The Missing X-Component of a Vector

Use the dot product identity

$\large{A_x \, B_x + A_y \, B_y = |A|\ |B| \cos \theta \\ x + 21 = \sqrt{50} \, \sqrt{x^2+9} \, \frac{1}{\sqrt{2}} \\ x^2 + 42x + 21^2 = 25 x^2 + 25(9) \\ 24 x^2 -42 x + 25(9) - 21^2 = 0}$

Solving the quadratic, the positive solution is $x = 4$

Consider another vector $(7, -1)$ . Since $(7, -1) \bullet (1, 7) = 0$ , it is perpendicular to vector $(1, 7)$ , and since $\sqrt{7^2 + (-1)^2} = \sqrt{1^2 + 7^2}$ , it has the same magnitude as vector $(1, 7)$ . This means we can construct a square with those vectors as sides, as pictured below, and the vector made by one of the diagonals of the square would be $(7 + 1, 7 - 1) = (8, 6)$ .

This means that this diagonal has a $\frac{y}{x}$ ratio of $\frac{4}{3}$ , and since a diagonal of a square makes a $45°$ to its side, and since the $y$ -component of our unknown vector is $3$ , its $x$ -component is $\frac{4}{3} \cdot 3 = \boxed{4}$ .

(Note that a square can also be formed on the other side of vector $(1, 7)$ with a side vector of $(-7, 1)$ and diagonal vector $(-6, 8)$ , but a $y$ -component of $3$ gives an $x$ -component of $-\frac{9}{4}$ , and therefore can be rejected as another solution since it is negative.)

Let $\alpha$ be the acute angle between the first vector and the positive $x$ -axis, and let $\beta$ be the acute angle between the second vector and the positive $x$ -axis. Because the slope of a line is equal to the tangent of the angle it makes with the positive $x$ -axis, we have

$\begin{aligned} \tan \alpha &= 7 \\ \tan \beta &= \dfrac{3}{x}. \end{aligned}$

The angle between the two vectors is $\alpha - \beta = 45^{\circ},$ because $x$ is positive. Taking the tangent, we have

$\begin{aligned} \tan(\alpha - \beta) &= \tan 45^{\circ} \\ \dfrac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} &= \tan 45^{\circ} \\ \dfrac{7 - \frac{3}{x}}{1 + \frac{21}{x}} &= 1 \\ 7x - 3 &= x + 21 \\ x &= \boxed{4}. \end{aligned}$