_0=(-3,-2)$, the point at origin is $P_0=(0,0)$ so then $(x,y)=(0,0)+t(-3,-2)$ is the vector equation.

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Can you find a vector orthogonal to $(4,0,-5)$? – rae306 Jun 29 '14 at 6:36
up vote 1 down vote accepted


$v = (4, 0 , -5) \tag{1}$

then clearly

$w_1 = (0, 1, 0) \tag{2}$

is orthogonal to $v$; so is

$w_2 = (5, 0, 4), \tag{3}$

so $w_1$ and $w_2$ must span the plane normal to $v$; I found $w_1$ and $w_2$ by simple inspection, that is, intelligent (I hope!) guesswork! The parametric equation of the plane is then

$(x, y, z) = sw_1 + tw_2 = s(0, 1, 0) + t(5, 0, 4) = (5t, s, 4t). \tag{4}$

We might also observe that $w_1$ is normal to $w_2$, so if we normalize $w_2$ to

$w_2' = (\dfrac{5}{\sqrt{41}}, 0, \dfrac{4}{\sqrt{41}}) \tag{5}$

we can express the plane in terms of the orthonormal pair $w_1$, $w_2'$:

$(x, y, z) = sw_1 + tw_2' = (\dfrac{5}{\sqrt{41}}t, s, \dfrac{4}{\sqrt{41}}t); \tag{6}$

and finally, we can always use the non-parametric vector form

$0 = v \cdot (x, y, z) = 4x - 5z. \tag{7}$

The vector $v$ given by (1) may be normalized to

$v' = (\dfrac{4}{\sqrt{41}}, 0, -\dfrac{5}{\sqrt{41}}) \tag{8}$

if so desired; then (7) becomes

$0 = v' \cdot (x, y, z) = \dfrac{4}{\sqrt{41}}x - \dfrac{5}{\sqrt{41}}z. \tag{9}$

Note that in the above I have been able to neglect inclusion of the point $P_0$ since here, as in the OP's $2$-dimensional example, $P_0 = 0$, the zero vector.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!

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$$ (1) \ \textbf{w} = (5,0,4), \ \textbf{x} = \textbf{v} \times \textbf{w} $$

$$ (2) \ \textbf{w} = (5,0,4), \ \textbf{x} \cdot w =0, \textbf{x} \cdot v =0$$

$\cdot$ The above gives two ways to determine your normal. I started by getting a vector orthogonal to the one we started with. In the first and second case once we have $\textbf{x}$ will be your normal and now to get the equation of the plane we do: $$\textbf{x} \cdot (X -v) = 0, \ \text{where}\ X = (x_1,x_2,x_3)$$

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If you know how to find an orthogonal vector to a given one in $\Bbb R^2$, you can do the same in $\Bbb R^3$ usually swapping two non-zero coordinates, switching one of the signs, and making the last coordinate zero. Surely you must be sure that other geometric aspects of the problem won't conflict with this zero we'll force.

Now, to your specific problem. Given the vector $v = (4,0,-5)$, some choices are: $(5,0,-4), (0,1,0)$. Also, any linear combination of these two guys will do the job (can you prove it?)

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