Exercise 10.3 · Q6: $|\vec a|$⁠, $|\vec b|$ when $(\vec a+\vec b)\cdot(\vec a-\vec b)=8$⁠, $|\vec a|=8|\vec b|$

Basics you need

Solution

1. Expand the scalar product

   $$\begin{aligned} &(\vec a + \vec b)\cdot(\vec a - \vec b) \\ &= \vec a\cdot\vec a - \vec a\cdot\vec b + \vec b\cdot\vec a - \vec b\cdot\vec b. \end{aligned}$$
   Distributivity lets us multiply out the brackets term by term, as with $(a + b)(a - b)$⁠: each term of the first bracket meets each term of the second, giving four scalar products, and the minus sign of $-\vec b$ stays with the two products it is in.
   Section 10.6.1, Properties 1 and 2. Basics: dot products obey the same rule

2. Simplify

   Since $\vec b\cdot\vec a = \vec a\cdot\vec b$⁠, the middle terms cancel, and $\vec a\cdot\vec a = |\vec a|^2$⁠, $\vec b\cdot\vec b = |\vec b|^2$⁠. So the given condition becomes

   $$|\vec a|^2 - |\vec b|^2 = 8.$$
   The scalar product of a vector with itself is the square of its length, because the angle of a vector with itself is 0 and $\cos 0 = 1$⁠. Nothing about the angle between $\vec a$ and $\vec b$ is needed: it cancelled with the middle terms.
   Section 10.6.1, Observations 3 and 7. Lesson: the dot product

3. Use $|\vec a| = 8|\vec b|$

   Squaring, $|\vec a|^2 = 64|\vec b|^2$⁠. Then

   $$\begin{aligned} 64|\vec b|^2 - |\vec b|^2 &= 8 \\ \Rightarrow 63|\vec b|^2 &= 8 \\ \Rightarrow |\vec b|^2 &= \frac{8}{63}. \end{aligned}$$
   The condition is about $|\vec a|$⁠, but the equation has $|\vec a|^2$⁠, so the factor 8 must be squared too: $(8|\vec b|)^2 = 64|\vec b|^2$⁠, not $8|\vec b|^2$⁠. Then there is one unknown, and $64 - 1 = 63$⁠.
   Substitution into $|\vec a|^2 - |\vec b|^2 = 8$⁠. Basics: dot products obey the same rule

4. Solve for $|\vec b|$

   $$|\vec b| = \sqrt{\frac{8}{63}} = \frac{\sqrt8}{\sqrt{63}} = \frac{2\sqrt2}{3\sqrt7}.$$
   The equation $t^2 = \frac{8}{63}$ has two roots, but $t = |\vec b|$ is a length, which is never negative, so only the positive root is kept. Then $\sqrt8 = \sqrt{4 \times 2} = 2\sqrt2$ and $\sqrt{63} = \sqrt{9 \times 7} = 3\sqrt7$⁠. Rationalised, this is $\frac{2\sqrt{14}}{21}$⁠, the same number.
   $|\vec b| \ge 0$ (Section 10.2). Lesson: the dot product

5. Find $|\vec a|$⁠, and check

   $$|\vec a| = 8|\vec b| = 8 \times \frac{2\sqrt2}{3\sqrt7} = \frac{16\sqrt2}{3\sqrt7}.$$

   So $|\vec a| = \frac{16\sqrt2}{3\sqrt7}$ and $|\vec b| = \frac{2\sqrt2}{3\sqrt7}$. Check: $|\vec a|^2 - |\vec b|^2 = \frac{512}{63} - \frac{8}{63} = \frac{504}{63} = 8$⁠.

   Putting both lengths back into $|\vec a|^2 - |\vec b|^2 = 8$ must return 8, and it does: $|\vec a|^2 = \frac{256 \times 2}{9 \times 7} = \frac{512}{63}$ and $504 = 8 \times 63$⁠. The condition $|\vec a| = 8|\vec b|$ holds by construction.
   $|\vec a|^2 - |\vec b|^2 = 8$ from Step 2. Lesson: Examples 13–21

Every fact used, and where it comes from