Q.Find the direction cosines of the sides of the triangle whose vertices are (3,5,−4), (−1,1,2) and (−5,−5,−2).
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Direction Cosines Properties
Direction Cosines and Their Properties
To describe which way a line points in 3D — ignoring its length — we give the angles it makes with the three coordinate axes. Call them α,β,γ (with the x-, y-, z-axis). Their cosines
l=cosα,m=cosβ,n=cosγ
are the direction cosines of the line.
Direction cosines are the cosines of the angles, not the angles themselves — a common slip.
For a point P(x,y,z) on a line through the origin at distance r=x2+y2+z2, right-triangle trigonometry gives
l=rx,m=ry,n=rz.
Property 1 — the squares sum to 1
l2+m2+n2=r2x2+y2+z2=r2r2=1.
This is the signature of direction cosines: any triple with l2+m2+n2=1 is the set of direction cosines of some line.
It is not l+m+n=1. Only the sum of squares equals 1.
Property 2 — they are a unit vector
Dividing OP=(x,y,z) by its length gives the unit vector u^=(l,m,n). So direction cosines are literally the components of a unit vector along the line — which is exactly why their squares sum to 1.
Property 3 — fixed up to sign
Reversing the line flips all three signs: a line has two sets, (l,m,n) and (−l,−m,−n).
Direction ratios
Any numbers (a,b,c) proportional to (l,m,n) are direction ratios. They are easier to read off, and you recover the cosines by normalising: …
The direction cosines of a side are the components of its unit vector: form each side vector, then divide by its length.
With A(3,5,−4), B(−1,1,2), C(−5,−5,−2):
AB=(−4,−4,6), ∣AB∣=16+16+36=68=217
⇒(−172,−172,173)
BC=(−4,−6,−4), ∣BC∣=16+36+16=68=217
⇒(−172,−173,−172)
CA=(8,10,−2), ∣CA∣=64+100+4=168=242 …
Form each side vector and divide by its length: AB gives (−172,−172,173), BC gives (−172,−173,−172), and CA gives (424,425,−421).
What direction cosines are
A directed segment in space makes angles α,β,γ with the x-, y-, z-axes. Its direction cosines l=cosα, m=cosβ, n=cosγ are exactly the components of the corresponding unit vector. So for a side vector (a,b,c) of length r,
l=ra,m=rb,n=rc.
A good sanity check: l2+m2+n2=1 every time. If your three numbers don't square-sum to 1, you have slipped.
Side AB
AB=B−A=(−1−3,1−5,2−(−4))=(−4,−4,6).
∣AB∣=(−4)2+(−4)2+62=68=217.
Dividing each component by 217:
(−172,−172,173).
Check: 174+4+9=1 ✓
Side BC
BC=C−B=(−5−(−1),−5−1,−2−2)=(−4,−6,−4).
∣BC∣=16+36+16=68=217.
(−172,−173,−172).
Check: 174+9+4=1 ✓
Side CA
CA=A−C=(3−(−5),5−(−5),−4−(−2))=(8,10,−2). …
Method: Direction cosines of a segment — vector, then normalise
The direction cosines of a side (a directed segment between two vertices) are just the direction cosines of the vector along it. So the task splits into: build the segment vector, then normalise — repeated once per side.
Steps
Step 1: Form each side vector as the difference of its endpoints. For a triangle A,B,C:
AB=B−A,BC=C−B,CA=A−C.
Never use a vertex's coordinates themselves as direction cosines — positions are not directions.
Step 2: Find each length. ∣AB∣=a2+b2+c2, and likewise for the others. …
Common Mistakes
Mistake 1: Using the vertex coordinates as direction cosines.
Why it's wrong: coordinates are positions; a side's direction is the difference of its endpoints. Correct approach: form AB=B−A first, then normalise.
Mistake 2: Dividing every side by the same length.
Why it's wrong: each side has its own magnitude — ∣AB∣=∣BC∣=217 here, but ∣CA∣=242. Correct approach: use each side's own length as its divisor. …
- KEAM 2026Set eng-2026-04204 marksMCQQ.If α,β,γ are the angles made by the straight line 2x−5=3y+1=7z−2 with the x-axis, y-axis and z-axis respectively, then cosα,cosβ,cosγ are respectively, (A) 51,257,257 (B) 51,253,2537 (C) 5−1,253,257 (D) 51,25−3,257 (E) 51,253,257
›Reveal solutionSolution
Direction ratios (2,3,7); divide each by the magnitude to get direction cosines.
Direction vector =(2,3,7), magnitude =4+9+7=20=25. …
- KEAM 2025Set eng-2025-04234 marksMCQQ.Let α, β and γ be the angles made by a straight line with the x-axis, y-axis and z-axis respectively. If cosα+cosβ+cosγ=35, then the value of cosαcosβ+cosβcosγ+cosγcosα is equal to (A) 311 (B) 98 (C) 911 (D) 37 (E) 97
›Reveal solutionSolution
The value is 98.
Concept and Intuition
Direction cosines satisfy cos2α+cos2β+cos2γ=1. Squaring the given sum links it to the pairwise products.
Step-by-Step Solution
- (cosα+cosβ+cosγ)2=∑cos2+2∑cosαcosβ.
- (35)2=1+2S⇒925=1+2S.
- 2S=925−1=916⇒S=98. …
- KEAM 2024Set eng-2024-06054 marksMCQQ.If α,β,γ are the angles made by 3x−1=2y+1=−z with the coordinate axes, then (cosα,cosβ,cosγ)= (A) (143,142,14−1) (B) (73,7−2,7−1) (C) (143,14−2,14−1) (D) (73,72,7−1) (E) (14−3,14−2,14−1)
›Reveal solutionSolution
Direction ratios (3,2,−1), magnitude 14; direction cosines divide each by 14.
Rewrite −z=−1z, so the line is 3x−1=2y+1=−1z with direction ratios (3,2,−1).
Magnitude =9+4+1=14. …
- KEAM 2024Set eng-2024-06064 marksMCQQ.If a line makes angle α, β and γ with positive direction of x, y, and z-axis respectively, then cos2α+cos2β+cos2γ= (A) 1 (B) −1 (C) 2 (D) −2 (E) 0
›Reveal solutionSolution
Direction cosines satisfy ∑cos2=1.
For direction angles, cos2α+cos2β+cos2γ=1. Using cos2θ=2cos2θ−1: …
- KEAM 2024Set eng-2024-06074 marksMCQQ.Let α,β,γ be the direction cosines of a vector a=xi^+yj^+zk^, where z<0. If α=105−4 and β=215, then γ is equal to (A) 105−8 (B) 105−8 (C) 105−5 (D) 21−5 (E) 21−8
›Reveal solutionSolution
Direction cosines satisfy α2+β2+γ2=1. Here α2=10516, β2=215=10525, so γ2=1−10541=10564. Since z<0, γ=−1058.
Using α2+β2+γ2=1:
α2=10516,β2=(215)2=215=10525. …
- KEAM 2024Set eng-2024-06084 marksMCQQ.A line makes angle α,β,γ with x, y and z-axis respectively. Then the value of sin2α+sin2β−cos2γ is (A) 3 (B) 2 (C) 1 (D) 23 (E) 0
›Reveal solutionSolution
Direction angles satisfy cos2α+cos2β+cos2γ=1. This reduces sin2α+sin2β−cos2γ to 1.
For a line making angles α,β,γ with the axes,
cos2α+cos2β+cos2γ=1.
Then
sin2α+sin2β=(1−cos2α)+(1−cos2β)=2−(cos2α+cos2β).
Since cos2α+cos2β=1−cos2γ, …
- KEAM 2023Set eng-2023-P2-B24 marksMCQQ.The direction cosines of the vector a=−2i^+j^−k^ are (A) (62,61,61) (B) (6−2,61,6−1) (C) (6−2,6−1,6−1) (D) b=0 (E) (6−2,6−1,61)
›Reveal solutionSolution
The direction cosines are (-2/\u221a6, 1/\u221a6, -1/\u221a6).
Concept and Intuition
Direction cosines are the components of a vector divided by its magnitude; they give the cosines of the angles the vector makes with the coordinate axes and satisfy l2+m2+n2=1.
Step-by-Step Solution
- The vector is a=−2i^+j^−k^, so its magnitude is ∣a∣=(−2)2+12+(−1)2=6.
- Divide each component by the magnitude: (6−2,61,6−1).
- Check: 64+1+1=1, confirming they are valid direction cosines. …
- KEAM 2023Set eng-2023-P2-B24 marksMCQQ.If θ1,θ2 and θ3 are the angles made by a line with the positive directions of the x, y, z axes, then the value of cos2θ1+cos2θ2+cos2θ3 is (A) −1 (B) 1 (C) 2 (D) −2 (E) 0
›Reveal solutionSolution
cos2θ1+cos2θ2+cos2θ3=−1.
Concept and Intuition
Direction cosines satisfy cos2θ1+cos2θ2+cos2θ3=1. Use cos2θ=2cos2θ−1.
Step-by-Step Solution
- ∑cos2θi=∑(2cos2θi−1)=2∑cos2θi−3.
- ∑cos2θi=1.
- =2(1)−3=−1. …
- KEAM 2021Set eng-2021-P2-B14 marksMCQQ.If α is the angle made by the vector a=5i^+3j^+4k^ with the positive x-axis, then cosα= (A) 125 (B) 21 (C) 22 (D) 55 (E) 102
›Reveal solutionSolution
cosα=22.
Concept and Intuition
The angle a vector makes with the positive x-axis has cosine equal to its x-component divided by its magnitude (its first direction cosine).
Step-by-Step Solution
- ∣a∣=52+32+42=50=52.
- cosα=525=21=22. …
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