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Chapter 10 · Vector Algebra

Example 18: ∣x⃗∣|\vec x| when (x⃗−a⃗)⋅(x⃗+a⃗)=8(\vec x-\vec a)\cdot(\vec x+\vec a)=8⁠, a⃗\vec a a unit vector

ExamplesExpanding scalar products · Section 10.6.1
The question

If a⃗\vec a is a unit vector and (x⃗−a⃗)⋅(x⃗+a⃗)=8(\vec x - \vec a)\cdot(\vec x + \vec a) = 8⁠, then find ∣x⃗∣|\vec x|⁠.

Exam answer

Since a⃗\vec a is a unit vector, ∣a⃗∣=1|\vec a| = 1⁠. Also,

(x⃗−a⃗)⋅(x⃗+a⃗)=8⇒x⃗⋅x⃗+x⃗⋅a⃗−a⃗⋅x⃗−a⃗⋅a⃗=8\begin{aligned} &(\vec x - \vec a)\cdot(\vec x + \vec a) = 8 \\ &\Rightarrow \vec x\cdot\vec x + \vec x\cdot\vec a \\ &\quad - \vec a\cdot\vec x - \vec a\cdot\vec a = 8 \end{aligned}

As a⃗⋅x⃗=x⃗⋅a⃗\vec a\cdot\vec x = \vec x\cdot\vec a and r⃗⋅r⃗=∣r⃗∣2\vec r\cdot\vec r = |\vec r|^2⁠,

∣x⃗∣2−∣a⃗∣2=8⇒∣x⃗∣2−1=8⇒∣x⃗∣2=9\begin{aligned} |\vec x|^2 - |\vec a|^2 &= 8 \\ \Rightarrow |\vec x|^2 - 1 &= 8 \\ \Rightarrow |\vec x|^2 &= 9 \end{aligned}

∴ ∣x⃗∣=3\therefore\ |\vec x| = 3 (a magnitude is non-negative).

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

The scalar product expands like ordinary algebra, so (x⃗−a⃗)⋅(x⃗+a⃗)(\vec x - \vec a)\cdot(\vec x + \vec a) behaves like (x−a)(x+a)=x2−a2(x - a)(x + a) = x^2 - a^2⁠:

(x⃗−a⃗)⋅(x⃗+a⃗)=∣x⃗∣2−∣a⃗∣2.(\vec x - \vec a)\cdot(\vec x + \vec a) = |\vec x|^2 - |\vec a|^2.

The middle terms cancel because the scalar product is commutative. "Unit vector" supplies ∣a⃗∣=1|\vec a| = 1⁠, and the rest is one line of arithmetic.

Given, to find, and the plan
Given
∣a⃗∣=1|\vec a| = 1 and (x⃗−a⃗)⋅(x⃗+a⃗)=8(\vec x - \vec a)\cdot(\vec x + \vec a) = 8⁠.
To find
∣x⃗∣|\vec x|⁠.
Plan
(1) Expand the scalar product. (2) Cancel the middle terms and use r⃗⋅r⃗=∣r⃗∣2\vec r\cdot\vec r = |\vec r|^2⁠. (3) Solve for ∣x⃗∣2|\vec x|^2⁠. (4) Take the root that a length can have, and check with a concrete example.

Basics you need

1Unit vector

A unit vector has magnitude 1 (Section 10.3). Lesson: six kinds of vectors

2Expanding scalar products

Distributivity (Property 1), commutativity a⃗⋅x⃗=x⃗⋅a⃗\vec a\cdot\vec x = \vec x\cdot\vec a (Observation 7) and r⃗⋅r⃗=∣r⃗∣2\vec r\cdot\vec r = |\vec r|^2 (Observation 3), Section 10.6.1. Basics: dot products obey the same rule

3Magnitudes are non-negative

∣x⃗∣≥0|\vec x| \ge 0 for every vector (Note, Section 10.2), so from ∣x⃗∣2=9|\vec x|^2 = 9 only ∣x⃗∣=3|\vec x| = 3 is possible. Lesson: a vector is a length with a direction

Solution

  1. Expand the scalar product

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

  2. Simplify

    Since a⃗⋅x⃗=x⃗⋅a⃗\vec a\cdot\vec x = \vec x\cdot\vec a⁠, the middle terms cancel, and x⃗⋅x⃗=∣x⃗∣2\vec x\cdot\vec x = |\vec x|^2⁠, a⃗⋅a⃗=∣a⃗∣2=1\vec a\cdot\vec a = |\vec a|^2 = 1⁠. So

    (x⃗−a⃗)⋅(x⃗+a⃗)=∣x⃗∣2−1.(\vec x - \vec a)\cdot(\vec x + \vec a) = |\vec x|^2 - 1.
    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\cos 0 = 1⁠. This is what turns an equation about vectors into an equation about the single number ∣x⃗∣|\vec x|⁠.
    Observations 3 and 7 (Section 10.6.1); ∣a⃗∣=1|\vec a| = 1 for a unit vector (Section 10.3). Lesson: the dot product

  3. Solve for the length

    ∣x⃗∣2−1=8  ⇒  ∣x⃗∣2=9  ⇒  ∣x⃗∣=3.|\vec x|^2 - 1 = 8 \;\Rightarrow\; |\vec x|^2 = 9 \;\Rightarrow\; |\vec x| = 3.

    So ∣x⃗∣=3|\vec x| = 3.

    The equation t2=9t^2 = 9 has two roots, t=3t = 3 and t=−3t = -3⁠. But t=∣x⃗∣t = |\vec x| is a length, which is never negative, so −3-3 is rejected. This is the book's remark "as magnitude of a vector is non negative".
    ∣x⃗∣≥0|\vec x| \ge 0⁠: a length is never negative (Note, Section 10.2). Lesson: Examples 13–21

  4. Check with a concrete pair

    Take a⃗=i^\vec a = \hat i (a unit vector) and x⃗=3j^\vec x = 3\hat j⁠, so ∣x⃗∣=3|\vec x| = 3⁠. Then x⃗−a⃗=−i^+3j^\vec x - \vec a = -\hat i + 3\hat j⁠, x⃗+a⃗=i^+3j^\vec x + \vec a = \hat i + 3\hat j⁠, and

    (x⃗−a⃗)⋅(x⃗+a⃗)=(−1)(1)+(3)(3)=−1+9=8.\begin{aligned} &(\vec x - \vec a)\cdot(\vec x + \vec a) \\ &= (-1)(1) + (3)(3) = -1 + 9 = 8. \end{aligned}
    The answer says nothing about the direction of x⃗\vec x⁠: any vector of length 3 works, whatever its angle with a⃗\vec a⁠, because x⃗⋅a⃗\vec x\cdot\vec a cancelled out in Step 2. A second example, x⃗=3i^\vec x = 3\hat i (along a⃗\vec a⁠), gives (2i^)⋅(4i^)=8(2\hat i)\cdot(4\hat i) = 8 as well.
    u⃗⋅v⃗=u1v1+u2v2+u3v3\vec u\cdot\vec v = u_1v_1 + u_2v_2 + u_3v_3 (Section 10.6.1). Lesson: the dot product
Result

∣x⃗∣=3|\vec x| = 3⁠.

(Textbook answer: x⃗⋅x⃗+x⃗⋅a⃗−a⃗⋅x⃗−a⃗⋅a⃗=8\vec x\cdot\vec x + \vec x\cdot\vec a - \vec a\cdot\vec x - \vec a\cdot\vec a = 8⁠, so ∣x⃗∣2−1=8|\vec x|^2 - 1 = 8⁠, i.e. ∣x⃗∣2=9|\vec x|^2 = 9⁠; therefore ∣x⃗∣=3|\vec x| = 3⁠, as magnitude of a vector is non negative.)

Check it with numbers

A pair in a general direction: a⃗=13(i^+2j^+2k^)\vec a = \frac13(\hat i + 2\hat j + 2\hat k)⁠, a unit vector since 1+4+49=1\frac{1 + 4 + 4}{9} = 1⁠, and x⃗=2i^+j^−2k^\vec x = 2\hat i + \hat j - 2\hat k⁠, with ∣x⃗∣=4+1+4=3|\vec x| = \sqrt{4 + 1 + 4} = 3⁠. Then x⃗⋅a⃗=2+2−43=0\vec x\cdot\vec a = \frac{2 + 2 - 4}{3} = 0⁠, and in decimals x⃗−a⃗≈1.666667i^\vec x - \vec a \approx 1.666667\hat i +0.333333j^−2.666667k^+ 0.333333\hat j - 2.666667\hat k⁠, x⃗+a⃗≈2.333333i^\vec x + \vec a \approx 2.333333\hat i +1.666667j^−1.333333k^+ 1.666667\hat j - 1.333333\hat k⁠.

Exactly,

(x⃗−a⃗)⋅(x⃗+a⃗)=53⋅73+13⋅53+(−83)(−43)=35+5+329=8.\begin{aligned} &(\vec x - \vec a)\cdot(\vec x + \vec a) \\ &= \frac{5}{3}\cdot\frac{7}{3} + \frac{1}{3}\cdot\frac{5}{3} + \left(-\frac{8}{3}\right)\left(-\frac{4}{3}\right) \\ &= \frac{35 + 5 + 32}{9} = 8. \end{aligned}
Common mistakes
  1. Treating a⃗⋅a⃗\vec a\cdot\vec a as a⃗\vec a⁠. a⃗⋅a⃗=∣a⃗∣2=1\vec a\cdot\vec a = |\vec a|^2 = 1 is a number; it is not the vector a⃗\vec a⁠.
  2. Keeping both roots. ∣x⃗∣=±3|\vec x| = \pm3 is wrong: a length is never negative.
  3. Forgetting ∣a⃗∣=1|\vec a| = 1⁠. Then ∣x⃗∣2=8+∣a⃗∣2|\vec x|^2 = 8 + |\vec a|^2 cannot be finished; "unit vector" is the missing number.
What to take away
  1. (x⃗−a⃗)⋅(x⃗+a⃗)=∣x⃗∣2−∣a⃗∣2(\vec x - \vec a)\cdot(\vec x + \vec a) = |\vec x|^2 - |\vec a|^2⁠, like (x−a)(x+a)=x2−a2(x - a)(x + a) = x^2 - a^2⁠.
  2. r⃗⋅r⃗=∣r⃗∣2\vec r\cdot\vec r = |\vec r|^2 turns vector equations into equations about lengths.
  3. When you solve for a length, keep only the non-negative root.

Every fact used, and where it comes from

FactUsed inSource
A unit vector has magnitude 1Step 2Section 10.3 (Lesson: six kinds of vectors)
A magnitude is never negative: ∣x⃗∣≥0|\vec x| \ge 0Step 3Note, Section 10.2
Distributivity of the scalar product; scalars come outStep 1Section 10.6.1, Properties 1 and 2 (Basics: dot products obey the same rule)
Commutativity; r⃗⋅r⃗=∣r⃗∣2\vec r\cdot\vec r = |\vec r|^2Step 2Section 10.6.1, Observations 7 and 3
Scalar product in componentsStep 4, CheckSection 10.6.1 (Lesson: the dot product)
The book's working and answerResultExample 18, Section 10.6.2 (Lesson: Examples 13–21)
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