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Chapter 10 · Vector Algebra

Exercise 10.2 · Q13: Direction cosines of AB→\overrightarrow{AB}⁠, A(1,2,−3)A(1,2,-3)⁠, B(−1,−2,1)B(-1,-2,1)

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

Find the direction cosines of the vector joining the points A (1, 2, −3-3⁠) and B (−1-1⁠, −2-2⁠, 1), directed from A to B.

Exam answer AB→=(−1−1)i^+(−2−2)j^+(1+3)k^=−2i^−4j^+4k^\begin{aligned} \overrightarrow{AB} &= (-1 - 1)\hat i + (-2 - 2)\hat j \\ &\quad + (1 + 3)\hat k \\ &= -2\hat i - 4\hat j + 4\hat k \end{aligned} ∣AB→∣=4+16+16=36=6\begin{aligned} |\overrightarrow{AB}| &= \sqrt{4 + 16 + 16} \\ &= \sqrt{36} = 6 \end{aligned}

So the direction cosines of AB→\overrightarrow{AB} are

l=−26=−13m=−46=−23n=46=23\begin{aligned} l &= \frac{-2}{6} = -\frac13 \\ m &= \frac{-4}{6} = -\frac23 \\ n &= \frac{4}{6} = \frac23 \end{aligned}

∴\therefore the direction cosines are −13,−23,23-\dfrac13, -\dfrac23, \dfrac23⁠.

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

The one idea this problem is built on

"Directed from A to B" means the vector AB→\overrightarrow{AB}⁠, which is B minus A. Its direction cosines are its components divided by its length. The signs matter: they record which way along each axis the vector goes.

Given, to find, and the plan
Given
A(1, 2, −3-3⁠) and B(−1-1⁠, −2-2⁠, 1).
To find
The direction cosines l,m,nl, m, n of AB→\overrightarrow{AB}⁠.
Plan
(1) Write AB→\overrightarrow{AB} in components. (2) Find ∣AB→∣|\overrightarrow{AB}|⁠. (3) Divide. (4) Check l2+m2+n2=1l^2 + m^2 + n^2 = 1⁠, and read the signs.

Basics you need

1Vector joining two points

P1P2→=(x2−x1)i^+(y2−y1)j^\overrightarrow{P_1P_2} = (x_2 - x_1)\hat i + (y_2 - y_1)\hat j +(z2−z1)k^+ (z_2 - z_1)\hat k⁠: terminal minus initial (Section 10.5.2). Basics: three numbers fix a point

2Direction cosines

For a vector with components x,y,zx, y, z and length rr⁠: l=xrl = \frac{x}{r}⁠, m=yrm = \frac{y}{r}⁠, n=zrn = \frac{z}{r}⁠, the cosines of the angles with OX, OY, OZ (Section 10.2). Basics: cos θ is a shadow, sin θ is a height

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. Write AB→\overrightarrow{AB} in components

    AB→=(−1−1)i^+(−2−2)j^+(1−(−3))k^=−2i^−4j^+4k^.\begin{aligned} \overrightarrow{AB} &= (-1 - 1)\hat i + (-2 - 2)\hat j \\ &\quad + (1 - (-3))\hat k \\ &= -2\hat i - 4\hat j + 4\hat k. \end{aligned}
    The vector starts at A and ends at B, so each component is B's coordinate minus A's. In the last one, subtracting −3-3 adds 3: 1−(−3)=1+3=41 - (-3) = 1 + 3 = 4⁠.
    P1P2→=OP2→−OP1→\overrightarrow{P_1P_2} = \overrightarrow{OP_2} - \overrightarrow{OP_1} (Section 10.5.2, Fig 10.15). Basics: three numbers fix a point

  2. Find its length

    ∣AB→∣=(−2)2+(−4)2+42=4+16+16=36=6.\begin{aligned} |\overrightarrow{AB}| &= \sqrt{(-2)^2 + (-4)^2 + 4^2} \\ &= \sqrt{4 + 16 + 16} = \sqrt{36} = 6. \end{aligned}
    This is the distance AB. The squares remove the signs, and 36 is a perfect square, so the cosines will be simple fractions.
    ∣r⃗∣=x2+y2+z2|\vec r| = \sqrt{x^2 + y^2 + z^2} (Section 10.5.1). Basics: use Pythagoras twice

  3. Divide to get the direction cosines

    l=−26=−13,m=−46=−23,n=46=23.\begin{aligned} l &= \frac{-2}{6} = -\frac13, \\ m &= \frac{-4}{6} = -\frac23, \\ n &= \frac{4}{6} = \frac23. \end{aligned}

    So the direction cosines are −13,−23,23-\frac13, -\frac23, \frac23.

    Each direction cosine is the component along that axis divided by the length, cos⁡α=xr\cos\alpha = \frac{x}{r} and so on. The components −2,−4,4-2, -4, 4 are direction ratios of AB→\overrightarrow{AB}⁠; dividing by 6 turns them into cosines.
    l=xrl = \frac{x}{r}⁠, m=yrm = \frac{y}{r}⁠, n=zrn = \frac{z}{r} (Section 10.2, Fig 10.3); components are direction ratios (Remarks (ii), Section 10.5.1). Basics: cos θ is a shadow

  4. Check, and read the signs

    l2+m2+n2=19+49+49=1.l^2 + m^2 + n^2 = \frac19 + \frac49 + \frac49 = 1.

    ll and mm are negative, so the angles with OX and OY are obtuse; nn is positive, so the angle with OZ is acute. Going from A to B, x and y decrease and z increases.

    X Y Z O A(1, 2, −3) B(−1, −2, 1) AB AB = −2î − 4ĵ + 4k̂ |AB| = 6 cosines: −1/3, −2/3, 2/3
    A(1, 2, −3-3⁠) and B(−1-1⁠, −2-2⁠, 1) in a true (orthographic) view with one scale on all three axes (negative half-axes dashed), with the vector AB→\overrightarrow{AB} of length 6 directed from A to B: it rises in z and moves back in x and y.
    Every set of direction cosines satisfies l2+m2+n2=1l^2 + m^2 + n^2 = 1⁠. The direction matters: from B to A the vector is 2i^+4j^−4k^2\hat i + 4\hat j - 4\hat k⁠, with direction cosines 13,23,−23\frac13, \frac23, -\frac23⁠, all signs reversed. That is why the question says "directed from A to B".
    l2+m2+n2=1l^2 + m^2 + n^2 = 1 (Note, Section 10.2). Compare Example 10 (vector joining two points) and Example 9 (direction cosines).
Result

AB→=−2i^−4j^+4k^\overrightarrow{AB} = -2\hat i - 4\hat j + 4\hat k⁠, ∣AB→∣=6|\overrightarrow{AB}| = 6⁠, and the direction cosines are −13,−23,23-\frac13, -\frac23, \frac23⁠.

(Textbook answer: −13,−23,23-\frac13, -\frac23, \frac23⁠.)

Check it with numbers

The cosines are about −0.333333-0.333333⁠, −0.666667-0.666667 and 0.6666670.666667⁠.

The angles: α=cos⁡−1(−13)≈109.471∘\alpha = \cos^{-1}\left(-\frac13\right) \approx 109.471^\circ⁠, β=cos⁡−1(−23)≈131.810∘\beta = \cos^{-1}\left(-\frac23\right) \approx 131.810^\circ and γ=cos⁡−1(23)≈48.190∘\gamma = \cos^{-1}\left(\frac23\right) \approx 48.190^\circ⁠: two obtuse, one acute, as the signs say.

Common mistakes
  1. Using A minus B. That is BA→\overrightarrow{BA}⁠, and every sign of the answer flips.
  2. Getting 1−3=−21 - 3 = -2 for the z-component. It is 1−(−3)=41 - (-3) = 4⁠.
  3. Dropping the signs. 13,23,23\frac13, \frac23, \frac23 has the right size but describes a different direction.
What to take away
  1. "Directed from A to B" means AB→\overrightarrow{AB} = B minus A.
  2. Direction cosines = components divided by the length; their squares add to 1.
  3. Reversing the direction reverses the sign of every direction cosine.

Every fact used, and where it comes from

FactUsed inSource
Vector joining two points: terminal minus initialStep 1Section 10.5.2, Fig 10.15 (Basics: three numbers fix a point)
∣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
l=xrl = \frac{x}{r}⁠, m=yrm = \frac{y}{r}⁠, n=zrn = \frac{z}{r}⁠; direction ratiosStep 3Section 10.2, Fig 10.3; Remarks (ii), Section 10.5.1
l2+m2+n2=1l^2 + m^2 + n^2 = 1Step 4Note, Section 10.2
The answerResultNCERT answer key, Exercise 10.2 Q13 (Lesson: components, lengths and unit vectors)
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