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Chapter 10 · Vector Algebra

Exercise 10.3 · Q9: ∣x⃗∣|\vec x| when (x⃗−a⃗)⋅(x⃗+a⃗)=12(\vec x-\vec a)\cdot(\vec x+\vec a)=12⁠, a⃗\vec a a unit vector

Exercise 10.3Expanding scalar products · Section 10.6.1
The question

Find ∣x⃗∣|\vec x|⁠, if for a unit vector a⃗\vec a⁠, (x⃗−a⃗)⋅(x⃗+a⃗)=12(\vec x - \vec a)\cdot(\vec x + \vec a) = 12⁠.

Exam answer

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

(x⃗−a⃗)⋅(x⃗+a⃗)=12⇒x⃗⋅x⃗+x⃗⋅a⃗−a⃗⋅x⃗−a⃗⋅a⃗=12\begin{aligned} &(\vec x - \vec a)\cdot(\vec x + \vec a) = 12 \\ &\Rightarrow \vec x\cdot\vec x + \vec x\cdot\vec a \\ &\quad - \vec a\cdot\vec x - \vec a\cdot\vec a = 12 \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=12⇒∣x⃗∣2−1=12⇒∣x⃗∣2=13\begin{aligned} |\vec x|^2 - |\vec a|^2 &= 12 \\ \Rightarrow |\vec x|^2 - 1 &= 12 \\ \Rightarrow |\vec x|^2 &= 13 \end{aligned}

∴ ∣x⃗∣=13\therefore\ |\vec x| = \sqrt{13} (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 does not depend on the order. "Unit vector" supplies ∣a⃗∣=1|\vec a| = 1⁠, and the rest is arithmetic.

Given, to find, and the plan
Given
∣a⃗∣=1|\vec a| = 1 and (x⃗−a⃗)⋅(x⃗+a⃗)=12(\vec x - \vec a)\cdot(\vec x + \vec a) = 12⁠.
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 a length can have, and check with concrete vectors.

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 (Section 10.2, Note), so from ∣x⃗∣2=13|\vec x|^2 = 13 only ∣x⃗∣=13|\vec x| = \sqrt{13} is possible. Lesson: the dot product

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.
    Section 10.6.1, Properties 1 and 2. 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 turns an equation about vectors into an equation about the single number ∣x⃗∣|\vec x|⁠.
    Section 10.6.1, Observations 3 and 7; ∣a⃗∣=1|\vec a| = 1 for a unit vector (Section 10.3). Lesson: the dot product

  3. Solve for the length

    ∣x⃗∣2−1=12⇒∣x⃗∣2=13⇒∣x⃗∣=13.\begin{aligned} |\vec x|^2 - 1 &= 12 \\ \Rightarrow |\vec x|^2 &= 13 \\ \Rightarrow |\vec x| &= \sqrt{13}. \end{aligned}

    So ∣x⃗∣=13|\vec x| = \sqrt{13}.

    The equation t2=13t^2 = 13 has two roots, t=13t = \sqrt{13} and t=−13t = -\sqrt{13}⁠. But t=∣x⃗∣t = |\vec x| is a length, which is never negative, so −13-\sqrt{13} is rejected. 13 is prime, so 13\sqrt{13} does not simplify.
    ∣x⃗∣≥0|\vec x| \ge 0 (Section 10.2, Note). Lesson: Examples 13–21

  4. Check with concrete vectors

    Take a⃗=k^\vec a = \hat k (a unit vector) and x⃗=2i^+3j^\vec x = 2\hat i + 3\hat j⁠, so ∣x⃗∣=4+9=13|\vec x| = \sqrt{4 + 9} = \sqrt{13}⁠. Then x⃗−a⃗=2i^+3j^−k^\vec x - \vec a = 2\hat i + 3\hat j - \hat k⁠, x⃗+a⃗=2i^+3j^+k^\vec x + \vec a = 2\hat i + 3\hat j + \hat k⁠, and

    (x⃗−a⃗)⋅(x⃗+a⃗)=(2)(2)+(3)(3)+(−1)(1)=4+9−1=12.\begin{aligned} &(\vec x - \vec a)\cdot(\vec x + \vec a) \\ &= (2)(2) + (3)(3) + (-1)(1) \\ &= 4 + 9 - 1 = 12. \end{aligned}
    The answer fixes only the length of x⃗\vec x⁠, not its direction: x⃗⋅a⃗\vec x\cdot\vec a cancelled in Step 2, so any vector of length 13\sqrt{13} works, whatever its angle with a⃗\vec a⁠. Here x⃗\vec x happens to be perpendicular to a⃗\vec a⁠; the Check box below uses a pair that is not.
    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∣=13|\vec x| = \sqrt{13}⁠.

(Textbook answer: 13\sqrt{13}⁠.)

Check it with numbers

13≈3.605551\sqrt{13} \approx 3.605551⁠, and 3.6055512≈13.0003.605551^2 \approx 13.000⁠, while 13−1=1213 - 1 = 12⁠.

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=3i^+2j^\vec x = 3\hat i + 2\hat j⁠, with ∣x⃗∣=9+4=13|\vec x| = \sqrt{9 + 4} = \sqrt{13}⁠. Here x⃗⋅a⃗=3+43=73≠0\vec x\cdot\vec a = \frac{3 + 4}{3} = \frac73 \ne 0⁠. In decimals x⃗−a⃗≈2.666667i^\vec x - \vec a \approx 2.666667\hat i +1.333333j^−0.666667k^+ 1.333333\hat j - 0.666667\hat k and x⃗+a⃗≈3.333333i^\vec x + \vec a \approx 3.333333\hat i +2.666667j^+0.666667k^+ 2.666667\hat j + 0.666667\hat k⁠. Exactly,

(x⃗−a⃗)⋅(x⃗+a⃗)=83⋅103+43⋅83+(−23)(23)=80+32−49=1089=12.\begin{aligned} &(\vec x - \vec a)\cdot(\vec x + \vec a) \\ &= \frac{8}{3}\cdot\frac{10}{3} + \frac{4}{3}\cdot\frac{8}{3} + \left(-\frac{2}{3}\right)\left(\frac{2}{3}\right) \\ &= \frac{80 + 32 - 4}{9} = \frac{108}{9} = 12. \end{aligned}
Common mistakes
  1. Forgetting ∣a⃗∣=1|\vec a| = 1⁠. Then ∣x⃗∣2=12+∣a⃗∣2|\vec x|^2 = 12 + |\vec a|^2 cannot be finished; "unit vector" is the missing number.
  2. Taking ∣x⃗∣=12|\vec x| = \sqrt{12} or 11\sqrt{11}⁠. The −∣a⃗∣2-|\vec a|^2 moves across as +1+1⁠: ∣x⃗∣2=12+1=13|\vec x|^2 = 12 + 1 = 13⁠.
  3. Keeping both roots. ∣x⃗∣=±13|\vec x| = \pm\sqrt{13} is wrong: a length is never negative.
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)
∣x⃗∣≥0|\vec x| \ge 0Step 3Section 10.2, Note
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 same method with 8 in place of 12Steps 1 to 3Example 18, Section 10.6.2 (Lesson: Examples 13–21)
The answerResultNCERT answer key, Exercise 10.3 Q9 (Lesson: the dot product)
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