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Chapter 10 · Vector Algebra

Example 9: Direction ratios and direction cosines of i^+j^−2k^\hat i+\hat j-2\hat k

ExamplesDirection ratios and direction cosines · Sections 10.2 and 10.5.1
The question

Write the direction ratio’s of the vector a⃗=i^+j^−2k^\vec a = \hat i + \hat j - 2\hat k and hence calculate its direction cosines.

Exam answer

For r⃗=xi^+yj^+zk^\vec r = x\hat i + y\hat j + z\hat k⁠, the direction ratios are its components x,y,zx, y, z⁠. For a⃗=i^+j^−2k^\vec a = \hat i + \hat j - 2\hat k⁠:

Direction ratios: a=1a = 1⁠, b=1b = 1⁠, c=−2c = -2⁠.

∣a⃗∣=12+12+(−2)2=6|\vec a| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt6

If l,m,nl, m, n are the direction cosines, then

l=a∣a⃗∣=16m=b∣a⃗∣=16n=c∣a⃗∣=−26\begin{aligned} l &= \frac{a}{|\vec a|} = \frac{1}{\sqrt6} \\ m &= \frac{b}{|\vec a|} = \frac{1}{\sqrt6} \\ n &= \frac{c}{|\vec a|} = \frac{-2}{\sqrt6} \end{aligned}

∴\therefore the direction cosines are 16,16,−26\dfrac{1}{\sqrt6}, \dfrac{1}{\sqrt6}, -\dfrac{2}{\sqrt6}⁠.

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

The one idea this problem is built on

The direction cosines l,m,nl, m, n of a vector are the cosines of the angles it makes with the positive x, y and z-axes. For r⃗=xi^+yj^+zk^\vec r = x\hat i + y\hat j + z\hat k they are l=x∣r⃗∣l = \frac{x}{|\vec r|}⁠, m=y∣r⃗∣m = \frac{y}{|\vec r|}⁠, n=z∣r⃗∣n = \frac{z}{|\vec r|}⁠: the components of the unit vector r^\hat r⁠. The components x,y,zx, y, z themselves are proportional to l,m,nl, m, n⁠, so they serve as direction ratios. So: read off the components, then divide them by the length.

Given, to find, and the plan
Given
a⃗=i^+j^−2k^\vec a = \hat i + \hat j - 2\hat k⁠.
To find
Its direction ratios, and from them its direction cosines.
Plan
(1) Direction ratios = components. (2) Compute ∣a⃗∣|\vec a|⁠. (3) Divide each direction ratio by ∣a⃗∣|\vec a|⁠. (4) Check l2+m2+n2=1l^2 + m^2 + n^2 = 1 and look at the angles.

Basics you need

1Direction cosines

If r⃗=OP→\vec r = \overrightarrow{OP} makes angles α,β,γ\alpha, \beta, \gamma with the positive x, y and z-axes, then l=cos⁡αl = \cos\alpha⁠, m=cos⁡βm = \cos\beta⁠, n=cos⁡γn = \cos\gamma are its direction cosines; from the right triangles of Fig 10.3, cos⁡α=xr\cos\alpha = \frac{x}{r}⁠, cos⁡β=yr\cos\beta = \frac{y}{r}⁠, cos⁡γ=zr\cos\gamma = \frac{z}{r} with r=∣r⃗∣r = |\vec r| (Section 10.2). Basics: cos θ is a shadow, sin θ is a height

2Direction ratios

Numbers proportional to the direction cosines, such as lr,mr,nrlr, mr, nr⁠, are direction ratios (Section 10.2). For a⃗=a1i^+a2j^+a3k^\vec a = a_1\hat i + a_2\hat j + a_3\hat k⁠, the components a1,a2,a3a_1, a_2, a_3 are direction ratios of a⃗\vec a (Remark (ii), Section 10.5.1). Lesson: components

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 the direction ratios

    The direction ratios a,b,ca, b, c of a vector are its components. For a⃗=1i^+1j^+(−2)k^\vec a = 1\hat i + 1\hat j + (-2)\hat k⁠:

    a=1a = 1⁠, b=1b = 1⁠, c=−2c = -2.

    Direction ratios only need to be proportional to the direction cosines, and the components are: x=l rx = l\,r⁠, y=m ry = m\,r⁠, z=n rz = n\,r with the same factor r=∣a⃗∣r = |\vec a|⁠. (Any nonzero multiple, such as 2,2,−42, 2, -4⁠, would also be a set of direction ratios; the components are the natural choice, as the book takes.) The letters a,b,ca, b, c here are numbers, not to be confused with the vector a⃗\vec a⁠.
    Direction ratios are proportional to direction cosines (Section 10.2); the components are direction ratios (Remark (ii), Section 10.5.1). Lesson: components

  2. Find the magnitude

    ∣a⃗∣=12+12+(−2)2=1+1+4=6.\begin{aligned} |\vec a| &= \sqrt{1^2 + 1^2 + (-2)^2} \\ &= \sqrt{1 + 1 + 4} = \sqrt6. \end{aligned}
    The direction cosines are the direction ratios divided by the length of the vector, so we need ∣a⃗∣|\vec a|⁠. The negative component still contributes +4+4⁠.
    ∣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=a∣a⃗∣=16,m=b∣a⃗∣=16,n=c∣a⃗∣=−26.\begin{aligned} l &= \frac{a}{|\vec a|} = \frac{1}{\sqrt6}, \\ m &= \frac{b}{|\vec a|} = \frac{1}{\sqrt6}, \\ n &= \frac{c}{|\vec a|} = \frac{-2}{\sqrt6}. \end{aligned}

    So the direction cosines are (16,16,−26)\left(\frac{1}{\sqrt6}, \frac{1}{\sqrt6}, -\frac{2}{\sqrt6}\right).

    In the right triangle formed by O, the tip P of the vector and the foot of the perpendicular from P to the x-axis, the side along the axis is xx and the hypotenuse is rr⁠, so cos⁡α=xr\cos\alpha = \frac{x}{r}⁠; the same holds for the other two axes. Equivalently, l,m,nl, m, n are the components of the unit vector a^=16(i^+j^−2k^)\hat a = \frac{1}{\sqrt6}(\hat i + \hat j - 2\hat k)⁠. (The book's working writes ∣r⃗∣|\vec r| for the length of the given vector a⃗\vec a⁠, and its first fraction prints "1 =" where "ll =" is meant; the values are as above.)
    cos⁡α=xr\cos\alpha = \frac{x}{r}⁠, cos⁡β=yr\cos\beta = \frac{y}{r}⁠, cos⁡γ=zr\cos\gamma = \frac{z}{r} (Section 10.2, Fig 10.3); li^+mj^+nk^l\hat i + m\hat j + n\hat k is the unit vector of the vector (Remark (iii), Section 10.5.1). Basics: cos θ is a shadow

  4. Check, and read the angles

    l2+m2+n2=16+16+46=1,l^2 + m^2 + n^2 = \frac16 + \frac16 + \frac46 = 1,

    as it must be for direction cosines, while a2+b2+c2=1+1+4=6≠1a^2 + b^2 + c^2 = 1 + 1 + 4 = 6 \ne 1⁠. Since n<0n < 0⁠, the angle γ\gamma with the z-axis is obtuse.

    X Y Z γ O P(1, 1, −2) −2 along z cos γ = −2/√6 < 0 so γ is obtuse
    a⃗=OP→\vec a = \overrightarrow{OP} with P(1, 1, −2-2⁠), its components dashed, in a true (orthographic) view with one scale on all three axes. The angle γ\gamma between OZ and OP is more than 90∘90^\circ because the z-component is negative: cos⁡γ=−26\cos\gamma = -\frac{2}{\sqrt6}⁠.
    l2+m2+n2=1l^2 + m^2 + n^2 = 1 holds for every vector, because it says ∣a^∣=1|\hat a| = 1⁠; direction ratios have no such rule (Note, Section 10.2). The sign of each direction cosine tells on which side of the corresponding axis-plane the vector points: a⃗\vec a goes towards positive x and y but negative z, so α\alpha and β\beta are acute and γ\gamma is obtuse.
    "l2+m2+n2=1l^2 + m^2 + n^2 = 1 but a2+b2+c2≠1a^2 + b^2 + c^2 \ne 1⁠, in general" (Note, Section 10.2). Lesson: Examples 4–10
Result

Direction ratios: 1,1,−21, 1, -2⁠. Direction cosines: (16,16,−26)\left(\frac{1}{\sqrt6}, \frac{1}{\sqrt6}, -\frac{2}{\sqrt6}\right)⁠, that is (66,66,−63)\left(\frac{\sqrt6}{6}, \frac{\sqrt6}{6}, -\frac{\sqrt6}{3}\right) after rationalising.

(Textbook answer: a=1a = 1⁠, b=1b = 1⁠, c=−2c = -2⁠; the direction cosines are (16,16,−26)\left(\frac{1}{\sqrt6}, \frac{1}{\sqrt6}, -\frac{2}{\sqrt6}\right)⁠.)

In the book's working the first fraction is printed "1 = a∣r⃗∣\frac{a}{|\vec r|}", where the letter ll is meant, and the length of the given vector a⃗\vec a is written ∣r⃗∣|\vec r|⁠. Neither affects the result.

Check it with numbers

16≈0.408248\frac{1}{\sqrt6} \approx 0.408248 and 26≈0.816497\frac{2}{\sqrt6} \approx 0.816497⁠. Then 0.4082482+0.4082482+0.8164972≈0.166667+0.166667+0.666667=1.0000000.408248^2 + 0.408248^2 + 0.816497^2 \approx 0.166667 + 0.166667 + 0.666667 = 1.000000⁠.

The angles: α=β=cos⁡−1(0.408248)≈65.905∘\alpha = \beta = \cos^{-1}(0.408248) \approx 65.905^\circ and γ=cos⁡−1(−0.816497)≈144.736∘\gamma = \cos^{-1}(-0.816497) \approx 144.736^\circ⁠, so γ\gamma is obtuse. The ratios l:m:n=0.408248:0.408248:−0.816497=1:1:−2l : m : n = 0.408248 : 0.408248 : -0.816497 = 1 : 1 : -2⁠, the direction ratios.

Common mistakes
  1. Giving the direction ratios as the direction cosines. 1,1,−21, 1, -2 are not cosines: their squares add to 6, not 1.
  2. Dropping the sign of nn⁠. n=−26n = -\frac{2}{\sqrt6}⁠: the vector points below the xy-plane.
  3. Dividing by 6 instead of 6\sqrt6⁠. Divide by the length ∣a⃗∣=6|\vec a| = \sqrt6⁠, not by its square.
What to take away
  1. Direction ratios of xi^+yj^+zk^x\hat i + y\hat j + z\hat k⁠: x,y,zx, y, z (any nonzero multiple also works).
  2. Direction cosines: xr,yr,zr\frac{x}{r}, \frac{y}{r}, \frac{z}{r} with r=x2+y2+z2r = \sqrt{x^2 + y^2 + z^2}⁠; they are the components of the unit vector.
  3. Check with l2+m2+n2=1l^2 + m^2 + n^2 = 1⁠; the sign of each cosine shows whether that angle is acute or obtuse.

Every fact used, and where it comes from

FactUsed inSource
Direction angles, direction cosines l=cos⁡αl = \cos\alpha⁠, …; cos⁡α=xr\cos\alpha = \frac{x}{r}⁠, …Step 3Section 10.2 and Fig 10.3 (Basics: cos θ is a shadow)
Direction ratios: numbers proportional to the direction cosinesStep 1Section 10.2
Components are direction ratiosStep 1Remark (ii), Section 10.5.1 (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
li^+mj^+nk^l\hat i + m\hat j + n\hat k is the unit vector in the direction of the vectorStep 3Remark (iii), Section 10.5.1
l2+m2+n2=1l^2 + m^2 + n^2 = 1⁠, but a2+b2+c2≠1a^2 + b^2 + c^2 \ne 1 in generalStep 4Note, Section 10.2
The book's working and answerResultExample 9, Section 10.5.1 (Lesson: Examples 4–10)
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