3.6.53D Geometry

Relation between direction cosines - l² + m² + n² = 1

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WHAT are direction cosines?


HOW to derive l2+m2+n2=1l^2+m^2+n^2=1 (from scratch)

Step 1 — Set up the unit vector. Take a unit vector u^\hat{u} pointing along the line. Write it in components: u^=(ux,uy,uz).\hat{u} = (u_x, u_y, u_z).

Why this step? A direction has no length of its own, so we standardise it to length 11. This removes the "how far" and keeps only "which way."

Step 2 — Identify each component as a direction cosine. The xx-component of u^\hat u is its projection onto the xx-axis: ux=u^i^=cosα=l.u_x = \hat{u}\cdot\hat{i} = \cos\alpha = l. Similarly uy=cosβ=mu_y = \cos\beta = m and uz=cosγ=nu_z = \cos\gamma = n.

Why this step? The projection of a unit vector onto an axis IS the cosine of the angle it makes with that axis — that's the geometric meaning of the dot product.

Step 3 — Use that u^\hat u has length 1. u^2=ux2+uy2+uz2=1.|\hat{u}|^2 = u_x^2 + u_y^2 + u_z^2 = 1.

Why this step? This is just the 3D Pythagoras theorem (distance formula) applied to a unit vector.

Step 4 — Substitute. l2+m2+n2=1\boxed{l^2 + m^2 + n^2 = 1}

Figure — Relation between direction cosines -  l² + m² + n² = 1

From Direction Ratios to Direction Cosines

If r=(a,b,c)\vec{r} = (a,b,c) with magnitude a2+b2+c2\sqrt{a^2+b^2+c^2}, then l=aa2+b2+c2,m=ba2+b2+c2,n=ca2+b2+c2.l = \frac{a}{\sqrt{a^2+b^2+c^2}}, \quad m = \frac{b}{\sqrt{a^2+b^2+c^2}}, \quad n = \frac{c}{\sqrt{a^2+b^2+c^2}}.

Why? Dividing a vector by its length always produces a unit vector, whose components automatically obey l2+m2+n2=1l^2+m^2+n^2=1.


Common Mistakes


Active Recall

Recall Quick self-test (hide and answer)
  1. State the relation between direction cosines. → l2+m2+n2=1l^2+m^2+n^2=1.
  2. Why does it equal exactly 1? → They are components of a unit vector.
  3. How do you get cosines from ratios (a,b,c)(a,b,c)? → Divide each by a2+b2+c2\sqrt{a^2+b^2+c^2}.
  4. What is sin2α+sin2β+sin2γ\sin^2\alpha+\sin^2\beta+\sin^2\gamma? → 22.
  5. How many independent direction cosines are there? → Two.
Recall Feynman: explain to a 12-year-old

Imagine a single straight arrow floating in a room. To tell a friend exactly which way it points, you ask: "How much does it lean toward the wall on your right? Toward the front wall? Toward the ceiling?" Each "lean" is a number between 1-1 and 11. Here's the magic: if you square all three leans and add them, you ALWAYS get exactly 11 — no matter which way the arrow points! That's because the arrow has a fixed length of "one step," and those three leans are just how much of that one step goes sideways, forward, and up. By the Pythagoras rule, the pieces must add back up to the whole step: 11.


80/20 — The 20% you must keep

  • Direction cosines = components of the unit direction vector.
  • l2+m2+n2=1\Rightarrow l^2+m^2+n^2=1 (Pythagoras on a unit vector).
  • Ratios → cosines: divide by magnitude.
  • Always include ±\pm; only 2 are independent.

Connections


What is the relation between direction cosines of a line?
l2+m2+n2=1l^2 + m^2 + n^2 = 1
Why does the sum of squares of direction cosines equal 1?
Because l,m,nl,m,n are the components of a unit vector along the line, and a unit vector has magnitude 1 (Pythagoras in 3D).
How do you convert direction ratios (a,b,c)(a,b,c) to direction cosines?
Divide each by the magnitude: l=aa2+b2+c2l=\frac{a}{\sqrt{a^2+b^2+c^2}}, similarly for m,nm,n.
What is sin2α+sin2β+sin2γ\sin^2\alpha+\sin^2\beta+\sin^2\gamma for direction angles?
22 (since cos2 ⁣α+cos2 ⁣β+cos2 ⁣γ=1\cos^2\!\alpha+\cos^2\!\beta+\cos^2\!\gamma=1 and subtracting from 3).
How many of the three direction cosines are independent?
Two — the third is fixed up to sign by n=±1l2m2n=\pm\sqrt{1-l^2-m^2}.
If l=12,m=12l=\frac1{\sqrt2}, m=\frac12, find nn.
n2=11214=14n=±12n^2 = 1-\frac12-\frac14=\frac14 \Rightarrow n=\pm\frac12.
Can a line make 3030^\circ with all three axes?
No, since 3cos230=2.2513\cos^2 30^\circ = 2.25 \ne 1.
Direction cosines of a line through (1,2,3)(1,2,3) and (4,6,3)(4,6,3)?
Ratios (3,4,0)(3,4,0), magnitude 5, so (35,45,0)(\frac35,\frac45,0).
Why do direction ratios NOT satisfy a2+b2+c2=1a^2+b^2+c^2=1 generally?
They are not normalised; only after dividing by the length do components form a unit vector.

Concept Map

cosine of

projected onto axes via

gives

are components of

via

yields

has

subtract from 3

not normalised so

produces

Angles alpha beta gamma with axes

Direction cosines l m n

Unit vector u along line

Dot product projection

Unit vector has length 1

3D Pythagoras

l2 + m2 + n2 = 1

sin2 sum = 2

Direction ratios a b c

Divide by magnitude

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, 3D mein koi bhi line ek direction point karti hai. Us direction ko number mein batane ke liye hum dekhte hai ki line x-axis, y-axis aur z-axis ke saath kitna angle banati hai — ye angles hai α,β,γ\alpha, \beta, \gamma. In angles ke cosine ko hum direction cosines kehte hai: l=cosαl=\cos\alpha, m=cosβm=\cos\beta, n=cosγn=\cos\gamma. Bas itni si baat hai.

Ab main secret formula: l2+m2+n2=1l^2 + m^2 + n^2 = 1. Ye yaad rakhne ki cheez nahi, samajhne ki cheez hai. Socho ek unit vector (length exactly 1 wala arrow) line ke along. Us arrow ke x, y, z components hi to l,m,nl, m, n hai (kyunki projection on axis = cosine of angle). Aur unit vector ki length 1 hoti hai, to Pythagoras lagao: components ke squares ka sum = length ka square = 11. Isliye l2+m2+n2=1l^2+m^2+n^2=1. Pure magic Pythagoras se aa gaya!

Ek important galti se bacho: agar tumhe direction ratios (a,b,c)(a,b,c) diye hai (jaise do points ka difference), to wo seedha identity satisfy nahi karte. Pehle unko length a2+b2+c2\sqrt{a^2+b^2+c^2} se divide karo, tab true direction cosines milte hai jo identity follow karte hai. Aur jab n2=1l2m2n^2 = 1 - l^2 - m^2 se nn nikalo, to ±\pm dono lena — kyunki line ke do opposite directions hote hai.

Practical fayda: agar koi tumhe teen angles de aur poochhe "aisi line possible hai kya?" — bas cos2\cos^2 ka sum check karo. Agar 1 aaye to possible, warna impossible. Ye identity ek filter ki tarah kaam karta hai. Yahi 20% concept pure 3D geometry mein baar-baar use hota hai — angle between lines, equation of line, sab mein.

Go deeper — visual, from zero

Test yourself — 3D Geometry

Connections