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Chapter 10 · Vector Algebra

Example 12: Right angled triangle with vertices 2i^−j^+k^2\hat i-\hat j+\hat k⁠, i^−3j^−5k^\hat i-3\hat j-5\hat k⁠, 3i^−4j^−4k^3\hat i-4\hat j-4\hat k

ExamplesVector joining two points, magnitudes · Sections 10.5.1 and 10.5.2
The question

Show that the points A(2i^−j^+k^)\text{A}(2\hat i - \hat j + \hat k)⁠, B(i^−3j^−5k^)\text{B}(\hat i - 3\hat j - 5\hat k)⁠, C(3i^−4j−4k^)\text{C}(3\hat i - 4j - 4\hat k) are the vertices of a right angled triangle.

Exam answer

The position vectors of A, B, C are 2i^−j^+k^2\hat i - \hat j + \hat k⁠, i^−3j^−5k^\hat i - 3\hat j - 5\hat k and 3i^−4j^−4k^3\hat i - 4\hat j - 4\hat k⁠.

AB→=(1−2)i^+(−3+1)j^+(−5−1)k^=−i^−2j^−6k^\begin{aligned} \overrightarrow{AB} &= (1 - 2)\hat i + (-3 + 1)\hat j \\ &\quad + (-5 - 1)\hat k \\ &= -\hat i - 2\hat j - 6\hat k \end{aligned} BC→=(3−1)i^+(−4+3)j^+(−4+5)k^=2i^−j^+k^\begin{aligned} \overrightarrow{BC} &= (3 - 1)\hat i + (-4 + 3)\hat j \\ &\quad + (-4 + 5)\hat k \\ &= 2\hat i - \hat j + \hat k \end{aligned} CA→=(2−3)i^+(−1+4)j^+(1+4)k^=−i^+3j^+5k^\begin{aligned} \overrightarrow{CA} &= (2 - 3)\hat i + (-1 + 4)\hat j \\ &\quad + (1 + 4)\hat k \\ &= -\hat i + 3\hat j + 5\hat k \end{aligned} ∣AB→∣2=1+4+36=41∣BC→∣2=4+1+1=6∣CA→∣2=1+9+25=35\begin{aligned} |\overrightarrow{AB}|^2 &= 1 + 4 + 36 = 41 \\ |\overrightarrow{BC}|^2 &= 4 + 1 + 1 = 6 \\ |\overrightarrow{CA}|^2 &= 1 + 9 + 25 = 35 \end{aligned} ∣BC→∣2+∣CA→∣2=6+35=41=∣AB→∣2\begin{aligned} &|\overrightarrow{BC}|^2 + |\overrightarrow{CA}|^2 \\ &= 6 + 35 = 41 = |\overrightarrow{AB}|^2 \end{aligned}

The three sides are nonzero and satisfy Pythagoras' theorem, so A, B, C are not collinear and the angle at C is 90∘90^\circ⁠.

Hence A, B, C are the vertices of a right angled triangle, right angled at C.

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

The one idea this problem is built on

Turn the three points into the three side vectors AB→\overrightarrow{AB}⁠, BC→\overrightarrow{BC}⁠, CA→\overrightarrow{CA} (terminal minus initial), find their lengths, and test Pythagoras: if the square of one side equals the sum of the squares of the other two, the angle opposite that side is a right angle.

Given, to find, and the plan
Given
A, B, C with position vectors 2i^−j^+k^2\hat i - \hat j + \hat k⁠, i^−3j^−5k^\hat i - 3\hat j - 5\hat k and 3i^−4j^−4k^3\hat i - 4\hat j - 4\hat k (the book prints the middle term of C as 4j4j without its hat; it is the unit vector j^\hat j⁠, as in A and B and in the book's own working). As points: A(2, −1-1⁠, 1), B(1, −3-3⁠, −5-5⁠), C(3, −4-4⁠, −4-4⁠).
To show
A, B, C are the vertices of a right angled triangle.
Plan
(1) Find AB→\overrightarrow{AB}⁠, BC→\overrightarrow{BC}⁠, CA→\overrightarrow{CA}⁠. (2) Find the squares of their lengths. (3) Test Pythagoras and name the right angle. (4) Make sure the points really form a triangle.

Basics you need

1Vector joining two points

P1P2→=(x2−x1)i^\overrightarrow{P_1P_2} = (x_2 - x_1)\hat i +(y2−y1)j^+(z2−z1)k^+ (y_2 - y_1)\hat j + (z_2 - z_1)\hat k⁠, and its magnitude is the distance P1P2P_1P_2 (Section 10.5.2). Basics: three numbers fix a point

2Magnitude from components

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

3Converse of Pythagoras' theorem

In a triangle with sides p,q,rp, q, r⁠, if r2=p2+q2r^2 = p^2 + q^2 then the angle opposite rr is a right angle (school geometry). Basics: use Pythagoras twice

Solution

  1. The three side vectors

    AB→=(1−2)i^+(−3+1)j^+(−5−1)k^=−i^−2j^−6k^,BC→=(3−1)i^+(−4+3)j^+(−4+5)k^=2i^−j^+k^,CA→=(2−3)i^+(−1+4)j^+(1+4)k^=−i^+3j^+5k^.\begin{aligned} \overrightarrow{AB} &= (1 - 2)\hat i + (-3 + 1)\hat j \\ &\quad + (-5 - 1)\hat k = -\hat i - 2\hat j - 6\hat k, \\ \overrightarrow{BC} &= (3 - 1)\hat i + (-4 + 3)\hat j \\ &\quad + (-4 + 5)\hat k = 2\hat i - \hat j + \hat k, \\ \overrightarrow{CA} &= (2 - 3)\hat i + (-1 + 4)\hat j \\ &\quad + (1 + 4)\hat k = -\hat i + 3\hat j + 5\hat k. \end{aligned}
    Each side vector is terminal minus initial: for AB→\overrightarrow{AB}⁠, B's coordinates minus A's. Subtracting a negative coordinate adds: −3−(−1)=−3+1-3 - (-1) = -3 + 1⁠, which is how the book writes it. As a check, going round the triangle returns to the start: AB→+BC→+CA→\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} has components −1+2−1=0-1 + 2 - 1 = 0⁠, −2−1+3=0-2 - 1 + 3 = 0 and −6+1+5=0-6 + 1 + 5 = 0⁠, so it is 0⃗\vec 0⁠.
    P1P2→=OP2→−OP1→\overrightarrow{P_1P_2} = \overrightarrow{OP_2} - \overrightarrow{OP_1} (Section 10.5.2). Lesson: components

  2. Squares of the lengths

    ∣AB→∣2=(−1)2+(−2)2+(−6)2=41,∣BC→∣2=22+(−1)2+12=6,∣CA→∣2=(−1)2+32+52=35.\begin{aligned} |\overrightarrow{AB}|^2 &= (-1)^2 + (-2)^2 + (-6)^2 = 41, \\ |\overrightarrow{BC}|^2 &= 2^2 + (-1)^2 + 1^2 = 6, \\ |\overrightarrow{CA}|^2 &= (-1)^2 + 3^2 + 5^2 = 35. \end{aligned}
    Pythagoras' theorem is about squares of sides, so we keep the squares and avoid the roots 41\sqrt{41}⁠, 6\sqrt6⁠, 35\sqrt{35}⁠. The sign of a component does not matter once it is squared.
    ∣r⃗∣2=x2+y2+z2|\vec r|^2 = x^2 + y^2 + z^2 (Section 10.5.1). Basics: use Pythagoras twice

  3. Test Pythagoras

    ∣BC→∣2+∣CA→∣2=6+35=41=∣AB→∣2.|\overrightarrow{BC}|^2 + |\overrightarrow{CA}|^2 = 6 + 35 = 41 = |\overrightarrow{AB}|^2.

    The longest side is AB, and its square is the sum of the squares of the other two. So the angle opposite AB, the angle at C, is a right angle.

    A(2, −1, 1) B(1, −3, −5) C(3, −4, −4) √35 √6 √41 BC² + CA² = AB² 6 + 35 = 41 right angle at C
    The triangle ABC drawn in its own plane, true to scale (one scale for both directions): BC=6\text{BC} = \sqrt6⁠, CA=35\text{CA} = \sqrt{35}⁠, AB=41\text{AB} = \sqrt{41}⁠, with the right angle at C.
    The converse of Pythagoras' theorem: in a triangle, if r2=p2+q2r^2 = p^2 + q^2⁠, the angle between the sides pp and qq is 90∘90^\circ⁠. Here p=BCp = \text{BC}⁠, q=CAq = \text{CA} and r=ABr = \text{AB}⁠, and BC and CA meet at C.
    Magnitude of P1P2→\overrightarrow{P_1P_2} is the distance P1P2P_1P_2 (Section 10.5.2). Lesson: Examples 11–12

  4. Make sure it is a triangle

    Three points form a triangle only if they are not collinear. If A, B, C were on one line, one distance would be the sum of the other two, for example AB=BC+CA\text{AB} = \text{BC} + \text{CA}⁠, and then

    AB2=BC2+CA2+2 BC⋅CA,\text{AB}^2 = \text{BC}^2 + \text{CA}^2 + 2\,\text{BC}\cdot\text{CA},

    which is more than BC2+CA2\text{BC}^2 + \text{CA}^2 because BC and CA are positive. The same happens whichever distance is the sum. Since AB2\text{AB}^2 is exactly BC2+CA2\text{BC}^2 + \text{CA}^2⁠, the points are not collinear: ABC is a genuine triangle, right angled at C.

    The book's solution goes straight from the Pythagoras test to the conclusion. This step closes the small gap: the Note after Example 21 in the book shows that AB→+BC→+CA→=0⃗\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = \vec 0 can hold for collinear points, so that sum alone does not prove there is a triangle.
    Points on a line: one distance is the sum of the other two (school geometry; compare the Remark after Example 20, Section 10.6.2). Basics: (x+y)2=x2+2xy+y2(x + y)^2 = x^2 + 2xy + y^2
Result

∣AB→∣2=41=6+35|\overrightarrow{AB}|^2 = 41 = 6 + 35 =∣BC→∣2+∣CA→∣2= |\overrightarrow{BC}|^2 + |\overrightarrow{CA}|^2⁠, so A, B, C are the vertices of a right angled triangle, with the right angle at C.

(Textbook answer: AB→=−i^−2j^−6k^\overrightarrow{AB} = -\hat i - 2\hat j - 6\hat k⁠, BC→=2i^−j^+k^\overrightarrow{BC} = 2\hat i - \hat j + \hat k⁠, CA→=−i^+3j^+5k^\overrightarrow{CA} = -\hat i + 3\hat j + 5\hat k⁠, and ∣AB→∣2=41=6+35|\overrightarrow{AB}|^2 = 41 = 6 + 35 =∣BC→∣2+∣CA→∣2= |\overrightarrow{BC}|^2 + |\overrightarrow{CA}|^2⁠; "Hence, the triangle is a right angled triangle." The book prints C as 3i^−4j−4k^3\hat i - 4j - 4\hat k⁠, without the hat on jj⁠; its working uses j^\hat j⁠.)

Check it with numbers

The side lengths: 41≈6.403124\sqrt{41} \approx 6.403124⁠, 6≈2.449490\sqrt6 \approx 2.449490⁠, 35≈5.916080\sqrt{35} \approx 5.916080⁠. The two shorter sides add to about 8.365578.36557⁠, more than 6.4031246.403124⁠, as the sides of a triangle must.

A second test with the scalar product (Section 10.6.1, met later in this chapter): BC→⋅CA→\overrightarrow{BC}\cdot\overrightarrow{CA} =(2)(−1)+(−1)(3)+(1)(5)= (2)(-1) + (-1)(3) + (1)(5) =−2−3+5=0= -2 - 3 + 5 = 0⁠, so BC and CA are perpendicular, which again puts the right angle at C.

Common mistakes
  1. Subtracting the wrong way round. Initial minus terminal only flips the signs of a side vector, so the lengths survive, but it is a bad habit: write terminal minus initial.
  2. Testing the wrong pair. The square that equals the sum of the other two must be the largest one, 41; checking 6+41=356 + 41 = 35 or 35+41=635 + 41 = 6 fails.
  3. Putting the right angle at the wrong vertex. The right angle is opposite the longest side AB, so it is at C, where BC and CA meet.
What to take away
  1. For a triangle given by position vectors, the sides are the vectors AB→\overrightarrow{AB}⁠, BC→\overrightarrow{BC}⁠, CA→\overrightarrow{CA}⁠.
  2. Compare squared lengths: r2=p2+q2r^2 = p^2 + q^2 gives a right angle opposite rr⁠.
  3. The Pythagoras relation with nonzero sides also rules out collinear points.

Every fact used, and where it comes from

FactUsed inSource
Position vector of a point; vector joining two pointsStep 1Sections 10.2 and 10.5.2 (Lesson: components)
∣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 (Basics: use Pythagoras twice)
Converse of Pythagoras' theoremStep 3School geometry
Points on a line: one distance is the sum of the other twoStep 4School geometry; compare the Remark after Example 20, Section 10.6.2
Scalar product in components; zero for perpendicular vectorsCheckSection 10.6.1, Observation 2
The book's working and answerResultExample 12, Section 10.5.3 (Lesson: Examples 11–12)
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