A line goes through point $A = (9, -7, 31)$ and the line is perpendicular to vectors $[7, 1, 2]$ and $[3, 0, 1]$.

What is the equation of the line?

The cross product of $[7, 1, 2]$ and $[3, 0, 1]$ is $[1, -1, -3]$.

I believe the line can be written as $x = [9, -7, 31] + t [1, -1, -3]$, where $t$ is a real number.

I'm somewhat skeptical of my own work. If you find the cross product (which will be perpendicular to the two given vectors); why does that mean it is automatically parallel with the line that is also perpendicular to these two vectors?

Given the vectors are in three-dimensional space; is it possible to produce a cross product of these two vectors and that vector not to be parallel with the line I found?

  • 1
    $\begingroup$ The answer to your last question is "no." Two vectors determine a plane. There is only one direction normal to that plane and the cross product gives it. $\endgroup$ – B. Goddard Jan 18 at 13:00
  • $\begingroup$ The people first think how to formulate a vector to be perpendicular to other two vectors in 3-dimensional space, then they invented the cross product. $\endgroup$ – kelvin hong 方 Jan 18 at 13:05
  • $\begingroup$ I believe two non-parallel vectors make a plane. That's true. That makes it easy to visualise. The cross product makes a vector that pierces the plane at right angles from above and below said plane. That's my understanding. $\endgroup$ – Madison Jan 18 at 13:15
  • $\begingroup$ I would slightly amend @B.Goddard's assertion to say that there are two opposite directions that are normal (perpendicular) to the plane, and if your vectors are $u$ and $v$, then these two directions are given by $u \times v$ and $v \times u$. $\endgroup$ – John Hughes Jan 18 at 13:20
  • $\begingroup$ Yes, above and below the plane. I believe. The opposite direction being the anti-commutative. $\endgroup$ – Madison Jan 18 at 13:27

When you find a cross product,

$$A=\begin{bmatrix}a&b&c\\7&1&2\\ 3& 0& 1\end{bmatrix}$$

you are finding a linearly independent vector, namely $\vec{v}=(a,b,c)$, to the other two vectors, namely $\vec{u}=(7,1,2),\vec{w}=(3,0,1)$ (and it is like that because you are finding a vector which gives you $\det(A)\neq 0$, hence $\dim(A)=3$, hence $\vec{u},\vec{v},\vec{w}$ generate $\mathbb{R}^3$).

That means that you will have a basis of the $3$-dimensional space (since you have $3$ linearly independent vectors). An that is why if you have a fourth vector, namely $\vec{\alpha}$ which is perpendicular to the first two vectors, it has to be proportional to the one you have find with the cross product.

Hence, what you have done is OK.

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    $\begingroup$ Okay, this is good. I was basically taught to just follow that it's true; but, it was lacking intuition. Soon to study linear independence. $\endgroup$ – Madison Jan 18 at 13:20

Consider the plane containing the point $A=(9,-7,31)$ and perpendicular to the vector $[7,1,2]$. Its equation is $7x + 1y + 2z = 7(9) + 1(-7) + 2(31)$, which simplifies to $7x + y + 2z = 118$.

Similarly the plane containing the point $A=(9,-7,31)$ and perpendicular to the vector $[3,0,1]$ has equation $3x+z=58$.

The intersection of these two planes will be the line that contains the point $A$ and is perpendicular to both vectors. It is not hard to check that the line $\ell(t) = [9+t, -7-t, 31-3t]$ is that line:

$$7x + y + 2z = 7(9+t)+(-7-t)+2(31-3t) = 118$$

and $$3x + z = 3(9+t)+(31-3t) = 58$$


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