Reference frames — body frame, inertial frame; rotation between them: Worked examples

3.5.1 · D3 · Physics › Guidance, Navigation & Control (GNC) › Reference frames — body frame, inertial frame; rotation betw

Intuition Yeh page kis liye hai

Parent note ne tumhe machine dikhaya tha: rotation matrix $R_{B I}$ , uski properties, aur teen quick examples. Yeh page us machine ko har us scenario se guzaarta hai jo exam ya flight software mein aa sakta hai — har sign, har quadrant of angle, degenerate cases (zero rotation, rotation about the axis a vector already lies on), limiting cases ( $18 0^{\circ}$ , tiny angles), ek real-world word problem, aur ek nasty exam twist jahan frame direction deliberately reverse ki gayi hai.

Throughout, GNC matlab Guidance, Navigation & Control — onboard software jo figure out karta hai ki vehicle kidhar point kar raha hai aur use kaise steer karna hai.

Shuru karne se pehle, notation ka ek reminder taaki koi lost na ho:

$v$ ek physical arrow hai. $v^{I}$ = uske teen numbers (components) jaise inertial frame mein dikhai dete hain; $v^{B}$ = same arrow ke numbers jaise body frame mein dikhai dete hain.
Kisi letter par hat, jaise $\overset{s}{^}$ , matlab ek unit vector — bilkul $1$ length ka arrow, sirf ek direction point karne ke liye use hota hai.
$R_{B I}$ = "body-from-inertial" table jo $v^{I}$ ko $v^{B}$ mein badalta hai: $v^{B} = R_{B I} v^{I}$ .
$R^{T}$ = transpose (rows ko columns mein flip karo). Rotation matrix ke liye yeh uska inverse bhi hai.
ECI = Earth-Centered Inertial — ek non-spinning frame jiska origin Earth ke center par hai aur axes door ke stars se locked hain; yeh ek concrete inertial frame $I$ hai. ECEF = Earth-Centered, Earth-Fixed — same origin lekin axes Earth ke saath rotate karte hain (toh ground par koi bhi point apne ECEF coordinates fixed rakhta hai). Inhe yahan sirf naam diya gaya hai; yeh fully ECI, ECEF and orbit reference frames mein build hote hain.

Scenario matrix

Is topic se aane wala har question in cells mein se ek (ya mix) hota hai. Neeche ke 9 worked examples mein se har ek us cell se tagged hai jise woh cover karta hai, aur saath milke yeh har row ko touch karte hain.

Cell	Kya tricky banata hai	Covered by
A. Positive angle, one axis	$sin$ ka sign sahi karo	Ex 1
B. Negative angle	kya $- θ$ transpose jaisa hai?	Ex 2
C. Quadrant II mein angle ( $9 0^{\circ} < θ < 18 0^{\circ}$ )	$cos$ negative ho jaata hai	Ex 3
D. Degenerate: zero rotation	matrix identity honi chahiye	Ex 4
E. Degenerate: vector rotation axis par	kuch change nahi hona chahiye	Ex 4
F. Limiting case: $θ = 18 0^{\circ}$	full flip, dono $\pm$ signs gayab	Ex 5
G. Reverse direction (inverse / transpose)	body→inertial instead of inertial→body	Ex 6
H. Do rotations ka composition	order matters, subscripts cancel hone chahiye	Ex 7
I. Real-world word problem	English → axes → matrix mein translate karo	Ex 9
J. Exam twist: galat-taraf subscripts	transpose trap dhundhna	Ex 9
K. Orthogonality se sanity check	$R^{T} R = I$ check karo	Ex 5 & 6
L. Quadrant III mein angle ( $18 0^{\circ} < θ < 27 0^{\circ}$ )	$sin$ aur $cos$ dono negative	Ex 8

Hum parent ka $z$ -rotation apna workhorse banate hain (yeh "frame-to-frame" convention hai, toh inertial components andar aate hain aur body components bahar nikalta hain):

$R_{z} (θ) = cos θ - sin θ 0 sin θ cos θ 0 001$

Har entry ko aise padho: "ek axis ka kitna hissa doosre mein leak hota hai". $z$ -row/column untouched hai kyunki hum $z$ ke around spin kar rahe hain.

Worked examples

Worked example Example 1 — Positive angle, one axis (Cell A)

Body $z$ ke around $+ 9 0^{\circ}$ ghuma hai. Ek inertial vector inertial $Y$ ki taraf point karta hai: $v^{I} = (0, 1, 0)$ . $v^{B}$ nikalo.

Forecast: compute karne se pehle guess karo — body $+ 9 0^{\circ}$ (counter-clockwise) ghuma, toh inertial $Y$ arrow ab body ke $+ x$ ke along appear hona chahiye. Guess $(1, 0, 0)$ ?

$θ = 9 0^{\circ}$ par matrix likho: $cos 9 0^{\circ} = 0, sin 9 0^{\circ} = 1$ . $R_{z} (9 0^{\circ}) = 0 - 1 0 100001$ Yeh step kyun? Numeric answer ke liye multiply karne se pehle symbol $θ$ ko numbers mein badalna zaroori hai — matrix andar unknowns ke saath kisi numeric answer ke liye useless hai.
Row-by-row $(0, 1, 0)$ se multiply karo: $v^{B} = 0 \cdot 0 + 1 \cdot 1 + 0 - 1 \cdot 0 + 0 \cdot 1 + 0 0 = 100$ Yeh step kyun? Har output number ek matrix row (inertial coords mein ek body axis) aur $v^{I}$ ka dot product hai — yeh literally hai "arrow ko us body axis par project karo".

Verify: length preserved: $∥ (0, 1, 0) ∥ = 1 = ∥ (1, 0, 0) ∥$ ✔. Aur answer hamare forecast $(1, 0, 0)$ se match karta hai. ✔

Figure mein kya dekhna hai: do blue arrows inertial $X, Y$ axes hain; do pink arrows $+ 9 0^{\circ}$ turn ke baad body axes hain. Yellow arrow hamara vector $v$ hai, inertial $Y$ ke along. Notice karo yeh exactly pink body- $x$ axis ke along land karta hai — isliye uske body components $(1, 0, 0)$ padhte hain.

Worked example Example 2 — Negative angle vs transpose (Cell B)

Same vector $v^{I} = (0, 1, 0)$ , lekin ab body $z$ ke around $- 3 0^{\circ}$ ghuma hai. $v^{B}$ nikalo, aur confirm karo ki $R_{z} (- 3 0^{\circ}) = R_{z} (3 0^{\circ})^{T}$ .

Forecast: clockwise turn — inertial $Y$ arrow body $+ x$ ki taraf thoda jhukna chahiye, toh ek chhota positive $x$ -component aur positive-ish $y$ -component expect karo.

$cos (- 3 0^{\circ}) = cos 3 0^{\circ} = \frac{3}{2} \approx 0.8660$ aur $sin (- 3 0^{\circ}) = - sin 3 0^{\circ} = - 0.5$ use karo. $R_{z} (- 3 0^{\circ}) = 0.8660 0.5 0 - 0.5 0.8660 0 001$ Yeh step kyun? $cos$ even hai (angle ka sign ignore karta hai) lekin $sin$ odd hai (sign flip karta hai). Yeh ek hi fact exactly isliye hai ki $R_{z} (- θ)$ equals $R_{z} (θ)^{T}$ — $θ$ ko negate karna sirf do $sin$ entries ko flip karta hai, jo yahan transpose karna karta hai.
Multiply karo: $v^{B} = (0.8660 \cdot 0 - 0.5 \cdot 1, 0.5 \cdot 0 + 0.8660 \cdot 1, 0) = (- 0.5, 0.8660, 0)$ . Yeh step kyun? Seedha row-times-vector; hum 4 decimals rakhte hain toh length check honest ho.

Verify: $(- 0.5)^{2} + 0.866 0^{2} = 0.25 + 0.75 = 1$ ✔. Parent ka $R_{z} (3 0^{\circ}) = 0.8660 - 0.5 0 0.5 0.8660 0 001$ transpose karo → uska transpose exactly step 1 ki matrix hai ✔. Toh "doosri taraf turn karo" = "transpose karo".

Worked example Example 3 — Quadrant II mein angle,

$cos$ negative ho jaata hai (Cell C) Body $z$ ke around $+ 13 5^{\circ}$ ghuma hai. $v^{I} = (1, 0, 0)$ lo. $v^{B}$ nikalo.

Forecast: $13 5^{\circ}$ quarter turn se zyada lekin half turn se kam hai. Inertial $X$ arrow body view mein "down-and-left" point karna chahiye — negative $x^{B}$ aur negative $y^{B}$ expect karo.

Quadrant-II values: $cos 13 5^{\circ} = - \frac{2}{2} \approx - 0.7071$ , $sin 13 5^{\circ} = + \frac{2}{2} \approx 0.7071$ . Yeh step kyun? Yahi is cell ka crux hai: $9 0^{\circ}$ ke baad, $cos$ negative ho jaata hai jabki $sin$ abhi positive hai. Jo students sirf first-quadrant values yaad karte hain unhe yahan galat sign milta hai.
Pehla body component: $v_{x}^{B} = cos 13 5^{\circ} \cdot 1 + sin 13 5^{\circ} \cdot 0 = - 0.7071$ .
Doosra: $v_{y}^{B} = - sin 13 5^{\circ} \cdot 1 + cos 13 5^{\circ} \cdot 0 = - 0.7071$ . Yeh step kyun? Hum sirf $R_{z}$ ka pehla column read karte hain kyunki input $(1, 0, 0)$ mein sirf $x$ -part hai — woh pehla column wahan hai jahan inertial $X$ body coords mein land karta hai. Top row $+ sin$ carry karta hai jabki middle row $- sin$ ; yeh built-in sign difference hai jo rotation ko correctly "wrap around" karta hai har quadrant mein, aur yahan yeh do negative components deliver karta hai.

Verify: length $0.707 1^{2} + 0.707 1^{2} = 1$ ✔; dono components negative, forecast (down-and-left) se match ✔.

Figure mein kya dekhna hai: yellow arrow $v$ hai inertial $X$ ke along; chalk arc $13 5^{\circ}$ body turn mark karta hai; pink body axes vertical se aage swing kar chuke hain. Yellow annotation dikhata hai ki $v$ ke body components lower-left corner mein land kar rahe hain — dono negative, exactly jaise compute kiya.

Worked example Example 4 — Ek saath do degenerate cases (Cells D & E)

(a) Body ne rotate nahi kiya ( $θ = 0$ ): $R_{z} (0)$ kya hai aur yeh kisi bhi $v$ par kya karta hai? (b) Body $z$ ke around $+ 5 0^{\circ}$ rotate karti hai, lekin vector axis par hai: $v^{I} = (0, 0, 7)$ . $v^{B}$ nikalo.

Forecast: (a) "no turn" kuch change nahi karna chahiye → identity matrix. (b) arrow exactly spin axis ke along hai, toh us ke around spinning kuch nahi hilata → $(0, 0, 7)$ unchanged.

(a) $cos 0 = 1, sin 0 = 0$ , toh $R_{z} (0) = 100010001 = I .$ Yeh step kyun? Zero ka rotation do-nothing operation hona chahiye; koi aur result matlab hoga ki dono frames perfectly aligned hone par bhi disagree karte hain — impossible.
(b) $R_{z} (5 0^{\circ})$ ko $(0, 0, 7)$ se multiply karo. $R_{z}$ ka poora teesra column $(0, 0, 1)$ hai, toh: $v^{B} = (0, 0, 1 \cdot 7) = (0, 0, 7) .$ Yeh step kyun? $z$ -rotation $z$ -component ko untouched chhodta hai (bottom row $(0, 0, 1)$ hai). Purely axis ke along ek vector woh ek arrow hai jise us axis ke around kisi bhi amount ki spin nahi hila sakti — degenerate/fixed case.

Verify: (a) $I v = v$ har $v$ ke liye ✔. (b) length $7$ andar, length $7$ bahar; answer $5 0^{\circ}$ se independent hai — koi bhi angle plug karo aur $z$ -part $7$ rahega ✔.

Worked example Example 5 — Limiting case

$θ = 18 0^{\circ}$ aur orthogonality check (Cells F & K) Body $z$ ke around $+ 18 0^{\circ}$ ghuma. $v^{I} = (3, 4, 0)$ lo. $v^{B}$ nikalo aur $R^{T} R = I$ verify karo.

Forecast: half turn left↔right aur front↔back flip karta hai lekin up/down rakhta hai. Expect karo $(3, 4, 0) \to (- 3, - 4, 0)$ .

$18 0^{\circ}$ par: $cos 18 0^{\circ} = - 1, sin 18 0^{\circ} = 0$ . $R_{z} (18 0^{\circ}) = - 1 00 0 - 1 0 001$ Yeh step kyun? Limit par dono $sin$ entries vanish ho jaati hain, toh $x$ aur $y$ ki koi cross-mixing nahi — matrix sirf unhe negate karta hai. Yeh quadrant II aur quadrant III ke beech ka clean boundary hai.
Multiply karo: $v^{B} = (- 1 \cdot 3, - 1 \cdot 4, 0) = (- 3, - 4, 0)$ . Yeh step kyun? Row-times-vector matrix apply karne ki actual definition hai; hum ise carry out karte hain geometric forecast (clean sign-flip of $x$ aur $y$ ) ko sirf intuition ke bajaye numbers se confirm karne ke liye.
Orthogonality check: kyunki $R$ ek aisi matrix hai jiske non-zero entries sirf main diagonal (top-left se bottom-right) par hain, yahan $- 1, - 1, 1$ , hame $R^{T} = R$ milta hai, aur $R^{T} R$ ki entries us diagonal par $(- 1)^{2}, (- 1)^{2}, 1^{2} = 1, 1, 1$ hain — yani $R^{T} R = I$ . Yeh step kyun? Har valid rotation matrix $R^{T} R = I$ satisfy karna chahiye; agar nahi karta, toh lengths change ho jaati. Yeh poore GNC (Guidance, Navigation & Control) mein sabse sasti correctness test hai.

Verify: $∥ (3, 4, 0) ∥ = 5$ aur $∥ (- 3, - 4, 0) ∥ = 5$ ✔; determinant $= (- 1) (- 1) (1) = + 1$ (proper rotation, no mirror) ✔.

Worked example Example 6 — Direction reverse karo: body→inertial (Cells G & K)

Gyro-integrated attitude deta hai $R_{B I} = R_{z} (6 0^{\circ})$ . Ek sensor body coords mein ek vector read karta hai, $v^{B} = (2, 0, 0)$ . Yeh inertial coords mein kya hai?

Forecast: hum "galat taraf" ja rahe hain (body→inertial), toh inverse chahiye. Kyunki yeh rotation hai, inverse = transpose = $R_{z} (- 6 0^{\circ})$ . Ek body- $x$ arrow, $6 0^{\circ}$ se un-rotated, positive- $X$ , positive- $Y$ ki taraf swing karna chahiye.

Transposing se invert karo (no Gaussian elimination): $v^{I} = R_{B I}^{T} v^{B} = R_{z} (6 0^{\circ})^{T} 200 .$ Yeh step kyun? Defining relation hai $v^{B} = R_{B I} v^{I}$ ; $v^{I}$ solve karne ke liye dono sides ko $R^{- 1} = R^{T}$ se multiply karo. Real inverse karna wasteful aur onboard kam accurate hoga.
$R_{z} (6 0^{\circ})^{T} = 0.5 0.8660 0 - 0.8660 0.5 0 001$ (do $sin$ signs swap karo). $(2, 0, 0)$ se multiply karo: $v^{I} = (2 \cdot 0.5, 2 \cdot 0.8660, 0) = (1, 1.7321, 0) .$ Yeh step kyun? Sirf pehla column matter karta hai kyunki $v^{B}$ mein sirf $x$ -part hai; $R^{T}$ ka pehla column $R$ ki pehli row hai — yani "body- $x$ inertial space mein kahan rehta hai".

Verify: length $1^{2} + 1.732 1^{2} = 1 + 3 = 2$ ✔ ( $∥ (2, 0, 0) ∥$ se match karta hai). Result inertial $X$ ke saath $6 0^{\circ}$ angle banata hai: $tan^{- 1} (1.7321/1) = 6 0^{\circ}$ ✔.

Worked example Example 7 — Composition, order matters (Cell H)

Vehicle ek intermediate orbit frame $O$ use karta hai — ek reference frame jo satellite ki orbit ke around position track karta hai (ECI, ECEF and orbit reference frames mein fully defined; yahan bas itna jaano ki yeh inertial aur body ke beech baithta hai). Chain: inertial→orbit $R_{O I} = R_{z} (9 0^{\circ})$ hai, orbit→body $R_{B O} = R_{z} (3 0^{\circ})$ hai. $R_{B I}$ build karo aur ise $v^{I} = (1, 0, 0)$ par apply karo.

Forecast: same axis par do turns bas add ho jaate hain: $9 0^{\circ} + 3 0^{\circ} = 12 0^{\circ}$ total. Toh answer $R_{z} (12 0^{\circ}) (1, 0, 0)$ ke barabar hona chahiye.

Parent ke rule se compose karo (adjacent subscripts cancel ho jaate hain): $R_{B I} = R_{B O} R_{O I} = R_{z} (3 0^{\circ}) R_{z} (9 0^{\circ}) = R_{z} (12 0^{\circ}) .$ Yeh step kyun? Sabse right wali matrix $R_{O I}$ $v^{I}$ par pehle act karti hai ( $I \to O$ ), phir $R_{B O}$ act karta hai ( $O \to B$ ). Inner $O$ 's touch karte hain aur cancel ho jaate hain, $B I$ bach jaata hai. Same axis ke around rotations ke liye, angles simply add ho jaate hain — ek fact jo hum check ke roop mein exploit kar sakte hain.
$12 0^{\circ}$ par apply karo: $cos 12 0^{\circ} = - 0.5, sin 12 0^{\circ} = 0.8660$ . $v^{B} = (cos 12 0^{\circ} \cdot 1, - sin 12 0^{\circ} \cdot 1, 0) = (- 0.5, - 0.8660, 0) .$ Yeh step kyun? Total angle maloom hone par, ek multiply kaam khatam karta hai; do matrices alag-alag apply karne ki zaroorat nahi.

Verify: length $0.25 + 0.75 = 1$ ✔. Independently matrices ek ek karke apply karo: $R_{z} (9 0^{\circ}) (1, 0, 0) = (0, - 1, 0)$ , phir $R_{z} (3 0^{\circ}) (0, - 1, 0) = (- 0.5, - 0.8660, 0)$ — same answer ✔, aur yeh confirm karta hai ki order/composition ne kaam kiya.

Worked example Example 8 — Quadrant III mein angle,

$sin$ aur $cos$ dono negative (Cell L) Body $z$ ke around $+ 22 5^{\circ}$ ghuma. $v^{I} = (1, 0, 0)$ lo. $v^{B}$ nikalo.

Forecast: $22 5^{\circ}$ half-turn ( $18 0^{\circ}$ ) ke baad lekin teen-chautha ( $27 0^{\circ}$ ) se pehle hai. $cos$ aur $sin$ yahan dono negative hain. Inertial $X$ arrow, $18 0^{\circ}$ se aage over-rotated, body view mein "up-and-left" point karna chahiye — negative $x^{B}$ aur positive $y^{B}$ expect karo.

Quadrant-III values: $cos 22 5^{\circ} = - \frac{2}{2} \approx - 0.7071$ , $sin 22 5^{\circ} = - \frac{2}{2} \approx - 0.7071$ . Yeh step kyun? Is cell ka poora point yahi hai: $18 0^{\circ}$ aur $27 0^{\circ}$ ke beech dono trig functions negative hain. Sirf quadrants I–II yaad karne se tum yahan helpless ho jaate ho.
Pehla body component: $v_{x}^{B} = cos 22 5^{\circ} \cdot 1 + sin 22 5^{\circ} \cdot 0 = - 0.7071$ .
Doosra: $v_{y}^{B} = - sin 22 5^{\circ} \cdot 1 + cos 22 5^{\circ} \cdot 0 = - (- 0.7071) = + 0.7071$ . Yeh step kyun? Hum phir sirf $R_{z}$ ka pehla column use karte hain (input pure $x$ hai). Middle row $- sin$ carry karta hai; kyunki $sin 22 5^{\circ}$ khud negative hai, do minus signs cancel ho jaate hain aur $y^{B}$ positive aata hai — exactly woh sign trap jise yeh cell expose karne ke liye bana hai.

Verify: length $0.707 1^{2} + 0.707 1^{2} = 1$ ✔; signs $(-, +)$ forecast (up-and-left) se match ✔.

Worked example Example 9 — Word problem + subscript trap (Cells I & J)

Ek satellite ka star tracker ek guide star ko inertial (ECI) direction $\overset{s}{^}^{I} = (1, 0, 0)$ par report karta hai (yaad karo hat matlab unit-length pointing direction). Mission control kehta hai vehicle inertial ke relative $z$ ke around $+ 4 0^{\circ}$ yaw kiya hai, toh $R_{B I} = R_{z} (4 0^{\circ})$ . Ek junior engineer star ki body direction $\overset{s}{^}^{B} = R_{B I}^{T} \overset{s}{^}^{I}$ compute karta hai. Kya yeh sahi hai? Correct $\overset{s}{^}^{B}$ do.

Forecast: defining rule hai $v^{B} = R_{B I} v^{I}$ — no transpose. Junior engineer ne transpose use kiya, toh woh reverse direction compute kar raha hai. Correct answer mein star negative body- $y$ ki taraf swing karna chahiye (positive yaw inertial $X$ ko body $- y$ ki taraf push karta hai).

Trap spot karo: $R_{B I}$ ke subscripts padhte hain "body from inertial". Inner subscript $I$ ko input $\overset{s}{^}^{I}$ ke frame se touch karna chahiye. $R_{B I} \overset{s}{^}^{I}$ mein $I$ 's milte hain aur cancel ho jaate hain, $B$ bachta hai. $R^{T} = R_{I B}$ use karne ke liye body-frame input chahiye hoga — galat. Yeh step kyun? Subscript-cancellation rule ek guardrail hai: yeh exactly is transpose mistake ko pakadta hai bina kisi geometry ke.
$4 0^{\circ}$ par correct computation: $cos 4 0^{\circ} = 0.7660, sin 4 0^{\circ} = 0.6428$ . $\overset{s}{^}^{B} = R_{z} (4 0^{\circ}) 100 = (cos 4 0^{\circ}, - sin 4 0^{\circ}, 0) = (0.7660, - 0.6428, 0) .$ Yeh step kyun? Hum real definition apply karte hain; $R_{z} (4 0^{\circ})$ ka pehla column batata hai ki inertial $X$ body coords mein kahan land karta hai.
Engineer ki galti dikhao: $R^{T} (1, 0, 0) = (cos 4 0^{\circ}, + sin 4 0^{\circ}, 0) = (0.7660, + 0.6428, 0)$ . Yeh step kyun? Wrong answer ko explicit banana ek transpose slip ki signature reveal karta hai — $y$ -component ka sign flip ho jaata hai jabki $x$ same rehta hai. Woh ek flipped sign is poore class ke exam traps ka tell-tale hai.

Verify: correct length $0.766 0^{2} + 0.642 8^{2} = 1$ ✔; $y$ ka sign negative hai jaisa $+ 4 0^{\circ}$ yaw ke liye forecast kiya ✔; engineer ka version sirf $y$ ke sign mein differ karta hai, confirm karta hai ki yeh reverse rotation (ek transpose) hai na ki requested wala ✔.

Takeaways — poora matrix, ek glance mein

Scenario matrix ka har cell ab worked ho gaya hai. Nau examples ke neeche ka pattern same tiny machine hai, $R_{z} (θ)$ , dhyan se padha gaya:

Signs $θ$ ke quadrant se aate hain. $cos$ $9 0^{\circ}$ ke baad negative flip karta hai (Ex 3); $sin$ $18 0^{\circ}$ ke baad negative flip karta hai (Ex 8); exactly $18 0^{\circ}$ par dono $sin$ entries vanish ho jaati hain aur pure sign-flip milta hai (Ex 5). Sirf first-quadrant instincts par kabhi trust mat karo.
"Doosri taraf turn karo" = transpose (Ex 2), aur transpose = inverse kisi bhi rotation ke liye (Ex 6) — kyunki $R^{T} R = I$ . Yeh onboard sabse sasta, sabse accurate reversal hai.
Degenerate cases tumhari sanity anchor karte hain: zero rotation $I$ deta hai (Ex 4a); spin axis par ek vector kabhi nahi hilta (Ex 4b).
Subscripts touch karke compose karo — inner labels kiss karte hain aur cancel ho jaate hain, aur same-axis angles simply add ho jaate hain (Ex 7).
Subscript notation ek guardrail hai sabse common exam trap ke against: ek transpose sneaking in karna (Ex 9).

Recall Kaun sa cell kaunsa hai?

Positive one-axis turn, $sin$ ka sign ::: Ex 1 (Cell A) "Doosri taraf turn karo" matlab kaun sa operation? ::: matrix transpose karo (Ex 2, Cell B) Quadrant II mein kiska sign flip hota hai? ::: $cos$ negative ho jaata hai (Ex 3, Cell C) $0^{\circ}$ rotation se kaun si matrix milti hai? ::: identity $I$ (Ex 4, Cell D) Spin axis par ek vector kitna change hota hai? ::: bilkul nahi (Ex 4, Cell E) $18 0^{\circ}$ par matrix $x, y$ ka kya karta hai? ::: unhe negate karta hai, koi cross-mixing nahi (Ex 5, Cell F) Rotation matrix ka sabse sasta inverse? ::: uska transpose $R^{T}$ (Ex 6, Cell G) Same-axis rotations compose kaise hote hain? ::: unke angles add ho jaate hain (Ex 7, Cell H) Quadrant III mein $sin$ aur $cos$ ke signs kya hain? ::: dono negative (Ex 8, Cell L) Subscript guardrail kaun si galti pakadta hai? ::: galat taraf transpose apply karna (Ex 9, Cell J) ECI aur ECEF ka full form kya hai? ::: Earth-Centered Inertial aur Earth-Centered Earth-Fixed

Yeh bhi dekho: Euler angles and gimbal lock · Quaternions for attitude representation · IMU and gyroscope sensing · Coriolis and fictitious forces in rotating frames · ECI, ECEF and orbit reference frames · Attitude determination — star trackers