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Chapter 10 · Vector Algebra

Exercise 10.2 · Q17: Right angled triangle with position vectors 3i^−4j^−4k^3\hat i-4\hat j-4\hat k⁠, 2i^−j^+k^2\hat i-\hat j+\hat k⁠, i^−3j^−5k^\hat i-3\hat j-5\hat k

Exercise 10.2Vector joining two points, magnitudes · Sections 10.5.1 and 10.5.2
The question

Show that the points A, B and C with position vectors, a⃗=3i^−4j^−4k^\vec a = 3\hat i - 4\hat j - 4\hat k⁠, b⃗=2i^−j^+k^\vec b = 2\hat i - \hat j + \hat k and c⃗=i^−3j^−5k^\vec c = \hat i - 3\hat j - 5\hat k⁠, respectively form the vertices of a right angled triangle.

Exam answer

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

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

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

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

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 the squares of 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 a⃗=3i^−4j^−4k^\vec a = 3\hat i - 4\hat j - 4\hat k⁠, b⃗=2i^−j^+k^\vec b = 2\hat i - \hat j + \hat k⁠, c⃗=i^−3j^−5k^\vec c = \hat i - 3\hat j - 5\hat k⁠. As points: A(3, −4-4⁠, −4-4⁠), B(2, −1-1⁠, 1), C(1, −3-3⁠, −5-5⁠).
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

AB→=b⃗−a⃗\overrightarrow{AB} = \vec b - \vec a⁠: the position vector of the end minus that of the start, component by component (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→=(2−3)i^+(−1+4)j^+(1+4)k^=−i^+3j^+5k^,BC→=(1−2)i^+(−3+1)j^+(−5−1)k^=−i^−2j^−6k^,CA→=(3−1)i^+(−4+3)j^+(−4+5)k^=2i^−j^+k^.\begin{aligned} \overrightarrow{AB} &= (2 - 3)\hat i + (-1 + 4)\hat j \\ &\quad + (1 + 4)\hat k = -\hat i + 3\hat j + 5\hat k, \\ \overrightarrow{BC} &= (1 - 2)\hat i + (-3 + 1)\hat j \\ &\quad + (-5 - 1)\hat k = -\hat i - 2\hat j - 6\hat k, \\ \overrightarrow{CA} &= (3 - 1)\hat i + (-4 + 3)\hat j \\ &\quad + (-4 + 5)\hat k = 2\hat i - \hat j + \hat k. \end{aligned}
    Each side vector is terminal minus initial: for AB→\overrightarrow{AB}⁠, B's components minus A's. Subtracting a negative adds: −1−(−4)=−1+4-1 - (-4) = -1 + 4⁠. Going round the triangle returns to the start: the components of AB→+BC→+CA→\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} are −1−1+2=0-1 - 1 + 2 = 0⁠, 3−2−1=03 - 2 - 1 = 0 and 5−6+1=05 - 6 + 1 = 0⁠.
    AB→=OB→−OA→\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} (Section 10.5.2). Lesson: components

  2. Squares of the lengths

    ∣AB→∣2=(−1)2+32+52=35,∣BC→∣2=(−1)2+(−2)2+(−6)2=41,∣CA→∣2=22+(−1)2+12=6.\begin{aligned} |\overrightarrow{AB}|^2 &= (-1)^2 + 3^2 + 5^2 = 35, \\ |\overrightarrow{BC}|^2 &= (-1)^2 + (-2)^2 + (-6)^2 = 41, \\ |\overrightarrow{CA}|^2 &= 2^2 + (-1)^2 + 1^2 = 6. \end{aligned}
    Pythagoras' theorem is about squares of sides, so we keep the squares and avoid the roots 35\sqrt{35}⁠, 41\sqrt{41}⁠, 6\sqrt6⁠. Signs disappear on squaring.
    ∣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

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

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

    A(3, −4, −4) B(2, −1, 1) C(1, −3, −5) √35 √6 √41 AB² + CA² = BC² 35 + 6 = 41 right angle at A
    The triangle ABC drawn in its own plane, true to scale (one scale for both directions): AB=35\text{AB} = \sqrt{35}⁠, CA=6\text{CA} = \sqrt6⁠, BC=41\text{BC} = \sqrt{41}⁠, with the right angle at A.
    The sides meeting at A are AB and CA; the side opposite A is BC. By the converse of Pythagoras' theorem, BC2=AB2+CA2\text{BC}^2 = \text{AB}^2 + \text{CA}^2 gives ∠A=90∘\angle A = 90^\circ⁠.
    Magnitude of AB→\overrightarrow{AB} is the distance AB (Section 10.5.2). These are the three points of Example 12 with the letters changed, so the right angle there, at C, is here at A. Lesson: Examples 11–12

  4. Make sure it is a triangle

    If A, B, C were on one line, one distance would be the sum of the other two, for example BC=AB+CA\text{BC} = \text{AB} + \text{CA}⁠, and then

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

    which is more than AB2+CA2\text{AB}^2 + \text{CA}^2 because AB and CA are positive. The other two cases are impossible too, since BC is the longest side. As BC2\text{BC}^2 is exactly AB2+CA2\text{AB}^2 + \text{CA}^2⁠, the points are not collinear: ABC is a genuine triangle.

    "Form the vertices of a triangle" needs three points not on one line. The sum AB→+BC→+CA→=0⃗\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = \vec 0 holds even for collinear points, so it proves nothing here; the Pythagoras relation with nonzero sides does.
    Points on a line: one distance is the sum of the other two (school geometry). Basics: (x+y)2=x2+2xy+y2(x + y)^2 = x^2 + 2xy + y^2
Result

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

(Textbook answer: the answer key has no entry for this "show that" question.)

Check it with numbers

The side lengths: 35≈5.916080\sqrt{35} \approx 5.916080⁠, 6≈2.449490\sqrt6 \approx 2.449490⁠, 41≈6.403124\sqrt{41} \approx 6.403124⁠. The two shorter sides add to about 8.3655708.365570⁠, 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): AC→=−CA→=−2i^+j^−k^\overrightarrow{AC} = -\overrightarrow{CA} = -2\hat i + \hat j - \hat k⁠, and AB→⋅AC→\overrightarrow{AB}\cdot\overrightarrow{AC} =(−1)(−2)+(3)(1)+(5)(−1)= (-1)(-2) + (3)(1) + (5)(-1) =2+3−5=0= 2 + 3 - 5 = 0⁠, so AB and AC are perpendicular, which again puts the right angle at A.

Common mistakes
  1. Mixing up the letters. Here A is 3i^−4j^−4k^3\hat i - 4\hat j - 4\hat k⁠; reading the points in another order moves the right angle to a different letter.
  2. Testing the wrong pair. The square that equals the sum of the other two must be the largest one, 41.
  3. Sign slips in the subtraction. −1−(−4)=3-1 - (-4) = 3 and 1−(−4)=51 - (-4) = 5⁠; writing −5-5 or −3-3 changes the lengths.
What to take away
  1. For a triangle given by position vectors, the sides are AB→\overrightarrow{AB}⁠, BC→\overrightarrow{BC}⁠, CA→\overrightarrow{CA}⁠, each "end minus start".
  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^∣2=x2+y2+z2|x\hat i + y\hat j + z\hat k|^2 = x^2 + y^2 + z^2Step 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
Scalar product in components; zero for perpendicular vectorsCheckSection 10.6.1, Observation 2
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