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Chapter 10 · Vector Algebra

Miscellaneous · Q12: d⃗\vec d perpendicular to a⃗\vec a and b⃗\vec b with c⃗⋅d⃗=15\vec c\cdot\vec d=15

Miscellaneous ExerciseVector product, scalar product · Sections 10.6.1 and 10.6.3
The question

Let a⃗=i^+4j^+2k^\vec a = \hat i + 4\hat j + 2\hat k⁠, b⃗=3i^−2j^+7k^\vec b = 3\hat i - 2\hat j + 7\hat k and c⃗=2i^−j^+4k^\vec c = 2\hat i - \hat j + 4\hat k⁠. Find a vector d⃗\vec d which is perpendicular to both a⃗\vec a and b⃗\vec b⁠, and c⃗⋅d⃗=15\vec c\cdot\vec d = 15⁠.

Exam answer

Since d⃗\vec d is perpendicular to both a⃗\vec a and b⃗\vec b⁠, d⃗\vec d is parallel to a⃗×b⃗\vec a\times\vec b⁠.

a⃗×b⃗=∣i^j^k^1423−27∣=i^(28+4)−j^(7−6)+k^(−2−12)=32i^−j^−14k^\begin{aligned} \vec a\times\vec b &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 4 & 2 \\ 3 & -2 & 7 \end{vmatrix} \\ &= \hat i(28 + 4) - \hat j(7 - 6) \\ &\quad + \hat k(-2 - 12) \\ &= 32\hat i - \hat j - 14\hat k \end{aligned}

Let d⃗=λ(32i^−j^−14k^)\vec d = \lambda(32\hat i - \hat j - 14\hat k)⁠.

Given c⃗⋅d⃗=15\vec c\cdot\vec d = 15⁠:

λ[2(32)+(−1)(−1)+4(−14)]=15⇒λ(64+1−56)=15⇒9λ=15⇒λ=53\begin{aligned} &\lambda[2(32) + (-1)(-1) \\ &\quad + 4(-14)] = 15 \\ &\Rightarrow \lambda(64 + 1 - 56) = 15 \\ &\Rightarrow 9\lambda = 15 \Rightarrow \lambda = \frac53 \end{aligned} d⃗=53(32i^−j^−14k^)\vec d = \frac53(32\hat i - \hat j - 14\hat k)

∴ d⃗=13(160i^−5j^−70k^)\therefore\ \vec d = \dfrac13(160\hat i - 5\hat j - 70\hat k)⁠.

The answer key prints 13(160i^−5j^+70k^)\frac13(160\hat i - 5\hat j + 70\hat k)⁠; with +70k^+70\hat k the vector is not perpendicular to a⃗\vec a or b⃗\vec b⁠, so the k^\hat k term is −70k^-70\hat k⁠.

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

The one idea this problem is built on

The vector product a⃗×b⃗\vec a\times\vec b is perpendicular to both a⃗\vec a and b⃗\vec b⁠, and every vector perpendicular to both is a multiple of it: d⃗=λ(a⃗×b⃗)\vec d = \lambda(\vec a\times\vec b)⁠. The extra condition c⃗⋅d⃗=15\vec c\cdot\vec d = 15 is one equation for the one unknown λ\lambda⁠.

Given, to find, and the plan
Given
a⃗=i^+4j^+2k^\vec a = \hat i + 4\hat j + 2\hat k⁠, b⃗=3i^−2j^+7k^\vec b = 3\hat i - 2\hat j + 7\hat k⁠, c⃗=2i^−j^+4k^\vec c = 2\hat i - \hat j + 4\hat k⁠.
To find
d⃗\vec d with d⃗⊥a⃗\vec d\perp\vec a⁠, d⃗⊥b⃗\vec d\perp\vec b and c⃗⋅d⃗=15\vec c\cdot\vec d = 15⁠.
Plan
(1) See why d⃗=λ(a⃗×b⃗)\vec d = \lambda(\vec a\times\vec b)⁠. (2) Compute a⃗×b⃗\vec a\times\vec b and check it. (3) Find λ\lambda from c⃗⋅d⃗=15\vec c\cdot\vec d = 15⁠. (4) Write d⃗\vec d and check all three conditions. (5) Compare with the answer key.

Basics you need

1Vector product in components

a⃗×b⃗\vec a\times\vec b is the determinant with rows i^,j^,k^\hat i, \hat j, \hat k⁠; a1,a2,a3a_1, a_2, a_3⁠; b1,b2,b3b_1, b_2, b_3⁠, and it is perpendicular to both a⃗\vec a and b⃗\vec b (Section 10.6.3). Basics: a determinant measures a signed area

2Scalar product in components

(a1i^+a2j^+a3k^)⋅(b1i^+b2j^+b3k^)(a_1\hat i + a_2\hat j + a_3\hat k)\cdot(b_1\hat i + b_2\hat j + b_3\hat k) =a1b1+a2b2+a3b3= a_1b_1 + a_2b_2 + a_3b_3⁠, and u⃗⋅v⃗=0\vec u\cdot\vec v = 0 means u⃗⊥v⃗\vec u\perp\vec v for nonzero vectors (Section 10.6.1). Lesson: the dot product

3Scalars come out of a scalar product

c⃗⋅(λn⃗)=λ(c⃗⋅n⃗)\vec c\cdot(\lambda\vec n) = \lambda(\vec c\cdot\vec n) (Section 10.6.1, Property 2). Basics: dot products obey the same rule

Solution

  1. Why d⃗\vec d must be a multiple of a⃗×b⃗\vec a\times\vec b

    a⃗\vec a and b⃗\vec b are not parallel (their components are not in proportion: 31≠−24\frac31 \ne \frac{-2}{4}⁠). So they lie along two different lines through O and fix one plane. A vector perpendicular to both is perpendicular to that plane, so it lies along the normal line of the plane, and a⃗×b⃗\vec a\times\vec b points along that normal. Therefore

    d⃗=λ(a⃗×b⃗)  for some scalar λ.\vec d = \lambda(\vec a\times\vec b)\ \text{ for some scalar } \lambda.
    The perpendicularity conditions alone do not fix d⃗\vec d⁠: any multiple of a⃗×b⃗\vec a\times\vec b satisfies them. They fix only the line of d⃗\vec d⁠. The third condition, c⃗⋅d⃗=15\vec c\cdot\vec d = 15⁠, then picks one vector on that line.
    a⃗×b⃗\vec a\times\vec b is perpendicular to both a⃗\vec a and b⃗\vec b (Definition 3, Section 10.6.3). Lesson: the cross product

  2. Compute a⃗×b⃗\vec a\times\vec b

    a⃗×b⃗=∣i^j^k^1423−27∣=i^(4⋅7−2⋅(−2))−j^(1⋅7−2⋅3)+k^(1⋅(−2)−4⋅3)=32i^−j^−14k^.\begin{aligned} \vec a\times\vec b &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 4 & 2 \\ 3 & -2 & 7 \end{vmatrix} \\ &= \hat i(4 \cdot 7 - 2 \cdot (-2)) \\ &\quad - \hat j(1 \cdot 7 - 2 \cdot 3) \\ &\quad + \hat k(1 \cdot (-2) - 4 \cdot 3) \\ &= 32\hat i - \hat j - 14\hat k. \end{aligned}

    Check: (a⃗×b⃗)⋅a⃗=32−4−28=0(\vec a\times\vec b)\cdot\vec a = 32 - 4 - 28 = 0 and (a⃗×b⃗)⋅b⃗=96+2−98=0(\vec a\times\vec b)\cdot\vec b = 96 + 2 - 98 = 0⁠.

    Expanding along the first row: the i^\hat i entry is the 2×22 \times 2 determinant left when its row and column are covered, and so on, with the sign pattern +,−,++, -, +⁠. The middle term carries the minus sign, which is where most slips happen. The two scalar products confirm that the result is perpendicular to both given vectors.
    Vector product in components (Section 10.6.3). Basics: a determinant measures a signed area

  3. Find λ\lambda from c⃗⋅d⃗=15\vec c\cdot\vec d = 15

    c⃗⋅d⃗=λ c⃗⋅(32i^−j^−14k^)=λ[2(32)+(−1)(−1)+4(−14)]=λ(64+1−56)=9λ.\begin{aligned} \vec c\cdot\vec d &= \lambda\,\vec c\cdot(32\hat i - \hat j - 14\hat k) \\ &= \lambda[2(32) + (-1)(-1) + 4(-14)] \\ &= \lambda(64 + 1 - 56) = 9\lambda. \end{aligned} 9λ=15⇒λ=159=53.9\lambda = 15 \Rightarrow \lambda = \frac{15}{9} = \frac53.
    The scalar λ\lambda comes out of the scalar product, so the condition becomes a linear equation in λ\lambda⁠. It has exactly one solution because c⃗⋅(a⃗×b⃗)=9≠0\vec c\cdot(\vec a\times\vec b) = 9 \ne 0⁠: c⃗\vec c is not perpendicular to the line of d⃗\vec d⁠.
    Scalar product in components; Property 2 (Section 10.6.1). Lesson: the dot product

  4. Write d⃗\vec d and check all three conditions

    d⃗=53(32i^−j^−14k^)=13(160i^−5j^−70k^).\begin{aligned} \vec d &= \frac53(32\hat i - \hat j - 14\hat k) \\ &= \frac13(160\hat i - 5\hat j - 70\hat k). \end{aligned}

    So d⃗=13(160i^−5j^−70k^)\vec d = \frac13(160\hat i - 5\hat j - 70\hat k). Check each condition with the bracket 160i^−5j^−70k^160\hat i - 5\hat j - 70\hat k and the factor 13\frac13⁠:

    d⃗⋅a⃗=13(160−20−140)=0,d⃗⋅b⃗=13(480+10−490)=0,c⃗⋅d⃗=13(320+5−280)=15.\begin{aligned} \vec d\cdot\vec a &= \tfrac13(160 - 20 - 140) = 0, \\ \vec d\cdot\vec b &= \tfrac13(480 + 10 - 490) = 0, \\ \vec c\cdot\vec d &= \tfrac13(320 + 5 - 280) = 15. \end{aligned}
    All three conditions of the question hold, so this d⃗\vec d is the answer. It is the only one: the line of d⃗\vec d was forced in Step 1, and λ\lambda was forced in Step 3.
    Scalar product in components; u⃗⋅v⃗=0\vec u\cdot\vec v = 0 for perpendicular vectors (Section 10.6.1, Observation 2). Lesson: the miscellaneous exercises

  5. Compare with the answer key

    The answer key prints 13(160i^−5j^+70k^)\frac13(160\hat i - 5\hat j + 70\hat k)⁠, with +70k^+70\hat k⁠. For that vector, with the same factor 13\frac13⁠:

    13(160−20+140)=2803≠0,13(480+10+490)=9803≠0,\begin{aligned} &\tfrac13(160 - 20 + 140) = \tfrac{280}{3} \ne 0, \\ &\tfrac13(480 + 10 + 490) = \tfrac{980}{3} \ne 0, \end{aligned}

    so it is perpendicular to neither a⃗\vec a nor b⃗\vec b⁠. With −70k^-70\hat k every condition holds (Step 4), so the sign of the k^\hat k term in the key should be minus.

    An answer key value can always be tested against the conditions of the question. Here only the k^\hat k sign differs, and the k^\hat k component of a⃗×b⃗\vec a\times\vec b is −14-14⁠, so the k^\hat k component of d⃗\vec d must be negative too.
    u⃗⋅v⃗=0\vec u\cdot\vec v = 0 if and only if u⃗⊥v⃗\vec u\perp\vec v⁠, for nonzero vectors (Section 10.6.1, Observation 2). Lesson: the dot product
Result

d⃗=13(160i^−5j^−70k^)\vec d = \frac13(160\hat i - 5\hat j - 70\hat k)⁠, that is 1603i^−53j^−703k^\frac{160}{3}\hat i - \frac53\hat j - \frac{70}{3}\hat k⁠.

(Textbook answer: printed as 13(160i^−5j^+70k^)\frac13(160\hat i - 5\hat j + 70\hat k)⁠. With +70k^+70\hat k the vector is not perpendicular to a⃗\vec a or b⃗\vec b⁠; the correct answer is 13(160i^−5j^−70k^)\frac13(160\hat i - 5\hat j - 70\hat k)⁠.)

Check it with numbers

In decimals, d⃗≈53.333333i^−1.666667j^−23.333333k^\vec d \approx 53.333333\hat i - 1.666667\hat j - 23.333333\hat k⁠.

∣a⃗×b⃗∣=1024+1+196=1221≈34.942810|\vec a\times\vec b| = \sqrt{1024 + 1 + 196} = \sqrt{1221} \approx 34.942810⁠. A second route to the same number: ∣a⃗×b⃗∣2=∣a⃗∣2∣b⃗∣2sin⁡2θ|\vec a\times\vec b|^2 = |\vec a|^2|\vec b|^2\sin^2\theta =∣a⃗∣2∣b⃗∣2−(a⃗⋅b⃗)2= |\vec a|^2|\vec b|^2 - (\vec a\cdot\vec b)^2⁠, with ∣a⃗∣2=21|\vec a|^2 = 21⁠, ∣b⃗∣2=62|\vec b|^2 = 62 and a⃗⋅b⃗=3−8+14=9\vec a\cdot\vec b = 3 - 8 + 14 = 9⁠; exactly, 21×62−92=1302−81=122121 \times 62 - 9^2 = 1302 - 81 = 1221⁠.

∣d⃗∣=531221≈58.238017|\vec d| = \frac53\sqrt{1221} \approx 58.238017⁠. The component of c⃗\vec c along d⃗\vec d is 15∣d⃗∣≈0.257564\frac{15}{|\vec d|} \approx 0.257564⁠, small, because c⃗\vec c is nearly perpendicular to d⃗\vec d⁠; that is why d⃗\vec d has to be so long.

Common mistakes
  1. Losing the minus sign of the j^\hat j term. The middle term of the determinant is −j^(7−6)=−j^-\hat j(7 - 6) = -\hat j⁠.
  2. Taking d⃗=a⃗×b⃗\vec d = \vec a\times\vec b itself. Then c⃗⋅d⃗=9\vec c\cdot\vec d = 9⁠, not 15; the multiple λ\lambda is needed.
  3. Not checking the answer. Testing d⃗⋅a⃗=0\vec d\cdot\vec a = 0⁠, d⃗⋅b⃗=0\vec d\cdot\vec b = 0⁠, c⃗⋅d⃗=15\vec c\cdot\vec d = 15 takes a minute and catches sign slips, including the one in the answer key.
What to take away
  1. Perpendicular to both a⃗\vec a and b⃗\vec b means d⃗=λ(a⃗×b⃗)\vec d = \lambda(\vec a\times\vec b)⁠.
  2. One more scalar condition fixes λ\lambda⁠.
  3. Always substitute the final vector back into every condition.

Every fact used, and where it comes from

FactUsed inSource
a⃗×b⃗\vec a\times\vec b is perpendicular to a⃗\vec a and b⃗\vec bStep 1Definition 3, Section 10.6.3 (Lesson: the cross product)
Vector product as a determinantStep 2Section 10.6.3 (Basics: a determinant measures a signed area)
Scalar product in componentsSteps 2 to 5, CheckSection 10.6.1
c⃗⋅(λn⃗)=λ(c⃗⋅n⃗)\vec c\cdot(\lambda\vec n) = \lambda(\vec c\cdot\vec n)Step 3Section 10.6.1, Property 2
u⃗⋅v⃗=0  ⟺  u⃗⊥v⃗\vec u\cdot\vec v = 0 \iff \vec u\perp\vec v (nonzero vectors)Steps 4, 5Section 10.6.1, Observation 2
∣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 Q12, with the k^\hat k sign corrected (Lesson: the miscellaneous exercises)
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