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Chapter 10 · Vector Algebra

Exercise 10.3 · Q18 (MCQ): When is λa⃗\lambda\vec a a unit vector?

Exercise 10.3Magnitude of a scalar multiple, unit vectors · Sections 10.3 and 10.5
The question

If a⃗\vec a is a nonzero vector of magnitude ‘aa’ and λ\lambda a nonzero scalar, then λa⃗\lambda\vec a is unit vector if

(A) λ=1\lambda = 1

(B) λ=−1\lambda = -1

(C) a=∣λ∣a = |\lambda|

(D) a=1/∣λ∣a = 1/|\lambda|

Exam answer

∣λa⃗∣=∣λ∣ ∣a⃗∣=∣λ∣ a|\lambda\vec a| = |\lambda|\,|\vec a| = |\lambda|\,a⁠.

λa⃗\lambda\vec a is a unit vector   ⟺  ∣λ∣ a=1  ⟺  a=1∣λ∣\iff |\lambda|\,a = 1 \iff a = \frac{1}{|\lambda|}⁠.

Answer: (D) a=1∣λ∣a = \dfrac{1}{|\lambda|}

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

The one idea this problem is built on

Multiplying a vector by λ\lambda multiplies its length by ∣λ∣|\lambda|: ∣λa⃗∣=∣λ∣ ∣a⃗∣|\lambda\vec a| = |\lambda|\,|\vec a|⁠. A unit vector has length 1, so the condition is ∣λ∣ a=1|\lambda|\,a = 1⁠. Each option must make this true for every vector it allows.

Given, to find, and the plan
Given
a⃗≠0⃗\vec a \ne \vec 0 with ∣a⃗∣=a|\vec a| = a (so a>0a > 0⁠), and a scalar λ≠0\lambda \ne 0⁠.
To find
The condition among (A) to (D) under which λa⃗\lambda\vec a is a unit vector.
Plan
(1) Find ∣λa⃗∣|\lambda\vec a|⁠. (2) Set it equal to 1 and solve. (3) Test every option. (4) Try a concrete vector.

Basics you need

1Length of a scalar multiple

∣λa⃗∣=∣λ∣ ∣a⃗∣|\lambda\vec a| = |\lambda|\,|\vec a| (Section 10.5). Lesson: components and scalar multiples

2Unit vector

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

3Absolute value

∣λ∣=λ|\lambda| = \lambda if λ>0\lambda > 0 and ∣λ∣=−λ|\lambda| = -\lambda if λ<0\lambda < 0⁠; lengths use ∣λ∣|\lambda| because a length is never negative (school algebra).

Solution

  1. The length of λa⃗\lambda\vec a

    ∣λa⃗∣=∣λ∣ ∣a⃗∣=∣λ∣ a.|\lambda\vec a| = |\lambda|\,|\vec a| = |\lambda|\,a.
    λ\lambda stretches (or shrinks) the arrow by the factor ∣λ∣|\lambda|⁠, and a negative λ\lambda also reverses it; reversing does not change the length. That is why the factor is ∣λ∣|\lambda|⁠, not λ\lambda⁠.
    Section 10.5. Lesson: components and scalar multiples

  2. The unit condition

    ∣λa⃗∣=1  ⟺  ∣λ∣ a=1  ⟺  a=1∣λ∣.|\lambda\vec a| = 1 \iff |\lambda|\,a = 1 \iff a = \frac{1}{|\lambda|}.
    ∣λ∣>0|\lambda| > 0 since λ≠0\lambda \ne 0⁠, so we may divide by it. Each step can be reversed, so this condition is both necessary and sufficient. It is option (D).
    Unit vector (Section 10.3). Lesson: six kinds of vectors

  3. Test every option

    Take a⃗=2i^\vec a = 2\hat i⁠, so a=2a = 2⁠.

    (A) λ=1\lambda = 1⁠: λa⃗=2i^\lambda\vec a = 2\hat i has length 2, not 1. (A) fails; it works only when aa happens to be 1.

    (B) λ=−1\lambda = -1⁠: λa⃗=−2i^\lambda\vec a = -2\hat i has length 2, not 1. (B) fails for the same reason.

    (C) a=∣λ∣a = |\lambda|⁠: with λ=2\lambda = 2⁠, λa⃗=4i^\lambda\vec a = 4\hat i has length ∣λ∣ a=a2=4|\lambda|\,a = a^2 = 4⁠. (C) fails; it gives a unit vector only when a=1a = 1⁠.

    (D) a=1∣λ∣a = \frac{1}{|\lambda|}⁠: with λ=12\lambda = \frac12 or λ=−12\lambda = -\frac12⁠, λa⃗=±i^\lambda\vec a = \pm\hat i⁠, length 1. In general ∣λ∣ a=∣λ∣⋅1∣λ∣=1|\lambda|\,a = |\lambda|\cdot\frac{1}{|\lambda|} = 1 always. (D) is correct.

    An option is the answer only if it makes λa⃗\lambda\vec a a unit vector for every nonzero a⃗\vec a it allows. One vector where it fails rules an option out; (D) is exactly the condition found in Step 2.
    ∣λa⃗∣=∣λ∣∣a⃗∣|\lambda\vec a| = |\lambda||\vec a| (Section 10.5). Lesson: the multiple-choice questions

  4. A concrete example

    a⃗=2i^−j^+2k^\vec a = 2\hat i - \hat j + 2\hat k has a=4+1+4=3a = \sqrt{4 + 1 + 4} = 3⁠. Option (D) asks for ∣λ∣=13|\lambda| = \frac13⁠:

    13a⃗=23i^−13j^+23k^,∣13a⃗∣=4+1+49=1.\begin{aligned} \frac13\vec a &= \frac23\hat i - \frac13\hat j + \frac23\hat k, \\ \left|\frac13\vec a\right| &= \sqrt{\frac{4 + 1 + 4}{9}} = 1. \end{aligned}

    λ=−13\lambda = -\frac13 gives the opposite unit vector, also of length 1.

    a, length a = 3 λ = 1/3: length 1 λ = −1/3: length 1 a = 1/|λ| = 3
    One scale throughout: a⃗\vec a of length a=3a = 3⁠, and λa⃗\lambda\vec a for λ=13\lambda = \frac13 and λ=−13\lambda = -\frac13⁠, both of length ∣λ∣ a=1|\lambda|\,a = 1⁠. Both values satisfy a=1∣λ∣a = \frac{1}{|\lambda|}⁠.
    With λ=1a\lambda = \frac1a⁠, λa⃗\lambda\vec a is the unit vector a^=a⃗∣a⃗∣\hat a = \frac{\vec a}{|\vec a|} in the direction of a⃗\vec a⁠; with λ=−1a\lambda = -\frac1a it is −a^-\hat a⁠. These are the only two unit vectors along a⃗\vec a⁠.
    Unit vector in the direction of a⃗\vec a⁠: a^=1∣a⃗∣a⃗\hat a = \frac{1}{|\vec a|}\vec a (Section 10.5). Lesson: components and scalar multiples
Result

λa⃗\lambda\vec a is a unit vector exactly when ∣λ∣ a=1|\lambda|\,a = 1⁠, that is (D) a=1∣λ∣a = \frac{1}{|\lambda|}⁠.

(Textbook answer: (D).)

Check it with numbers

For a⃗=2i^−j^+2k^\vec a = 2\hat i - \hat j + 2\hat k and λ=13\lambda = \frac13⁠: the components of λa⃗\lambda\vec a are 23≈0.666667\frac23 \approx 0.666667⁠, −13≈−0.333333-\frac13 \approx -0.333333⁠, 23≈0.666667\frac23 \approx 0.666667⁠, and exactly 49+19+49=1\frac49 + \frac19 + \frac49 = 1⁠.

A wrong value, λ=2\lambda = 2 with the same a⃗\vec a⁠: ∣λa⃗∣=2×3=6|\lambda\vec a| = 2 \times 3 = 6⁠, not 1, and indeed a=3≠1∣2∣=0.5a = 3 \ne \frac{1}{|2|} = 0.5⁠.

Common mistakes
  1. Choosing (A). λ=1\lambda = 1 leaves a⃗\vec a unchanged, and a⃗\vec a need not be a unit vector.
  2. Writing a=1λa = \frac1\lambda⁠. For negative λ\lambda that makes the length negative; the condition needs ∣λ∣|\lambda|⁠.
  3. Choosing (C). a=∣λ∣a = |\lambda| gives length a2a^2⁠, which is 1 only in the special case a=1a = 1⁠.
What to take away
  1. ∣λa⃗∣=∣λ∣ ∣a⃗∣|\lambda\vec a| = |\lambda|\,|\vec a|⁠.
  2. λa⃗\lambda\vec a is a unit vector exactly when ∣λ∣=1∣a⃗∣|\lambda| = \frac{1}{|\vec a|}⁠; then λa⃗=±a^\lambda\vec a = \pm\hat a⁠.
  3. In an MCQ, one counterexample is enough to reject an option.

Every fact used, and where it comes from

FactUsed inSource
∣λa⃗∣=∣λ∣∣a⃗∣|\lambda\vec a| = |\lambda||\vec a|Steps 1, 3Section 10.5 (Lesson: components and scalar multiples)
Unit vector: magnitude 1Step 2Section 10.3 (Lesson: six kinds of vectors)
a^=1∣a⃗∣a⃗\hat a = \frac{1}{|\vec a|}\vec aStep 4Section 10.5
∣xi^+yj^+zk^∣=x2+y2+z2|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2}Step 4, CheckSection 10.5.1
The answerResultNCERT answer key, Exercise 10.3 Q18 (Lesson: the multiple-choice questions)
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