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Chapter 10 · Vector Algebra

Exercise 10.2 · Q12: Direction cosines of i^+2j^+3k^\hat i+2\hat j+3\hat k

Exercise 10.2Direction cosines · Sections 10.2 and 10.5.1
The question

Find the direction cosines of the vector i^+2j^+3k^\hat i + 2\hat j + 3\hat k⁠.

Exam answer

Let a⃗=i^+2j^+3k^\vec a = \hat i + 2\hat j + 3\hat k⁠. Its direction ratios are 1, 2, 3.

∣a⃗∣=12+22+32=14\begin{aligned} |\vec a| &= \sqrt{1^2 + 2^2 + 3^2} \\ &= \sqrt{14} \end{aligned}

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

l=1∣a⃗∣=114m=2∣a⃗∣=214n=3∣a⃗∣=314\begin{aligned} l &= \frac{1}{|\vec a|} = \frac{1}{\sqrt{14}} \\ m &= \frac{2}{|\vec a|} = \frac{2}{\sqrt{14}} \\ n &= \frac{3}{|\vec a|} = \frac{3}{\sqrt{14}} \end{aligned}

∴\therefore the direction cosines are 114,214,314\dfrac{1}{\sqrt{14}}, \dfrac{2}{\sqrt{14}}, \dfrac{3}{\sqrt{14}}⁠.

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 OX, OY and OZ. For xi^+yj^+zk^x\hat i + y\hat j + z\hat k they are xr,yr,zr\frac{x}{r}, \frac{y}{r}, \frac{z}{r} with rr the length: the components of the unit vector. So we find the length and divide each component by it.

Given, to find, and the plan
Given
a⃗=i^+2j^+3k^\vec a = \hat i + 2\hat j + 3\hat k⁠.
To find
Its direction cosines l,m,nl, m, n⁠.
Plan
(1) Read the direction ratios 1, 2, 3. (2) Compute ∣a⃗∣|\vec a|⁠. (3) Divide each 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⁡α=xrl = \cos\alpha = \frac{x}{r}⁠, m=cos⁡β=yrm = \cos\beta = \frac{y}{r}⁠, n=cos⁡γ=zrn = \cos\gamma = \frac{z}{r}⁠, where r=∣r⃗∣r = |\vec r| (Section 10.2, Fig 10.3). Basics: cos θ is a shadow, sin θ is a height

2Direction ratios

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: they are proportional to l,m,nl, m, n (Section 10.2; Remarks (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. Read the direction ratios

    For a⃗=1i^+2j^+3k^\vec a = 1\hat i + 2\hat j + 3\hat k⁠, the direction ratios are 1, 2, 3.

    The components are proportional to the direction cosines (x=lrx = lr⁠, y=mry = mr⁠, z=nrz = nr with the same rr⁠), so they are direction ratios. They are not yet the cosines: 12+22+32=141^2 + 2^2 + 3^2 = 14⁠, not 1.
    Components are direction ratios (Remarks (ii), Section 10.5.1). Lesson: components

  2. Find the length

    ∣a⃗∣=12+22+32=1+4+9=14.\begin{aligned} |\vec a| &= \sqrt{1^2 + 2^2 + 3^2} \\ &= \sqrt{1 + 4 + 9} = \sqrt{14}. \end{aligned}
    The direction cosines are the direction ratios divided by the length of the vector, so we need ∣a⃗∣|\vec a|⁠.
    ∣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=114,m=214,n=314.\begin{aligned} l &= \frac{1}{\sqrt{14}}, \\ m &= \frac{2}{\sqrt{14}}, \\ n &= \frac{3}{\sqrt{14}}. \end{aligned}

    So the direction cosines are 114,214,314\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}.

    In the right triangle formed by O, the tip P(1, 2, 3) and the foot of the perpendicular from P to the x-axis, the side along the axis is 1 and the hypotenuse is 14\sqrt{14}⁠, so cos⁡α=114\cos\alpha = \frac{1}{\sqrt{14}}⁠; the y and z-axes give the other two the same way. These are also the components of the unit vector 114(i^+2j^+3k^)\frac{1}{\sqrt{14}}(\hat i + 2\hat j + 3\hat k)⁠.
    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 (Remarks (iii), Section 10.5.1). Basics: cos θ is a shadow

  4. Check, and read the angles

    l2+m2+n2=114+414+914=1.l^2 + m^2 + n^2 = \frac{1}{14} + \frac{4}{14} + \frac{9}{14} = 1.

    All three cosines are positive, so all three angles are acute: the vector points into the region where x, y and z are all positive.

    X Y Z 1 2 3 O P(1, 2, 3) unit |OP| = √14 ≈ 3.74 unit vector: (1/√14, 2/√14, 3/√14) = (l, m, n)
    The vector OP→\overrightarrow{OP} to P(1, 2, 3), with its components dashed along the edges of the box, and the unit vector along it (short thick arrow). The components of that unit vector are the direction cosines. Drawn in a true (orthographic) view with one scale on all three axes.
    l2+m2+n2=1l^2 + m^2 + n^2 = 1 holds for every vector, because it says the unit vector li^+mj^+nk^l\hat i + m\hat j + n\hat k has length 1. The largest cosine, nn⁠, belongs to the smallest angle: the vector leans most towards OZ.
    "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). Compare Example 9. Lesson: Examples 4–10
Result

The direction cosines are 114,214,314\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}⁠, that is 1414,147,31414\frac{\sqrt{14}}{14}, \frac{\sqrt{14}}{7}, \frac{3\sqrt{14}}{14} after rationalising.

(Textbook answer: 114,214,314\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}⁠.)

Check it with numbers

14≈3.741657\sqrt{14} \approx 3.741657⁠, so l≈0.267261l \approx 0.267261⁠, m≈0.534522m \approx 0.534522 and n≈0.801784n \approx 0.801784⁠.

The angles: α≈74.50∘\alpha \approx 74.50^\circ⁠, β≈57.69∘\beta \approx 57.69^\circ and γ≈36.70∘\gamma \approx 36.70^\circ⁠, all acute, with γ\gamma the smallest.

Common mistakes
  1. Giving 1, 2, 3 as the direction cosines. Those are direction ratios; their squares add to 14, not 1.
  2. Dividing by 14 instead of 14\sqrt{14}⁠. Then the squares add to 14196=114\frac{14}{196} = \frac{1}{14}⁠, not 1.
  3. Mixing up the order. ll goes with i^\hat i (OX), mm with j^\hat j (OY), nn with k^\hat k (OZ).
What to take away
  1. Direction cosines of xi^+yj^+zk^x\hat i + y\hat j + z\hat k⁠: xr,yr,zr\frac{x}{r}, \frac{y}{r}, \frac{z}{r} with r=x2+y2+z2r = \sqrt{x^2 + y^2 + z^2}⁠.
  2. They are the components of the unit vector, so l2+m2+n2=1l^2 + m^2 + n^2 = 1⁠.
  3. Positive cosine, acute angle; negative cosine, obtuse angle.

Every fact used, and where it comes from

FactUsed inSource
Direction ratios: proportional to the direction cosines; the components are direction ratiosStep 1Section 10.2; Remarks (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
l=cos⁡α=xrl = \cos\alpha = \frac{x}{r}⁠, m=yrm = \frac{y}{r}⁠, n=zrn = \frac{z}{r}Step 3Section 10.2, Fig 10.3 (Basics: cos θ is a shadow)
li^+mj^+nk^l\hat i + m\hat j + n\hat k is the unit vector of the vectorSteps 3, 4Remarks (iii), 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 Q12 (Lesson: components, lengths and unit vectors)
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