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Chapter 10 · Vector Algebra

Exercise 10.3 · Q15: ∠ABC\angle ABC for A(1,2,3)A(1,2,3)⁠, B(−1,0,0)B(-1,0,0)⁠, C(0,1,2)C(0,1,2)

Exercise 10.3Angle between two vectors · Section 10.6.1
The question

If the vertices A, B, C of a triangle ABC are (1, 2, 3), (−1-1⁠, 0, 0), (0, 1, 2), respectively, then find ∠ABC\angle ABC⁠. [∠ABC\angle ABC is the angle between the vectors BA→\overrightarrow{BA} and BC→\overrightarrow{BC}].

Exam answer BA→=(1+1)i^+(2−0)j^+(3−0)k^=2i^+2j^+3k^\begin{aligned} \overrightarrow{BA} &= (1 + 1)\hat i + (2 - 0)\hat j \\ &\quad + (3 - 0)\hat k \\ &= 2\hat i + 2\hat j + 3\hat k \end{aligned} BC→=(0+1)i^+(1−0)j^+(2−0)k^=i^+j^+2k^\begin{aligned} \overrightarrow{BC} &= (0 + 1)\hat i + (1 - 0)\hat j \\ &\quad + (2 - 0)\hat k \\ &= \hat i + \hat j + 2\hat k \end{aligned} ∣BA→∣=4+4+9=17∣BC→∣=1+1+4=6BA→⋅BC→=2(1)+2(1)+3(2)=10\begin{aligned} |\overrightarrow{BA}| &= \sqrt{4 + 4 + 9} = \sqrt{17} \\ |\overrightarrow{BC}| &= \sqrt{1 + 1 + 4} = \sqrt6 \\ \overrightarrow{BA}\cdot\overrightarrow{BC} &= 2(1) + 2(1) + 3(2) \\ &= 10 \end{aligned}

∠ABC\angle ABC is the angle between BA→\overrightarrow{BA} and BC→\overrightarrow{BC}⁠, so

cos⁡∠ABC=BA→⋅BC→∣BA→∣∣BC→∣=1017 6=10102\begin{aligned} \cos\angle ABC &= \frac{\overrightarrow{BA}\cdot\overrightarrow{BC}}{|\overrightarrow{BA}||\overrightarrow{BC}|} \\ &= \frac{10}{\sqrt{17}\,\sqrt6} = \frac{10}{\sqrt{102}} \end{aligned}

∴ ∠ABC=cos⁡−1(10102)\therefore\ \angle ABC = \cos^{-1}\left(\dfrac{10}{\sqrt{102}}\right)⁠.

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

The one idea this problem is built on

The angle at B is the angle between the two sides that start at B, BA→\overrightarrow{BA} and BC→\overrightarrow{BC}⁠. The scalar product gives its cosine:

cos⁡∠ABC=BA→⋅BC→∣BA→∣ ∣BC→∣.\cos\angle ABC = \frac{\overrightarrow{BA}\cdot\overrightarrow{BC}}{|\overrightarrow{BA}|\,|\overrightarrow{BC}|}.
Given, to find, and the plan
Given
A(1, 2, 3), B(−1-1⁠, 0, 0), C(0, 1, 2).
To find
∠ABC\angle ABC⁠, the angle between BA→\overrightarrow{BA} and BC→\overrightarrow{BC}⁠.
Plan
(1) Find BA→\overrightarrow{BA} and BC→\overrightarrow{BC}⁠. (2) Find their lengths. (3) Find their scalar product. (4) Divide to get the cosine, and take cos⁡−1\cos^{-1}⁠. (5) Check with the third side.

Basics you need

1Vector joining two points

PQ→\overrightarrow{PQ} = (coordinates of Q) −- (coordinates of P), component by component (Section 10.5.2). Basics: three numbers fix a point

2Scalar product and length 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) and ∣p⃗∣=p12+p22+p32|\vec p| = \sqrt{p_1^2 + p_2^2 + p_3^2} (Section 10.5.1). Lesson: the dot product

3Angle between two vectors

cos⁡θ=p⃗⋅q⃗∣p⃗∣∣q⃗∣\cos\theta = \frac{\vec p\cdot\vec q}{|\vec p||\vec q|} with 0≤θ≤π0 \le \theta \le \pi (Section 10.6.1, Observation 6), so θ=cos⁡−1(… )\theta = \cos^{-1}(\dots) is the principal value. Basics: cos θ is a shadow

Solution

  1. The two sides starting at B

    BA→=(1−(−1))i^+(2−0)j^+(3−0)k^=2i^+2j^+3k^,BC→=(0−(−1))i^+(1−0)j^+(2−0)k^=i^+j^+2k^.\begin{aligned} \overrightarrow{BA} &= (1 - (-1))\hat i + (2 - 0)\hat j \\ &\quad + (3 - 0)\hat k = 2\hat i + 2\hat j + 3\hat k, \\ \overrightarrow{BC} &= (0 - (-1))\hat i + (1 - 0)\hat j \\ &\quad + (2 - 0)\hat k = \hat i + \hat j + 2\hat k. \end{aligned}
    The angle at B is between arrows that both leave B, so B is the starting point of each: end minus start, with B's coordinates subtracted. Subtracting −1-1 adds 1.
    Vector joining two points (Section 10.5.2). Basics: three numbers fix a point

  2. Their lengths

    ∣BA→∣=22+22+32=17,∣BC→∣=12+12+22=6.\begin{aligned} |\overrightarrow{BA}| &= \sqrt{2^2 + 2^2 + 3^2} = \sqrt{17}, \\ |\overrightarrow{BC}| &= \sqrt{1^2 + 1^2 + 2^2} = \sqrt6. \end{aligned}
    The length of a vector is the square root of the sum of the squares of its components (Pythagoras in three dimensions). These are the side lengths BA and BC.
    Magnitude from components (Section 10.5.1). Basics: use Pythagoras twice

  3. Their scalar product

    BA→⋅BC→=2(1)+2(1)+3(2)=2+2+6=10.\begin{aligned} \overrightarrow{BA}\cdot\overrightarrow{BC} &= 2(1) + 2(1) + 3(2) \\ &= 2 + 2 + 6 = 10. \end{aligned}
    Multiply matching components and add. The result is positive, so the angle at B is acute.
    Scalar product in components (Section 10.6.1). Basics: dot products obey the same rule

  4. The cosine and the angle

    cos⁡∠ABC=1017 6=10102,\cos\angle ABC = \frac{10}{\sqrt{17}\,\sqrt6} = \frac{10}{\sqrt{102}}, ∠ABC=cos⁡−1(10102).\angle ABC = \cos^{-1}\left(\frac{10}{\sqrt{102}}\right).

    So ∠ABC=cos⁡−1(10102)\angle ABC = \cos^{-1}\left(\frac{10}{\sqrt{102}}\right), about 8.05∘8.05^\circ⁠.

    θ B(−1, 0, 0) C(0, 1, 2) A(1, 2, 3) √17 √6 √3 θ = ∠ABC = cos⁻¹(10/√102) BA·BC = 10, θ ≈ 8.05°
    The triangle ABC drawn in its own plane, true to scale (one scale for both directions): BA=17\text{BA} = \sqrt{17}⁠, BC=6\text{BC} = \sqrt6⁠, AC=3\text{AC} = \sqrt3⁠. The angle at B is small, about 8.05∘8.05^\circ⁠.
    17 6=102\sqrt{17}\,\sqrt6 = \sqrt{102}⁠. The angle between two vectors lies in [0,π][0, \pi]⁠, which is exactly the principal range of cos⁡−1\cos^{-1}⁠, so cos⁡−1\cos^{-1} gives the angle itself, with no second candidate. As 10102\frac{10}{\sqrt{102}} is close to 1, the angle is small.
    Section 10.6.1, Observation 6. Same method as Exercise 10.3 Q2. Lesson: Examples 13–21

  5. Check with the third side

    CA→=(1−0)i^+(2−1)j^+(3−2)k^=i^+j^+k^\overrightarrow{CA} = (1 - 0)\hat i + (2 - 1)\hat j + (3 - 2)\hat k = \hat i + \hat j + \hat k⁠, so AC2=3\text{AC}^2 = 3⁠. The cosine rule at B predicts

    BA2+BC2−2 BA→⋅BC→=17+6−20=3.\begin{aligned} &\text{BA}^2 + \text{BC}^2 - 2\,\overrightarrow{BA}\cdot\overrightarrow{BC} \\ &= 17 + 6 - 20 = 3. \end{aligned}
    Since CA→=BA→−BC→\overrightarrow{CA} = \overrightarrow{BA} - \overrightarrow{BC}⁠, expanding ∣BA→−BC→∣2|\overrightarrow{BA} - \overrightarrow{BC}|^2 gives exactly this expression: it is the cosine rule b2=a2+c2−2accos⁡Bb^2 = a^2 + c^2 - 2ac\cos B in vector form. Both ways give AC2=3\text{AC}^2 = 3⁠, so the scalar product 10 is right.
    Distributivity of the scalar product (Section 10.6.1, Property 1). Basics: dot products obey the same rule
Result

cos⁡∠ABC=10102\cos\angle ABC = \frac{10}{\sqrt{102}}⁠, so ∠ABC=cos⁡−1(10102)\angle ABC = \cos^{-1}\left(\frac{10}{\sqrt{102}}\right) (about 8.05∘8.05^\circ⁠). Rationalised, 10102=510251\frac{10}{\sqrt{102}} = \frac{5\sqrt{102}}{51}⁠.

(Textbook answer: cos⁡−1(10102)\cos^{-1}\left(\frac{10}{\sqrt{102}}\right)⁠.)

Check it with numbers

102≈10.099505\sqrt{102} \approx 10.099505⁠, so cos⁡∠ABC=1010.099505≈0.990148\cos\angle ABC = \frac{10}{10.099505} \approx 0.990148⁠, and ∠ABC≈8.05∘\angle ABC \approx 8.05^\circ⁠. For the angle in radians to six places, the cosine is needed to eight, 0.990147540.99014754⁠: then ∠ABC≈0.140490\angle ABC \approx 0.140490 radian. (A small angle changes its cosine very little, so the last digits of the cosine matter.)

The other two angles, by the same formula: at A, cos⁡A=751≈0.980196\cos A = \frac{7}{\sqrt{51}} \approx 0.980196⁠, so A≈11.42∘A \approx 11.42^\circ⁠; at C, cos⁡C=−418≈−0.942809\cos C = -\frac{4}{\sqrt{18}} \approx -0.942809⁠, so C≈160.53∘C \approx 160.53^\circ⁠. Then 8.05+11.42+160.53=180.008.05 + 11.42 + 160.53 = 180.00⁠, as the angles of a triangle must add to 180∘180^\circ⁠.

Common mistakes
  1. Using AB→\overrightarrow{AB} instead of BA→\overrightarrow{BA}⁠. AB→⋅BC→=−10\overrightarrow{AB}\cdot\overrightarrow{BC} = -10 gives the outside angle, cos⁡−1(−10102)≈171.95∘\cos^{-1}\left(-\frac{10}{\sqrt{102}}\right) \approx 171.95^\circ⁠. Both arrows must start at B.
  2. Using position vectors. OA→⋅OC→\overrightarrow{OA}\cdot\overrightarrow{OC} gives the angle at the origin O, not at B.
  3. Sign slips with B's −1-1⁠. 1−(−1)=21 - (-1) = 2 and 0−(−1)=10 - (-1) = 1⁠.
What to take away
  1. For the angle at a vertex, use the two side vectors that start at that vertex.
  2. cos⁡θ=p⃗⋅q⃗∣p⃗∣∣q⃗∣\cos\theta = \frac{\vec p\cdot\vec q}{|\vec p||\vec q|}⁠, and cos⁡−1\cos^{-1} of it is the angle, since 0≤θ≤π0 \le \theta \le \pi⁠.
  3. A positive scalar product means an acute angle; a negative one, an obtuse angle.

Every fact used, and where it comes from

FactUsed inSource
Vector joining two pointsSteps 1, 5Section 10.5.2 (Basics: three numbers fix a point)
∣p⃗∣=p12+p22+p32|\vec p| = \sqrt{p_1^2 + p_2^2 + p_3^2}Step 2Section 10.5.1 (Basics: use Pythagoras twice)
Scalar product in componentsStep 3Section 10.6.1 (Lesson: the dot product)
cos⁡θ=p⃗⋅q⃗∣p⃗∣∣q⃗∣\cos\theta = \frac{\vec p\cdot\vec q}{|\vec p||\vec q|}⁠, 0≤θ≤π0 \le \theta \le \piStep 4, CheckSection 10.6.1, Definition 2 and Observation 6 (Basics: cos θ is a shadow)
Distributivity of the scalar productStep 5Section 10.6.1, Property 1
Angles of a triangle add to 180∘180^\circCheckSchool geometry
The answerResultNCERT answer key, Exercise 10.3 Q15 (Lesson: the dot product exercises)
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