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Chapter 10 · Vector Algebra

Exercise 10.3 · Q10: λ\lambda making a⃗+λb⃗\vec a+\lambda\vec b perpendicular to c⃗\vec c

Exercise 10.3Perpendicular vectors, scalar product in components · Section 10.6.1
The question

If a⃗=2i^+2j^+3k^\vec a = 2\hat i + 2\hat j + 3\hat k⁠, b⃗=−i^+2j^+k^\vec b = -\hat i + 2\hat j + \hat k and c⃗=3i^+j^\vec c = 3\hat i + \hat j are such that a⃗+λb⃗\vec a + \lambda\vec b is perpendicular to c⃗\vec c⁠, then find the value of λ\lambda⁠.

Exam answer a⃗+λb⃗=(2−λ)i^+(2+2λ)j^+(3+λ)k^\begin{aligned} \vec a + \lambda\vec b &= (2 - \lambda)\hat i + (2 + 2\lambda)\hat j \\ &\quad + (3 + \lambda)\hat k \end{aligned}

Since a⃗+λb⃗\vec a + \lambda\vec b is perpendicular to c⃗\vec c⁠, (a⃗+λb⃗)⋅c⃗=0(\vec a + \lambda\vec b)\cdot\vec c = 0⁠.

⇒3(2−λ)+1(2+2λ)+0(3+λ)=0⇒6−3λ+2+2λ=0⇒8−λ=0\begin{aligned} &\Rightarrow 3(2 - \lambda) + 1(2 + 2\lambda) \\ &\quad + 0(3 + \lambda) = 0 \\ &\Rightarrow 6 - 3\lambda + 2 + 2\lambda = 0 \\ &\Rightarrow 8 - \lambda = 0 \end{aligned}

∴ λ=8\therefore\ \lambda = 8⁠.

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

The one idea this problem is built on

For nonzero vectors, perpendicular means scalar product zero. So write a⃗+λb⃗\vec a + \lambda\vec b in components, take its scalar product with c⃗\vec c⁠, and set the result equal to 0. That gives one linear equation in λ\lambda⁠.

Given, to find, and the plan
Given
a⃗=2i^+2j^+3k^\vec a = 2\hat i + 2\hat j + 3\hat k⁠, b⃗=−i^+2j^+k^\vec b = -\hat i + 2\hat j + \hat k⁠, c⃗=3i^+j^\vec c = 3\hat i + \hat j⁠, and a⃗+λb⃗\vec a + \lambda\vec b perpendicular to c⃗\vec c⁠.
To find
The value of λ\lambda⁠.
Plan
(1) Write a⃗+λb⃗\vec a + \lambda\vec b in components. (2) Turn "perpendicular" into (a⃗+λb⃗)⋅c⃗=0(\vec a + \lambda\vec b)\cdot\vec c = 0⁠. (3) Work out the scalar product and solve for λ\lambda⁠. (4) See the same equation a second way, and check by substituting back.

Basics you need

1Sums and multiples in components

λ(b1i^+b2j^+b3k^)=λb1i^+λb2j^+λb3k^\lambda(b_1\hat i + b_2\hat j + b_3\hat k) = \lambda b_1\hat i + \lambda b_2\hat j + \lambda b_3\hat k⁠, and vectors are added component by component (Section 10.5.1). Lesson: components

2Perpendicular means scalar product zero

For nonzero vectors, p⃗⋅q⃗=0\vec p\cdot\vec q = 0 exactly when p⃗⊥q⃗\vec p \perp \vec q (Section 10.6.1, Observation 2). Lesson: the dot product

3Scalar product in components

p⃗⋅q⃗=p1q1+p2q2+p3q3\vec p\cdot\vec q = p_1q_1 + p_2q_2 + p_3q_3 (Section 10.6.1). A missing component counts as 0. Basics: dot products obey the same rule

Solution

  1. Write a⃗+λb⃗\vec a + \lambda\vec b in components

    a⃗+λb⃗=(2i^+2j^+3k^)+λ(−i^+2j^+k^)=(2−λ)i^+(2+2λ)j^+(3+λ)k^.\begin{aligned} \vec a + \lambda\vec b &= (2\hat i + 2\hat j + 3\hat k) \\ &\quad + \lambda(-\hat i + 2\hat j + \hat k) \\ &= (2 - \lambda)\hat i + (2 + 2\lambda)\hat j \\ &\quad + (3 + \lambda)\hat k. \end{aligned}
    Multiplying b⃗\vec b by λ\lambda multiplies every component: λb⃗=−λi^+2λj^+λk^\lambda\vec b = -\lambda\hat i + 2\lambda\hat j + \lambda\hat k⁠. Then matching components are added: 2+(−λ)=2−λ2 + (-\lambda) = 2 - \lambda⁠, 2+2λ2 + 2\lambda and 3+λ3 + \lambda⁠.
    Sum and scalar multiple in components (Section 10.5.1). Lesson: components

  2. Turn "perpendicular" into an equation

    (a⃗+λb⃗)⋅c⃗=0.(\vec a + \lambda\vec b)\cdot\vec c = 0.
    For two nonzero vectors the scalar product is 0 exactly when they are perpendicular. Both vectors here are nonzero: ∣c⃗∣=9+1=10|\vec c| = \sqrt{9 + 1} = \sqrt{10}⁠, and a⃗+λb⃗\vec a + \lambda\vec b can never be 0⃗\vec 0⁠, because its i^\hat i and j^\hat j components would need λ=2\lambda = 2 and λ=−1\lambda = -1 at the same time. So "perpendicular" and "scalar product 0" say exactly the same thing.
    Section 10.6.1, Observation 2. Lesson: the dot product

  3. Work out the scalar product and solve

    Write c⃗=3i^+1j^+0k^\vec c = 3\hat i + 1\hat j + 0\hat k⁠. Then

    (a⃗+λb⃗)⋅c⃗=3(2−λ)+1(2+2λ)+0(3+λ)=6−3λ+2+2λ=8−λ.\begin{aligned} (\vec a + \lambda\vec b)\cdot\vec c &= 3(2 - \lambda) + 1(2 + 2\lambda) \\ &\quad + 0(3 + \lambda) \\ &= 6 - 3\lambda + 2 + 2\lambda \\ &= 8 - \lambda. \end{aligned} 8−λ=0  ⇒  λ=8.8 - \lambda = 0 \;\Rightarrow\; \lambda = 8.

    So λ=8\lambda = 8.

    The scalar product multiplies matching components and adds. c⃗\vec c has no k^\hat k term, so its k^\hat k component is 0 and the third product drops out. The λ\lambda terms combine as −3λ+2λ=−λ-3\lambda + 2\lambda = -\lambda⁠, leaving a linear equation with exactly one solution.
    p⃗⋅q⃗=p1q1+p2q2+p3q3\vec p\cdot\vec q = p_1q_1 + p_2q_2 + p_3q_3 (Section 10.6.1). Basics: dot products obey the same rule

  4. The same equation, a second way

    Distributivity splits the scalar product before any components are combined:

    (a⃗+λb⃗)⋅c⃗=a⃗⋅c⃗+λ (b⃗⋅c⃗),(\vec a + \lambda\vec b)\cdot\vec c = \vec a\cdot\vec c + \lambda\,(\vec b\cdot\vec c),

    with a⃗⋅c⃗=2(3)+2(1)+3(0)=8\vec a\cdot\vec c = 2(3) + 2(1) + 3(0) = 8 and b⃗⋅c⃗=(−1)(3)+2(1)+1(0)=−1\vec b\cdot\vec c = (-1)(3) + 2(1) + 1(0) = -1⁠. So again (a⃗+λb⃗)⋅c⃗=8−λ(\vec a + \lambda\vec b)\cdot\vec c = 8 - \lambda⁠.

    2 4 6 10 2 4 6 8 λ (a + λb)·c a·c = 8 λ = 8 (a + λb)·c = 8 − λ starts at a·c = 8 slope b·c = −1 zero only at λ = 8
    The scalar product (a⃗+λb⃗)⋅c⃗=8−λ(\vec a + \lambda\vec b)\cdot\vec c = 8 - \lambda plotted against λ\lambda (one scale on both axes): a straight line that starts at a⃗⋅c⃗=8\vec a\cdot\vec c = 8⁠, falls by ∣b⃗⋅c⃗∣=1|\vec b\cdot\vec c| = 1 for each unit of λ\lambda⁠, and crosses 0 only at λ=8\lambda = 8⁠.
    The scalar product is linear in λ\lambda⁠, so its graph is a straight line: the value at λ=0\lambda = 0 is a⃗⋅c⃗\vec a\cdot\vec c and the slope is b⃗⋅c⃗\vec b\cdot\vec c⁠. A line with nonzero slope crosses 0 exactly once, which is why there is exactly one λ\lambda⁠.
    Distributivity and scalars coming out (Section 10.6.1, Properties 1 and 2). Basics: dot products obey the same rule

  5. Check by substituting back

    a⃗+8b⃗=(2−8)i^+(2+16)j^+(3+8)k^=−6i^+18j^+11k^,\begin{aligned} \vec a + 8\vec b &= (2 - 8)\hat i + (2 + 16)\hat j \\ &\quad + (3 + 8)\hat k \\ &= -6\hat i + 18\hat j + 11\hat k, \end{aligned} (−6i^+18j^+11k^)⋅(3i^+j^)=−18+18+0=0.\begin{aligned} &(-6\hat i + 18\hat j + 11\hat k)\cdot(3\hat i + \hat j) \\ &= -18 + 18 + 0 = 0. \end{aligned}
    With λ=8\lambda = 8 the scalar product is exactly 0, and both vectors are nonzero, so a⃗+8b⃗\vec a + 8\vec b is indeed perpendicular to c⃗\vec c⁠.
    Section 10.6.1, Observation 2. Lesson: the dot product exercises
Result

λ=8\lambda = 8⁠, and then a⃗+8b⃗=−6i^+18j^+11k^\vec a + 8\vec b = -6\hat i + 18\hat j + 11\hat k is perpendicular to c⃗=3i^+j^\vec c = 3\hat i + \hat j⁠.

(Textbook answer: 8.)

Check it with numbers

Values of λ\lambda on either side of 8: λ=7\lambda = 7 gives a⃗+7b⃗=−5i^+16j^+10k^\vec a + 7\vec b = -5\hat i + 16\hat j + 10\hat k and a scalar product −15+16=1-15 + 16 = 1⁠; λ=9\lambda = 9 gives −7i^+20j^+12k^-7\hat i + 20\hat j + 12\hat k and −21+20=−1-21 + 20 = -1⁠. The value changes sign at λ=8\lambda = 8⁠, as 8−λ8 - \lambda says.

At λ=8\lambda = 8⁠: ∣a⃗+8b⃗∣=36+324+121=481≈21.931712|\vec a + 8\vec b| = \sqrt{36 + 324 + 121} = \sqrt{481} \approx 21.931712 and ∣c⃗∣=10≈3.162278|\vec c| = \sqrt{10} \approx 3.162278⁠, both nonzero, and cos⁡θ=0481 10=0\cos\theta = \frac{0}{\sqrt{481}\,\sqrt{10}} = 0⁠, so θ=90∘\theta = 90^\circ⁠.

Common mistakes
  1. Losing the minus sign of −i^-\hat i⁠. λ(−i^)=−λi^\lambda(-\hat i) = -\lambda\hat i⁠, so the i^\hat i component is 2−λ2 - \lambda⁠, not 2+λ2 + \lambda⁠. With 2+λ2 + \lambda one gets 8+5λ=08 + 5\lambda = 0 and the wrong λ=−85\lambda = -\frac85⁠.
  2. Giving c⃗\vec c a k^\hat k component. c⃗=3i^+j^\vec c = 3\hat i + \hat j has k^\hat k component 0, not 1.
  3. Multiplying only part of b⃗\vec b by λ\lambda⁠. Every component of b⃗\vec b is multiplied: λb⃗=−λi^+2λj^+λk^\lambda\vec b = -\lambda\hat i + 2\lambda\hat j + \lambda\hat k⁠.
What to take away
  1. "Perpendicular" (for nonzero vectors) becomes the equation "scalar product = 0".
  2. Build the combination a⃗+λb⃗\vec a + \lambda\vec b in components first, then take the scalar product.
  3. (a⃗+λb⃗)⋅c⃗=a⃗⋅c⃗+λ(b⃗⋅c⃗)(\vec a + \lambda\vec b)\cdot\vec c = \vec a\cdot\vec c + \lambda(\vec b\cdot\vec c) is linear in λ\lambda⁠, so there is one answer whenever b⃗⋅c⃗≠0\vec b\cdot\vec c \ne 0⁠.

Every fact used, and where it comes from

FactUsed inSource
Scalar multiple and sum in componentsSteps 1, 5Section 10.5.1 (Lesson: components)
Nonzero p⃗\vec p⁠, q⃗\vec q⁠: p⃗⋅q⃗=0\vec p\cdot\vec q = 0 exactly when p⃗⊥q⃗\vec p \perp \vec qSteps 2, 5Section 10.6.1, Observation 2 (Lesson: the dot product)
p⃗⋅q⃗=p1q1+p2q2+p3q3\vec p\cdot\vec q = p_1q_1 + p_2q_2 + p_3q_3Steps 3, 4, 5Section 10.6.1 (Basics: dot products obey the same rule)
Distributivity; scalars come out of a scalar productStep 4Section 10.6.1, Properties 1 and 2
cos⁡θ=p⃗⋅q⃗∣p⃗∣∣q⃗∣\cos\theta = \frac{\vec p\cdot\vec q}{|\vec p||\vec q|}CheckSection 10.6.1, Observation 6
The answerResultNCERT answer key, Exercise 10.3 Q10 (Lesson: the dot product exercises)
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