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Chapter 10 · Vector Algebra

Miscellaneous · Q13: λ\lambda from the scalar product of i^+j^+k^\hat i+\hat j+\hat k with a unit vector along a sum

Miscellaneous ExerciseUnit vector, scalar product, projection · Sections 10.5 and 10.6
The question

The scalar product of the vector i^+j^+k^\hat i + \hat j + \hat k with a unit vector along the sum of vectors 2i^+4j^−5k^2\hat i + 4\hat j - 5\hat k and λi^+2j^+3k^\lambda\hat i + 2\hat j + 3\hat k is equal to one. Find the value of λ\lambda⁠.

Exam answer

Let b⃗=2i^+4j^−5k^\vec b = 2\hat i + 4\hat j - 5\hat k and c⃗=λi^+2j^+3k^\vec c = \lambda\hat i + 2\hat j + 3\hat k⁠.

b⃗+c⃗=(2+λ)i^+6j^−2k^∣b⃗+c⃗∣=(2+λ)2+36+4=λ2+4λ+44\begin{aligned} \vec b + \vec c &= (2 + \lambda)\hat i + 6\hat j - 2\hat k \\ |\vec b + \vec c| &= \sqrt{(2 + \lambda)^2 + 36 + 4} \\ &= \sqrt{\lambda^2 + 4\lambda + 44} \end{aligned}

The unit vector along b⃗+c⃗\vec b + \vec c is

n^=(2+λ)i^+6j^−2k^λ2+4λ+44\hat n = \frac{(2 + \lambda)\hat i + 6\hat j - 2\hat k}{\sqrt{\lambda^2 + 4\lambda + 44}}

Given (i^+j^+k^)⋅n^=1(\hat i + \hat j + \hat k)\cdot\hat n = 1⁠:

(2+λ)+6−2λ2+4λ+44=1⇒λ+6=λ2+4λ+44\begin{aligned} &\frac{(2 + \lambda) + 6 - 2}{\sqrt{\lambda^2 + 4\lambda + 44}} = 1 \\ &\Rightarrow \lambda + 6 = \sqrt{\lambda^2 + 4\lambda + 44} \end{aligned}

Squaring both sides (a root must have λ+6≥0\lambda + 6 \ge 0⁠),

λ2+12λ+36=λ2+4λ+44⇒8λ=8\begin{aligned} &\lambda^2 + 12\lambda + 36 = \lambda^2 + 4\lambda + 44 \\ &\Rightarrow 8\lambda = 8 \end{aligned}

∴ λ=1\therefore\ \lambda = 1⁠, and λ+6=7≥0\lambda + 6 = 7 \ge 0⁠.

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

The one idea this problem is built on

The scalar product with a unit vector is a projection: (i^+j^+k^)⋅n^(\hat i + \hat j + \hat k)\cdot\hat n is the length of the shadow of i^+j^+k^\hat i + \hat j + \hat k on the line of n^\hat n⁠. To make the unit vector we divide the sum by its length, which contains λ\lambda⁠; setting the result equal to 1 gives one equation for λ\lambda⁠.

Given, to find, and the plan
Given
b⃗=2i^+4j^−5k^\vec b = 2\hat i + 4\hat j - 5\hat k⁠, c⃗=λi^+2j^+3k^\vec c = \lambda\hat i + 2\hat j + 3\hat k⁠, and (i^+j^+k^)⋅n^=1(\hat i + \hat j + \hat k)\cdot\hat n = 1⁠, where n^\hat n is the unit vector along b⃗+c⃗\vec b + \vec c⁠.
To find
λ\lambda⁠.
Plan
(1) Add the two vectors. (2) Find the length of the sum and the unit vector n^\hat n⁠. (3) Write the scalar product condition. (4) Solve, watching the squaring step. (5) Check, and see the projection.

Basics you need

1Unit vector along a vector

n^=s⃗∣s⃗∣\hat n = \frac{\vec s}{|\vec s|}⁠, with ∣xi^+yj^+zk^∣=x2+y2+z2|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2} (Sections 10.5 and 10.5.1). Lesson: components and unit vectors

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 (Section 10.6.1). Lesson: the dot product

3Squaring both sides of an equation

p=qp = \sqrt q implies p2=qp^2 = q⁠, but p2=qp^2 = q also allows p=−qp = -\sqrt q⁠; a root found after squaring must be checked in the equation before squaring (school algebra). Basics: (x+y)2=x2+2xy+y2(x + y)^2 = x^2 + 2xy + y^2

Solution

  1. Add the two vectors

    s⃗=(2i^+4j^−5k^)+(λi^+2j^+3k^)=(2+λ)i^+6j^−2k^.\begin{aligned} \vec s &= (2\hat i + 4\hat j - 5\hat k) + (\lambda\hat i + 2\hat j + 3\hat k) \\ &= (2 + \lambda)\hat i + 6\hat j - 2\hat k. \end{aligned}
    Vectors in components are added component by component: 2+λ2 + \lambda along i^\hat i⁠, 4+2=64 + 2 = 6 along j^\hat j⁠, −5+3=−2-5 + 3 = -2 along k^\hat k⁠. Only the i^\hat i component depends on λ\lambda⁠.
    Addition in components (Section 10.5.1). Lesson: components

  2. The length of the sum and the unit vector

    ∣s⃗∣=(2+λ)2+62+(−2)2=λ2+4λ+4+36+4=λ2+4λ+44,\begin{aligned} |\vec s| &= \sqrt{(2 + \lambda)^2 + 6^2 + (-2)^2} \\ &= \sqrt{\lambda^2 + 4\lambda + 4 + 36 + 4} \\ &= \sqrt{\lambda^2 + 4\lambda + 44}, \end{aligned} n^=(2+λ)i^+6j^−2k^λ2+4λ+44.\hat n = \frac{(2 + \lambda)\hat i + 6\hat j - 2\hat k}{\sqrt{\lambda^2 + 4\lambda + 44}}.
    "A unit vector along the sum" is the sum divided by its own length. The length is never 0, since λ2+4λ+44=(λ+2)2+40≥40\lambda^2 + 4\lambda + 44 = (\lambda + 2)^2 + 40 \ge 40⁠, so n^\hat n exists for every λ\lambda⁠.
    n^=s⃗∣s⃗∣\hat n = \frac{\vec s}{|\vec s|} (Section 10.5). Lesson: components and unit vectors

  3. Write the condition

    (i^+j^+k^)⋅n^=(2+λ)+6−2λ2+4λ+44=λ+6λ2+4λ+44=1.\begin{aligned} (\hat i + \hat j + \hat k)\cdot\hat n &= \frac{(2 + \lambda) + 6 - 2}{\sqrt{\lambda^2 + 4\lambda + 44}} \\ &= \frac{\lambda + 6}{\sqrt{\lambda^2 + 4\lambda + 44}} = 1. \end{aligned}
    Each component of i^+j^+k^\hat i + \hat j + \hat k is 1, so the scalar product is the sum of the components of n^\hat n⁠. The common denominator comes out of the sum.
    Scalar product in components (Section 10.6.1). Lesson: the dot product

  4. Solve

    λ+6=λ2+4λ+44λ2+12λ+36=λ2+4λ+448λ=8λ=1.\begin{aligned} \lambda + 6 &= \sqrt{\lambda^2 + 4\lambda + 44} \\ \lambda^2 + 12\lambda + 36 &= \lambda^2 + 4\lambda + 44 \\ 8\lambda &= 8 \\ \lambda &= 1. \end{aligned}

    Before squaring, the equation needs λ+6≥0\lambda + 6 \ge 0⁠, because a square root is never negative. For λ=1\lambda = 1⁠, λ+6=7\lambda + 6 = 7 and 1+4+44=49=7\sqrt{1 + 4 + 44} = \sqrt{49} = 7⁠: the equation holds. So λ=1\lambda = 1.

    Multiplying by the positive denominator and then squaring removes the square root; the λ2\lambda^2 terms cancel, leaving a linear equation with one root. Squaring could have introduced a false root (one with λ+6<0\lambda + 6 < 0⁠), which is why the root is put back into the unsquared equation.
    Squaring both sides, then checking (school algebra). Basics: (x+y)2=x2+2xy+y2(x + y)^2 = x^2 + 2xy + y^2

  5. Check, and see the projection

    With λ=1\lambda = 1⁠, s⃗=3i^+6j^−2k^\vec s = 3\hat i + 6\hat j - 2\hat k and ∣s⃗∣=9+36+4=7|\vec s| = \sqrt{9 + 36 + 4} = 7⁠, so

    n^=37i^+67j^−27k^,(i^+j^+k^)⋅n^=3+6−27=1.\begin{aligned} \hat n &= \tfrac37\hat i + \tfrac67\hat j - \tfrac27\hat k, \\ (\hat i + \hat j + \hat k)\cdot\hat n &= \tfrac{3 + 6 - 2}{7} = 1. \end{aligned}
    O u = i + j + k, |u| = √3 s = 3i + 6j − 2k, |s| = 7 1 projection u·n̂ = 1 θ cos θ = 1/√3 θ ≈ 54.74°
    The plane of u⃗=i^+j^+k^\vec u = \hat i + \hat j + \hat k and s⃗=3i^+6j^−2k^\vec s = 3\hat i + 6\hat j - 2\hat k⁠, drawn true to scale with one scale. The foot of the perpendicular from the tip of u⃗\vec u lies 1 unit along s⃗\vec s⁠: the projection u⃗⋅n^=1\vec u\cdot\hat n = 1⁠. Since ∣u⃗∣=3|\vec u| = \sqrt3⁠, the angle between them has cosine 13\frac{1}{\sqrt3}⁠.
    The scalar product with a unit vector is the projection of i^+j^+k^\hat i + \hat j + \hat k on the line of s⃗\vec s⁠. The question says this shadow has length exactly 1, and the figure shows it. As ∣i^+j^+k^∣=3|\hat i + \hat j + \hat k| = \sqrt3⁠, the angle θ\theta between the two vectors satisfies 3cos⁡θ=1\sqrt3\cos\theta = 1⁠.
    Projection of a⃗\vec a on b⃗\vec b is a⃗⋅b^\vec a\cdot\hat b (Section 10.6.2, Observation 2). Basics: cos θ is a shadow
Result

λ=1\lambda = 1⁠; the sum is then 3i^+6j^−2k^3\hat i + 6\hat j - 2\hat k⁠, of length 7.

(Textbook answer: λ=1\lambda = 1⁠.)

Check it with numbers

n^=37i^+67j^−27k^\hat n = \frac37\hat i + \frac67\hat j - \frac27\hat k has components ≈0.428571\approx 0.428571⁠, 0.8571430.857143 and −0.285714-0.285714⁠, and exactly 9+36+449=1\frac{9 + 36 + 4}{49} = 1⁠, so it is a unit vector.

The angle between i^+j^+k^\hat i + \hat j + \hat k and n^\hat n is cos⁡−113≈54.7356∘\cos^{-1}\frac{1}{\sqrt3} \approx 54.7356^\circ⁠, with 13≈0.577350\frac{1}{\sqrt3} \approx 0.577350⁠.

A wrong value, λ=2\lambda = 2⁠: the sum is 4i^+6j^−2k^4\hat i + 6\hat j - 2\hat k of length 56\sqrt{56}⁠, and the scalar product is 856≈1.069045\frac{8}{\sqrt{56}} \approx 1.069045⁠, not 1.

Common mistakes
  1. Forgetting to divide by the length. (i^+j^+k^)⋅s⃗=λ+6=1(\hat i + \hat j + \hat k)\cdot\vec s = \lambda + 6 = 1 gives λ=−5\lambda = -5⁠, but then s⃗=−3i^+6j^−2k^\vec s = -3\hat i + 6\hat j - 2\hat k has length 7 and the scalar product with the unit vector is 17\frac17⁠, not 1.
  2. Expanding (2+λ)2(2 + \lambda)^2 as 4+λ24 + \lambda^2⁠. The middle term 4λ4\lambda is what leaves 8λ=88\lambda = 8 after the λ2\lambda^2 terms cancel.
  3. Not checking after squaring. Here the root survives, but the check λ+6=7=49\lambda + 6 = 7 = \sqrt{49} is part of a complete answer.
What to take away
  1. "Unit vector along s⃗\vec s" means s⃗∣s⃗∣\frac{\vec s}{|\vec s|}⁠.
  2. A scalar product with a unit vector is a projection.
  3. After squaring an equation with a square root, check the root in the original equation.

Every fact used, and where it comes from

FactUsed inSource
Addition in componentsStep 1Section 10.5.1 (Lesson: components)
∣xi^+yj^+zk^∣=x2+y2+z2|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2}Steps 2, 5, CheckSection 10.5.1
Unit vector n^=s⃗∣s⃗∣\hat n = \frac{\vec s}{|\vec s|}Steps 2, 5Section 10.5
Scalar product in componentsSteps 3, 5Section 10.6.1 (Lesson: the dot product)
Projection of a⃗\vec a on b⃗\vec b is a⃗⋅b^\vec a\cdot\hat bStep 5Section 10.6.2, Observation 2
Check a root after squaringStep 4School algebra
The answerResultNCERT answer key, Miscellaneous Exercise Q13 (Lesson: the miscellaneous exercises)
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