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Chapter 10 · Vector Algebra

Miscellaneous · Q18 (MCQ): Value of i^⋅(j^×k^)\hat i\cdot(\hat j\times\hat k) + j^⋅(i^×k^)+\,\hat j\cdot(\hat i\times\hat k) + k^⋅(i^×j^)+\,\hat k\cdot(\hat i\times\hat j)

Miscellaneous ExerciseVector products of i, j, k · Sections 10.6.1 and 10.6.3
The question

The value of i^.(j^×k^)+j^⋅(i^×k^)+k^⋅(i^×j^)\hat i.(\hat j\times\hat k) + \hat j\cdot(\hat i\times\hat k) + \hat k\cdot(\hat i\times\hat j) is

(A) 00

(B) −1-1

(C) 11

(D) 33

Exam answer

j^×k^=i^\hat j\times\hat k = \hat i⁠, i^×k^=−j^\hat i\times\hat k = -\hat j⁠, i^×j^=k^\hat i\times\hat j = \hat k⁠.

i^⋅(j^×k^)+j^⋅(i^×k^)+k^⋅(i^×j^)=i^⋅i^+j^⋅(−j^)+k^⋅k^=1−1+1=1\begin{aligned} &\hat i\cdot(\hat j\times\hat k) + \hat j\cdot(\hat i\times\hat k) \\ &\quad + \hat k\cdot(\hat i\times\hat j) \\ &= \hat i\cdot\hat i + \hat j\cdot(-\hat j) + \hat k\cdot\hat k \\ &= 1 - 1 + 1 = 1 \end{aligned}

Answer: (C) 1

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

The one idea this problem is built on

Each bracket is a vector product of two of i^,j^,k^\hat i, \hat j, \hat k⁠, which is the third one with a sign: ++ when the pair follows the cycle i^→j^→k^→i^\hat i \to \hat j \to \hat k \to \hat i⁠, −- when it goes against it. After that, each term is a scalar product of a unit vector with ±\pm itself, so it is ±1\pm 1⁠.

Given, to find, and the plan
Given
The mutually perpendicular unit vectors i^,j^,k^\hat i, \hat j, \hat k of a right handed system.
To find
i^⋅(j^×k^)+j^⋅(i^×k^)+k^⋅(i^×j^)\hat i\cdot(\hat j\times\hat k) + \hat j\cdot(\hat i\times\hat k) + \hat k\cdot(\hat i\times\hat j)⁠, among (A) to (D).
Plan
(1) Find the three vector products. (2) Take the scalar products and add. (3) Examine every option. (4) Check with the determinant formula.

Basics you need

1Vector products of i^,j^,k^\hat i, \hat j, \hat k

i^×j^=k^\hat i\times\hat j = \hat k⁠, j^×k^=i^\hat j\times\hat k = \hat i⁠, k^×i^=j^\hat k\times\hat i = \hat j (Observation 4), and reversing the order changes the sign: i^×k^=−j^\hat i\times\hat k = -\hat j (Observation 7), Section 10.6.3. Lesson: the cross product

2Scalar products of i^,j^,k^\hat i, \hat j, \hat k

i^⋅i^=j^⋅j^=k^⋅k^=1\hat i\cdot\hat i = \hat j\cdot\hat j = \hat k\cdot\hat k = 1 (Section 10.6.1, Observation 5). Lesson: the dot product

3Scalars come out of a scalar product

a⃗⋅(−b⃗)=−(a⃗⋅b⃗)\vec a\cdot(-\vec b) = -(\vec a\cdot\vec b) (Section 10.6.1, Property 2). Basics: some answers need a "which way"

Solution

  1. The three vector products

    j^×k^=i^,i^×k^=−(k^×i^)=−j^,i^×j^=k^.\begin{aligned} \hat j\times\hat k &= \hat i, \\ \hat i\times\hat k &= -(\hat k\times\hat i) = -\hat j, \\ \hat i\times\hat j &= \hat k. \end{aligned}
    î ĵ k̂ along the arrows: + î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ against the arrows: − î × k̂ = −ĵ sum = 1 − 1 + 1 = 1
    The cycle i^→j^→k^→i^\hat i \to \hat j \to \hat k \to \hat i⁠. A product of two of them taken along the arrows gives the third with a plus sign; taken against the arrows, with a minus sign. In this question j^×k^\hat j\times\hat k and i^×j^\hat i\times\hat j go along the cycle, and i^×k^\hat i\times\hat k goes against it.
    j^→k^\hat j \to \hat k and i^→j^\hat i \to \hat j follow the cycle, so those products are +i^+\hat i and +k^+\hat k⁠. The pair i^,k^\hat i, \hat k goes against it (the cycle runs k^→i^\hat k \to \hat i⁠), and the vector product is not commutative: i^×k^=−k^×i^=−j^\hat i\times\hat k = -\hat k\times\hat i = -\hat j⁠.
    Section 10.6.3, Observations 4, 6 and 7. Lesson: the cross product

  2. The scalar products, and the sum

    i^⋅(j^×k^)=i^⋅i^=1,j^⋅(i^×k^)=j^⋅(−j^)=−1,k^⋅(i^×j^)=k^⋅k^=1,\begin{aligned} \hat i\cdot(\hat j\times\hat k) &= \hat i\cdot\hat i = 1, \\ \hat j\cdot(\hat i\times\hat k) &= \hat j\cdot(-\hat j) = -1, \\ \hat k\cdot(\hat i\times\hat j) &= \hat k\cdot\hat k = 1, \end{aligned} 1+(−1)+1=1.1 + (-1) + 1 = 1.

    So the value is 1.

    Each unit vector dotted with itself gives its length squared, 1. The minus sign of −j^-\hat j comes out of the scalar product, so the middle term is −1-1⁠. The book prints the first dot as "." and the others as "⋅\cdot"; both mean the scalar product.
    Section 10.6.1, Observation 5 and Property 2. Lesson: the dot product

  3. Examine every option

    (A) 0 and (B) −1-1⁠: the three terms are 1,−1,11, -1, 1⁠, whose sum is 1, not 0 or −1-1⁠. (A) and (B) fail.

    (C) 1: the value found in Step 2. (C) is correct.

    (D) 3: this is what comes out if i^×k^\hat i\times\hat k is taken as +j^+\hat j⁠, making all three terms 1. (D) fails.

    Only one of the three terms can go wrong in an interesting way: the middle one, whose pair is against the cycle. Getting its sign right is the whole question.
    Section 10.6.3, Observation 7. Lesson: the multiple-choice questions

  4. Check with the determinant formula

    Writing i^=(1,0,0)\hat i = (1, 0, 0)⁠, j^=(0,1,0)\hat j = (0, 1, 0)⁠, k^=(0,0,1)\hat k = (0, 0, 1) in the component formula,

    i^×k^=∣i^j^k^100001∣=0i^−1j^+0k^=−j^.\begin{aligned} \hat i\times\hat k &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{vmatrix} \\ &= 0\hat i - 1\hat j + 0\hat k = -\hat j. \end{aligned}

    In the same way j^×k^=i^\hat j\times\hat k = \hat i and i^×j^=k^\hat i\times\hat j = \hat k⁠, which confirms Step 1 and the value 1.

    The determinant formula does not depend on remembering the cycle, so it is an independent check of the one sign that matters: the j^\hat j entry is −(1⋅1−0⋅0)=−1-(1 \cdot 1 - 0 \cdot 0) = -1⁠.
    Vector product in components (Section 10.6.3). Basics: a determinant measures a signed area
Result

The value is 1−1+1=11 - 1 + 1 = 1⁠, option (C) 1.

(Textbook answer: (C).)

Check it with numbers

Sizes first: each bracket is the vector product of two perpendicular unit vectors, of length 1×1×sin⁡90∘=11 \times 1 \times \sin 90^\circ = 1⁠, along the third axis. So each term is +1+1 or −1-1⁠, and a sum of three such terms is odd: it can never be 0, which rules out (A) before any sign is found.

Signs: +1+1⁠, −1-1⁠, +1+1 from Step 2, so the sum is 1. All values here are exact integers; no decimals are involved.

Common mistakes
  1. Taking i^×k^=j^\hat i\times\hat k = \hat j⁠. The order is against the cycle, so i^×k^=−j^\hat i\times\hat k = -\hat j⁠; this slip gives 3.
  2. Treating every term as 0. i^⋅(j^×k^)\hat i\cdot(\hat j\times\hat k) is not i^⋅j^×k^\hat i\cdot\hat j\times\hat k read as (i^⋅j^)×k^(\hat i\cdot\hat j)\times\hat k⁠; the bracket is computed first, and a scalar cannot be crossed with a vector.
  3. Losing the sign in the scalar product. j^⋅(−j^)=−(j^⋅j^)=−1\hat j\cdot(-\hat j) = -(\hat j\cdot\hat j) = -1⁠.
What to take away
  1. i^×j^=k^\hat i\times\hat j = \hat k⁠, j^×k^=i^\hat j\times\hat k = \hat i⁠, k^×i^=j^\hat k\times\hat i = \hat j⁠; reversing the order changes the sign.
  2. Read the cycle i^→j^→k^→i^\hat i \to \hat j \to \hat k \to \hat i⁠: along it ++⁠, against it −-⁠.
  3. In u⃗⋅(v⃗×w⃗)\vec u\cdot(\vec v\times\vec w)⁠, work out the bracket first.

Every fact used, and where it comes from

FactUsed inSource
i^×j^=k^\hat i\times\hat j = \hat k⁠, j^×k^=i^\hat j\times\hat k = \hat i⁠, k^×i^=j^\hat k\times\hat i = \hat jStep 1Section 10.6.3, Observation 4 (Lesson: the cross product)
b⃗×a⃗=−a⃗×b⃗\vec b\times\vec a = -\vec a\times\vec b⁠; i^×k^=−j^\hat i\times\hat k = -\hat jSteps 1, 3Section 10.6.3, Observations 6 and 7
i^⋅i^=j^⋅j^=k^⋅k^=1\hat i\cdot\hat i = \hat j\cdot\hat j = \hat k\cdot\hat k = 1Step 2Section 10.6.1, Observation 5
a⃗⋅(λb⃗)=λ(a⃗⋅b⃗)\vec a\cdot(\lambda\vec b) = \lambda(\vec a\cdot\vec b)Step 2Section 10.6.1, Property 2
Vector product as a determinantStep 4Section 10.6.3 (Basics: a determinant measures a signed area)
∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡θ|\vec a\times\vec b| = |\vec a||\vec b|\sin\thetaCheckDefinition 3, Section 10.6.3
The answerResultNCERT answer key, Miscellaneous Exercise Q18 (Lesson: the multiple-choice questions)
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