Miscellaneous · Q14: Mutually perpendicular vectors of equal magnitude: $\vec a+\vec b+\vec c$ is equally inclined to each

Basics you need

Solution

1. Read the question as intended

   The printed text asks about "the vector $\vec c\cdot\vec d = 15$". That expression is the last condition of Question 12, where a vector $\vec d$ is defined. Here no $\vec d$ is given, and $\vec c\cdot\vec d$ would be a number in any case, so it cannot be "equally inclined" to anything. The only vector built from the given $\vec a, \vec b, \vec c$ that the symmetry of the data points to is their sum, and the question is solved for

   $$\vec r = \vec a + \vec b + \vec c.$$
   "Equally inclined" compares angles, and angles are defined between vectors. The data treat $\vec a, \vec b, \vec c$ in exactly the same way, so the vector to test is the one that also treats them the same way: $\vec a + \vec b + \vec c$⁠.
   A scalar product is a number (Section 10.6.1, Observation 1). Lesson: the miscellaneous exercises

2. The length of $\vec a + \vec b + \vec c$

   $$\begin{aligned} |\vec r|^2 &= (\vec a + \vec b + \vec c)\cdot(\vec a + \vec b + \vec c) \\ &= |\vec a|^2 + |\vec b|^2 + |\vec c|^2 \\ &\quad + 2(\vec a\cdot\vec b + \vec b\cdot\vec c + \vec c\cdot\vec a) \\ &= k^2 + k^2 + k^2 + 2(0) = 3k^2, \end{aligned}$$

   so $|\vec r| = \sqrt3\,k$⁠.

   Expanding gives nine products: three squares of lengths and six mixed products, which pair up by commutativity. "Mutually perpendicular" makes every mixed product 0, and "equal magnitudes" makes each square $k^2$⁠. The length is positive, so we take the positive square root.
   Section 10.6.1, Property 1, Observations 2, 3 and 7. Basics: dot products obey the same rule

3. The three scalar products

   $$\begin{aligned} \vec r\cdot\vec a &= |\vec a|^2 + \vec b\cdot\vec a + \vec c\cdot\vec a = k^2, \\ \vec r\cdot\vec b &= \vec a\cdot\vec b + |\vec b|^2 + \vec c\cdot\vec b = k^2, \\ \vec r\cdot\vec c &= \vec a\cdot\vec c + \vec b\cdot\vec c + |\vec c|^2 = k^2. \end{aligned}$$
   In each line, distributivity splits the product into three terms. Two of them are products of different given vectors, which are 0, and the third is a length squared, $k^2$⁠.
   Distributivity (Section 10.6.1, Property 1). Lesson: the dot product

4. The three cosines, and the conclusion

   Let $\vec r$ make angles $\alpha, \beta, \gamma$ with $\vec a, \vec b, \vec c$⁠. Each given vector has length $k$⁠, so

   $$\begin{aligned} \cos\alpha &= \frac{\vec r\cdot\vec a}{|\vec r||\vec a|} = \frac{k^2}{\sqrt3\,k \cdot k} = \frac{1}{\sqrt3}, \\ \cos\beta &= \frac{\vec r\cdot\vec b}{|\vec r||\vec b|} = \frac{1}{\sqrt3}, \\ \cos\gamma &= \frac{\vec r\cdot\vec c}{|\vec r||\vec c|} = \frac{1}{\sqrt3}. \end{aligned}$$

   The three angles lie in $[0, \pi]$⁠, where the cosine takes each value once, so $\alpha = \beta = \gamma = \cos^{-1}\frac{1}{\sqrt3}$⁠. Hence $\vec a + \vec b + \vec c$ is equally inclined to $\vec a$⁠, $\vec b$ and $\vec c$.

   The three fractions have the same numerator $k^2$ and the same denominator $\sqrt3\,k^2$⁠, and that is exactly where both hypotheses are used: perpendicularity made the numerators equal to $k^2$⁠, and equal magnitudes made the denominators equal. The value $\frac{1}{\sqrt3}$ does not depend on $k$⁠.
   $\cos\theta = \frac{\vec u\cdot\vec v}{|\vec u||\vec v|}$ (Section 10.6.1, Observation 6). Basics: cos θ is a shadow

5. The picture: the diagonal of a cube

   Placed tail to tail, $\vec a, \vec b, \vec c$ are three edges of a cube of side $k$ meeting at a corner, and $\vec a + \vec b + \vec c$ is the diagonal of the cube from that corner. In the plane of $\vec a$ and the diagonal, the diagonal splits into $\vec a$ and $\vec b + \vec c$⁠, which are perpendicular, with $|\vec b + \vec c| = \sqrt2\,k$⁠:

   This is the cube version of Miscellaneous Q11: when $\vec a, \vec b, \vec c$ are $k\hat i, k\hat j, k\hat k$⁠, the vector $\vec a + \vec b + \vec c = k(\hat i + \hat j + \hat k)$ is equally inclined to the three axes. The argument above shows that the axes need not be $\hat i, \hat j, \hat k$⁠: any three mutually perpendicular vectors of equal length behave the same way.
   Triangle law; Pythagoras in the right triangle (Section 10.4). Basics: use Pythagoras twice

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