Find the coordinates of the vector x

Question

Find the coordinates of the vector $x$, if it is known that is perpendicular to the vectors $a_1 = (4, -2, -3)$ and $a_2 = (0, 1, 3)$, forms an acute angle with a unit j and $|x| = 26$ (values must be an integer)

I tried but I can't understand what I should do with the angle.

I have that

$a_{11}x_1 + a_{12}x_2 + a_{13}x_3 = 0$

$a_{21}x_1 + a_{22}x_2 + a_{23}x_3 = 0$

$4x_1 - 2x_2 + 3x_2 = 0$

$x_2 + 3x_2 = 0$

$x_1 = -3/4x_3$

$x_2 = -3x_3$

$x_3 = -1/3x_2$

$26^2 = x_1^2 + x_2^2 + x_3^2$

and then solve this but my answers are not correct because of the misunderstanding what to do with the angle and how to solve it

Answer 1

So, you start by solving \begin{align*} (x_1, x_2, x_3) \cdot (4, -2, -3) &= 0 \\ (x_1, x_2, x_3) \cdot (0, 1, 3) &= 0. \end{align*} This will give you ever vector orthogonal to both $(4, -2, -3)$ and $(0, 1, 3)$. In expanding this, you make a small error, forgetting a minus sign: \begin{align*} 4x_1 - 2x_2 \color{red}- 3x_3 &= 0 \\ x_2 + 3x_3 &= 0. \end{align*} Despite the small error, you seem to have arrived at the correct solution. The second equation comes to $$x_2 = -3x_3$$ as you got, and the first equation therefore becomes $$x_1 = \frac{1}{4}(2x_2 + 3x_3) = \frac{1}{4}(2(-3x_3) + 3x_3) = -\frac{3}{4}x_3,$$ again agreeing with your working.

We don't need to solve for $x_3$ in terms of $x_2$. We have let $x_3$ be a free variable; it can be anything it wants, and the other variables are along for the ride. In particular, we get $$(x_1, x_2, x_3) = \left(-\frac{3}{4}x_3, -3x_3, x_3\right) = x_3\left(-\frac{3}{4}, -3, 1\right).$$ I just want a single non-zero solution to this, and I don't care about the length. I could choose $x_3 = 1$ and get a perfectly fine solution, but I'm going to choose $x_3 = -4$, just to give us some integers! When $x_3 = -4$, we get the solution: $$(x_1, x_2, x_3) = (3, 12, -4).$$ Take some time to demonstrate for yourself that this is perpendicular to the given vectors.

Next, find the length of this vector. If it has length $26$, we've lucked out! Otherwise, we may need to scale this vector, changing its length, but not its direction. Doing so will still mean the vector is orthogonal to the two given vectors. $$\|(3, 12, -4)\| = \sqrt{3^2 + 12^2 + 4^2} = \sqrt{169} = 13.$$ Pretty close! This vector is precisely half the length we want, so I'm going to double its length: $(6, 24, -8)$.

Lastly, we want an acute angle with $j$. Two vectors have an acute angle if their dot product is positive. So, we compute $(6, 24, -8) \cdot j = 24 > 0$. We therefore have the vector we wanted. If we had the dot product be negative, then I would simply have negated the whole vector (i.e. $(-6, -24, 8)$) to get a parallel vector of the same length, but pointing in the opposite direction. We didn't have to do this, so the final answer is $(6, 24, -8)$.

Answer 2

There are two choices of vector which are perpendicular to $a_1$ and $a_2$. To see why, think of the coordinate vectors $i$, $j$, and $k$. If you want a vector of length $3$ perpendicular to both $i$ and $j$, notice both $3k$ and $-3k$ work!

You correctly identified that the system you are supposed to solve is:

• $x \cdot a_1 = 0$
• $x \cdot a_2 = 0$
• $|x| = 26$

But there will be two choices of $x$ that work, which will be perfectly opposite each other. In the same way that you might take the positive answer of a quadratic equation, you want to take the vector $x$ that makes an acute angle with the $j$ axis.

Recalling the definition of the angle between two vectors $v$ and $w$ is $\cos \theta = \dfrac{v \cdot w}{|v||w|}$, we see we need the following additional constraint:

• $\arccos \left ( \dfrac{x \cdot j}{|x||j|} \right ) < \dfrac{\pi}{2}$

Obviously you would instead ask for that to be $< 90^{\circ}$ if you are working with degrees.

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I hope this helps ^_^

Answer 3

First cross the two vectors to get a vector perpendicular to both. The cross product will yield $(-3, - 12, 4)$. Note that the vector you're trying to determine is a scalar multiple of this vector, in either direction, so it can be written as $\vec x=k(-3, - 12, 4)$ (where $k$ can be positive or negative (but not zero as that would be a trivial case).

Now recall that the dot product allows you to determine the angle between two vectors. $\vec a \cdot \vec b = |\vec a||\vec b| \cos \theta$ where $\theta$ is the angle between the two vectors. Note (from the sign of the cosine in the first and second quadrants) that an acute angle will give a positive dot product (scalar) whereas an obtuse angle will give a negative dot product (scalar). (A right angle would give you a zero dot product).

In this case, your first vector is $\vec x$ and the second is the unit vector $(0,1,0)=\vec j$. The dot product is simply $-12k$, and this needs to be positive so $k<0$.

Also, the magnitude of $\vec x$ is $|k|\sqrt {(-3)^2 +(-12)^2 +4^2} = 13|k|$.

Setting $13|k| = 26$, we get $k = \pm 2$. Recalling that $k<0$, we get $k=-2$ and the final answer of $\vec x = (6, 24, - 8)$.