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Chapter 10 · Vector Algebra

Example 14: Angle between i^+j^−k^\hat i+\hat j-\hat k and i^−j^+k^\hat i-\hat j+\hat k

ExamplesScalar product in components, angle between vectors · Section 10.6.1
The question

Find angle ‘θ\theta’ between the vectors a⃗=i^+j^−k^\vec a = \hat i + \hat j - \hat k and b⃗=i^−j^+k^\vec b = \hat i - \hat j + \hat k⁠.

Exam answer

The angle θ\theta between a⃗\vec a and b⃗\vec b is given by cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣\cos\theta = \frac{\vec a\cdot\vec b}{|\vec a||\vec b|}⁠.

a⃗⋅b⃗=(i^+j^−k^)⋅(i^−j^+k^)=(1)(1)+(1)(−1)+(−1)(1)=1−1−1=−1\begin{aligned} \vec a\cdot\vec b &= (\hat i + \hat j - \hat k)\cdot(\hat i - \hat j + \hat k) \\ &= (1)(1) + (1)(-1) + (-1)(1) \\ &= 1 - 1 - 1 = -1 \end{aligned} ∣a⃗∣=12+12+(−1)2=3|\vec a| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt3 ∣b⃗∣=12+(−1)2+12=3|\vec b| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt3 cos⁡θ=−13×3=−13\cos\theta = \frac{-1}{\sqrt3 \times \sqrt3} = -\frac13

Since 0≤θ≤π0 \le \theta \le \pi⁠,

∴ θ=cos⁡−1(−13)\therefore\ \theta = \cos^{-1}\Big(-\dfrac13\Big)⁠.

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

The one idea this problem is built on

When the vectors are given in components, the scalar product is the sum of the products of matching components:

a⃗⋅b⃗=a1b1+a2b2+a3b3.\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3.

Divide it by ∣a⃗∣∣b⃗∣|\vec a||\vec b| to get cos⁡θ\cos\theta⁠, and take the one angle in [0,π][0, \pi] with that cosine. A negative scalar product means an obtuse angle.

Given, to find, and the plan
Given
a⃗=i^+j^−k^\vec a = \hat i + \hat j - \hat k⁠, b⃗=i^−j^+k^\vec b = \hat i - \hat j + \hat k⁠.
To find
The angle θ\theta between a⃗\vec a and b⃗\vec b⁠.
Plan
(1) Compute a⃗⋅b⃗\vec a\cdot\vec b from components. (2) Compute ∣a⃗∣|\vec a| and ∣b⃗∣|\vec b|⁠. (3) Form cos⁡θ\cos\theta and take cos⁡−1\cos^{-1}⁠. (4) Read off whether the angle is acute or obtuse.

Basics you need

1Scalar product in components

For a⃗=a1i^+a2j^+a3k^\vec a = a_1\hat i + a_2\hat j + a_3\hat k and b⃗=b1i^+b2j^+b3k^\vec b = b_1\hat i + b_2\hat j + b_3\hat k⁠, a⃗⋅b⃗=a1b1+a2b2+a3b3\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3 (Section 10.6.1, from i^⋅i^=1\hat i\cdot\hat i = 1⁠, i^⋅j^=0\hat i\cdot\hat j = 0⁠, …). Lesson: the dot product

2The angle formula

cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣\cos\theta = \frac{\vec a\cdot\vec b}{|\vec a||\vec b|}⁠, with 0≤θ≤π0 \le \theta \le \pi (Section 10.6.1, Observation 6). Lesson: Examples 13–21

3Magnitude from components

∣xi^+yj^+zk^∣=x2+y2+z2|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2} (Section 10.5.1). Basics: use Pythagoras twice

Solution

  1. The scalar product

    a⃗⋅b⃗=(i^+j^−k^)⋅(i^−j^+k^)=(1)(1)+(1)(−1)+(−1)(1)=1−1−1=−1.\begin{aligned} \vec a\cdot\vec b &= (\hat i + \hat j - \hat k)\cdot(\hat i - \hat j + \hat k) \\ &= (1)(1) + (1)(-1) + (-1)(1) \\ &= 1 - 1 - 1 = -1. \end{aligned}
    Multiplying out the brackets gives nine products such as i^⋅j^\hat i\cdot\hat j⁠; the six mixed ones are 0 because the unit vectors are perpendicular, and i^⋅i^=j^⋅j^=k^⋅k^=1\hat i\cdot\hat i = \hat j\cdot\hat j = \hat k\cdot\hat k = 1⁠. What remains is the product of the i^\hat i components, plus that of the j^\hat j components, plus that of the k^\hat k components. Keep each sign with its component: the j^\hat j components are 1 and −1-1⁠, the k^\hat k components −1-1 and 1.
    a⃗⋅b⃗=a1b1+a2b2+a3b3\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3 (Section 10.6.1, using Properties 1 and 2 and Observation 5). Lesson: the dot product

  2. The two magnitudes

    ∣a⃗∣=12+12+(−1)2=3,∣b⃗∣=12+(−1)2+12=3.\begin{aligned} |\vec a| &= \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt3, \\ |\vec b| &= \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt3. \end{aligned}
    Both vectors have components ±1\pm1⁠, so both have length 3\sqrt3 (each is a long diagonal of a unit cube). Their product is 3×3=3\sqrt3 \times \sqrt3 = 3⁠, which is the 3 in the book's cos⁡θ=−13\cos\theta = \frac{-1}{3}⁠.
    ∣r⃗∣=x2+y2+z2|\vec r| = \sqrt{x^2 + y^2 + z^2} (Section 10.5.1). Basics: use Pythagoras twice

  3. Cosine of the angle, then the angle

    cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣=−13×3=−13,\cos\theta = \frac{\vec a\cdot\vec b}{|\vec a||\vec b|} = \frac{-1}{\sqrt3 \times \sqrt3} = -\frac13, θ=cos⁡−1(−13).\theta = \cos^{-1}\Big(-\frac13\Big).

    So θ=cos⁡−1(−13)\theta = \cos^{-1}\big(-\frac13\big).

    The angle between two vectors lies in [0,π][0, \pi]⁠, the principal range of cos⁡−1\cos^{-1}⁠, and on that interval there is exactly one angle with cosine −13-\frac13⁠. It is not a standard angle, so the exact answer is left as cos⁡−1(−13)\cos^{-1}\big(-\frac13\big)⁠, as in the book. An equivalent form is π−cos⁡−113\pi - \cos^{-1}\frac13⁠, because cos⁡−1(−x)=π−cos⁡−1x\cos^{-1}(-x) = \pi - \cos^{-1}x⁠.
    Observation 6 (Section 10.6.1); principal range of cos⁡−1\cos^{-1}⁠. Basics: cos θ is a shadow

  4. Acute or obtuse?

    cos⁡θ=−13<0\cos\theta = -\frac13 < 0⁠, so π2<θ≤π\frac{\pi}{2} < \theta \le \pi⁠: the angle is obtuse, about 109.47∘109.47^\circ⁠.

    θ ≈ 109.47° O a, length √3 b, length √3 −1/√3 cos θ = −1/3 < 0 so θ is obtuse
    The two vectors drawn in their own plane, true to scale: both have length 3\sqrt3⁠, and the angle between them is cos⁡−1(−13)≈109.47∘\cos^{-1}\big(-\frac13\big) \approx 109.47^\circ⁠. Because the angle is obtuse, the shadow of b⃗\vec b on the line of a⃗\vec a falls behind O, at signed length ∣b⃗∣cos⁡θ=−13|\vec b|\cos\theta = -\frac{1}{\sqrt3}⁠.
    The sign of a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta is the sign of cos⁡θ\cos\theta⁠, since the lengths are positive. Positive means acute, zero means perpendicular, negative means obtuse. Here the vectors agree in the i^\hat i direction but disagree in the j^\hat j and k^\hat k directions, so on balance they point away from each other.
    Sign of the scalar product (Section 10.6.1, Definition 2 and Observation 2). Lesson: the dot product
Result

cos⁡θ=−13\cos\theta = -\frac13⁠, so θ=cos⁡−1(−13)\theta = \cos^{-1}\big(-\frac13\big)⁠, an obtuse angle of about 109.47∘109.47^\circ⁠.

(Textbook answer: cos⁡θ=−13\cos\theta = \frac{-1}{3}⁠, hence the required angle is θ=cos⁡−1(−13)\theta = \cos^{-1}\left(-\frac13\right)⁠.)

Check it with numbers

−13≈−0.333333-\frac13 \approx -0.333333⁠, and cos⁡−1(−0.333333)≈1.910633\cos^{-1}(-0.333333) \approx 1.910633 radians ≈109.4712∘\approx 109.4712^\circ⁠, between 90∘90^\circ and 180∘180^\circ⁠.

The same angle from the other form: cos⁡−113≈70.528779∘\cos^{-1}\frac13 \approx 70.528779^\circ⁠, and 180∘−70.528779∘=109.471221∘180^\circ - 70.528779^\circ = 109.471221^\circ⁠.

Common mistakes
  1. Sign slips in the scalar product. (1)(−1)+(−1)(1)=−2(1)(-1) + (-1)(1) = -2⁠, not 0; the scalar product is 1−2=−11 - 2 = -1⁠.
  2. Dropping the minus sign. cos⁡−113≈70.53∘\cos^{-1}\frac13 \approx 70.53^\circ is the acute angle between a⃗\vec a and −b⃗-\vec b⁠, not between a⃗\vec a and b⃗\vec b⁠.
  3. Dividing by ∣a⃗∣+∣b⃗∣|\vec a| + |\vec b|⁠. The denominator is the product 3×3=3\sqrt3 \times \sqrt3 = 3⁠, not the sum 232\sqrt3⁠.
What to take away
  1. a⃗⋅b⃗=a1b1+a2b2+a3b3\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3⁠: multiply matching components and add.
  2. cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣\cos\theta = \frac{\vec a\cdot\vec b}{|\vec a||\vec b|}⁠, and θ=cos⁡−1(…)\theta = \cos^{-1}(\ldots) in [0,π][0, \pi]⁠.
  3. Negative scalar product, obtuse angle.

Every fact used, and where it comes from

FactUsed inSource
a⃗⋅b⃗=a1b1+a2b2+a3b3\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3Step 1Section 10.6.1 (Lesson: the dot product)
i^⋅i^=1\hat i\cdot\hat i = 1⁠, i^⋅j^=0\hat i\cdot\hat j = 0⁠, and so onStep 1Section 10.6.1, Observation 5
∣xi^+yj^+zk^∣=x2+y2+z2|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2}Step 2Section 10.5.1
cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣\cos\theta = \frac{\vec a\cdot\vec b}{|\vec a||\vec b|}⁠, 0≤θ≤π0 \le \theta \le \piSteps 3, 4Section 10.6.1, Definition 2 and Observation 6
cos⁡−1(−x)=π−cos⁡−1x\cos^{-1}(-x) = \pi - \cos^{-1}xStep 3Inverse trigonometric functions (Basics: cos θ is a shadow)
The book's working and answerResultExample 14, Section 10.6.2 (Lesson: Examples 13–21)
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