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Chapter 10 · Vector Algebra

Exercise 10.3 · Q17: 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

Exercise 10.3Perpendicular vectors, scalar product in components · Section 10.6.1
The question

Show that the 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 form the vertices of a right angled triangle.

Exam answer

Let A, B, C have 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.

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} BC→⋅CA→=2(−1)+(−1)(3)+1(5)=−2−3+5=0\begin{aligned} &\overrightarrow{BC}\cdot\overrightarrow{CA} \\ &= 2(-1) + (-1)(3) + 1(5) \\ &= -2 - 3 + 5 = 0 \end{aligned}

BC→\overrightarrow{BC} and CA→\overrightarrow{CA} are nonzero, so BC→⊥CA→\overrightarrow{BC} \perp \overrightarrow{CA} and ∠C=90∘\angle C = 90^\circ⁠. Being perpendicular, they are not parallel, so A, B, C are not collinear.

Hence the given vectors form 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

Treat the three vectors as the position vectors of points A, B, C. A triangle has a right angle exactly when two of its sides are perpendicular, and for nonzero sides that means their scalar product is 0. So find the three side vectors and test the three pairs.

Given, to show, and the plan
Given
Points 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, 3i^−4j^−4k^3\hat i - 4\hat j - 4\hat k⁠; as coordinates 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) Take the scalar product of each pair. (3) The zero one names the right angle. (4) Make sure the points really form a triangle, and check with Pythagoras.

Basics you need

1Vector joining two points

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

2Scalar product 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). Basics: dot products obey the same rule

3Perpendicular means scalar product zero

For nonzero vectors, p⃗⋅q⃗=0\vec p\cdot\vec q = 0 exactly when p⃗⊥q⃗\vec p \perp \vec q (Section 10.6.1, Observation 2). Lesson: the dot product

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 is end minus start. Subtracting a negative adds: −3−(−1)=−3+1-3 - (-1) = -3 + 1⁠. As a check, the three add to 0⃗\vec 0⁠: −1+2−1=0-1 + 2 - 1 = 0⁠, −2−1+3=0-2 - 1 + 3 = 0⁠, −6+1+5=0-6 + 1 + 5 = 0⁠, since going round the triangle returns to A.
    Vector joining two points (Section 10.5.2). Basics: three numbers fix a point

  2. The three scalar products

    AB→⋅BC→=−2+2−6=−6,BC→⋅CA→=−2−3+5=0,CA→⋅AB→=1−6−30=−35.\begin{aligned} \overrightarrow{AB}\cdot\overrightarrow{BC} &= -2 + 2 - 6 = -6, \\ \overrightarrow{BC}\cdot\overrightarrow{CA} &= -2 - 3 + 5 = 0, \\ \overrightarrow{CA}\cdot\overrightarrow{AB} &= 1 - 6 - 30 = -35. \end{aligned}
    Multiply matching components and add, for example BC→⋅CA→=2(−1)+(−1)(3)+1(5)\overrightarrow{BC}\cdot\overrightarrow{CA} = 2(-1) + (-1)(3) + 1(5)⁠. Only BC→⋅CA→\overrightarrow{BC}\cdot\overrightarrow{CA} is zero, so only one pair of sides is perpendicular: the triangle has exactly one right angle (a triangle can never have two).
    Scalar product in components (Section 10.6.1). Basics: dot products obey the same rule

  3. The right angle is at C

    ∣BC→∣=4+1+1=6≠0|\overrightarrow{BC}| = \sqrt{4 + 1 + 1} = \sqrt6 \ne 0 and ∣CA→∣=1+9+25=35≠0|\overrightarrow{CA}| = \sqrt{1 + 9 + 25} = \sqrt{35} \ne 0⁠. Their scalar product is 0, so BC→⊥CA→\overrightarrow{BC} \perp \overrightarrow{CA}⁠: the sides CB and CA meet at C at a right angle, ∠ACB=90∘\angle ACB = 90^\circ.

    C(3, −4, −4) A(2, −1, 1) B(1, −3, −5) √35 √6 √41 BC·CA = 0 6 + 35 = 41 right angle at C
    The triangle ABC drawn in its own plane, true to scale (one scale for both directions): CB=6\text{CB} = \sqrt6 and CA=35\text{CA} = \sqrt{35} meet at a right angle at C, and the hypotenuse AB has length 41\sqrt{41}⁠.
    The angle at C is between CA→\overrightarrow{CA} and CB→=−BC→\overrightarrow{CB} = -\overrightarrow{BC}⁠. Reversing one vector only changes the sign of the scalar product, and −0=0-0 = 0⁠, so CA→⋅CB→=0\overrightarrow{CA}\cdot\overrightarrow{CB} = 0 as well.
    Section 10.6.1, Observation 2. Lesson: the dot product

  4. It is a genuine triangle, and Pythagoras agrees

    Two nonzero perpendicular vectors are not parallel, so CA→\overrightarrow{CA} and CB→\overrightarrow{CB} point along different lines through C, and A, B, C are not collinear. The squared lengths are

    ∣AB→∣2=1+4+36=41,∣BC→∣2=6,∣CA→∣2=35,\begin{aligned} |\overrightarrow{AB}|^2 &= 1 + 4 + 36 = 41, \\ |\overrightarrow{BC}|^2 &= 6, \quad |\overrightarrow{CA}|^2 = 35, \end{aligned}

    and 6+35=416 + 35 = 41⁠: the square on AB, the side opposite C, equals the sum of the other two.

    "Form the vertices of a triangle" needs three points not on one line; the perpendicular sides settle that. Pythagoras' theorem is the same right angle seen through lengths, which is how Example 12 proves it for these very points.
    Converse of Pythagoras' theorem (school geometry). The same three points, relabelled, appear in Exercise 10.2 Q17. Lesson: Examples 11–12
Result

BC→⋅CA→=0\overrightarrow{BC}\cdot\overrightarrow{CA} = 0 with both sides nonzero, so the angle at C is 90∘90^\circ and A, B, C are not collinear: the three vectors form the vertices of a right angled triangle, right angled at C. Hence proved.

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

Check it with numbers

Side lengths: 6≈2.449490\sqrt6 \approx 2.449490⁠, 35≈5.916080\sqrt{35} \approx 5.916080⁠, 41≈6.403124\sqrt{41} \approx 6.403124⁠.

The other two angles: AB→⋅AC→=35\overrightarrow{AB}\cdot\overrightarrow{AC} = 35⁠, so cos⁡A=3541 35=3541≈0.923936\cos A = \frac{35}{\sqrt{41}\,\sqrt{35}} = \frac{\sqrt{35}}{\sqrt{41}} \approx 0.923936 and A≈22.49∘A \approx 22.49^\circ⁠; BA→⋅BC→=6\overrightarrow{BA}\cdot\overrightarrow{BC} = 6⁠, so cos⁡B=641≈0.382546\cos B = \frac{\sqrt6}{\sqrt{41}} \approx 0.382546 and B≈67.51∘B \approx 67.51^\circ⁠. Then 22.49+67.51=90.0022.49 + 67.51 = 90.00⁠, leaving 90∘90^\circ for C.

Common mistakes
  1. Dotting the position vectors. (2i^−j^+k^)⋅(i^−3j^−5k^)=2+3−5=0(2\hat i - \hat j + \hat k)\cdot(\hat i - 3\hat j - 5\hat k) = 2 + 3 - 5 = 0 too, but that is the angle at the origin O between OA→\overrightarrow{OA} and OB→\overrightarrow{OB}⁠, not an angle of triangle ABC.
  2. Stopping at the first pair. AB→⋅BC→=−6≠0\overrightarrow{AB}\cdot\overrightarrow{BC} = -6 \ne 0 does not rule out a right angle; test all three pairs.
  3. Skipping "not collinear". A right angle needs two nonzero, non-parallel sides; say so.
What to take away
  1. Right angle test: some pair of side vectors has scalar product 0 (both nonzero).
  2. The zero pair names the vertex: BC→\overrightarrow{BC} and CA→\overrightarrow{CA} share the vertex C.
  3. Pythagoras with squared lengths is an independent check of the same fact.

Every fact used, and where it comes from

FactUsed inSource
Position vector; vector joining two pointsStep 1Sections 10.2 and 10.5.2 (Basics: three numbers fix a point)
Scalar product in componentsStep 2, CheckSection 10.6.1 (Basics: dot products obey the same rule)
Nonzero vectors: scalar product 0 exactly when perpendicularStep 3Section 10.6.1, Observation 2 (Lesson: the dot product)
∣xi^+yj^+zk^∣2=x2+y2+z2|x\hat i + y\hat j + z\hat k|^2 = x^2 + y^2 + z^2Steps 3, 4Section 10.5.1
Converse of Pythagoras' theoremStep 4School geometry
cos⁡θ=p⃗⋅q⃗∣p⃗∣∣q⃗∣\cos\theta = \frac{\vec p\cdot\vec q}{|\vec p||\vec q|}CheckSection 10.6.1, Observation 6 (Lesson: the dot product exercises)
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