Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)

Let the three sides of a triangle ABC be given by the vectors 2hatmathbfi - hatmathbfj + hatmathbfk , hatmathbfi - 3hatmathbfj - 5hatmathbfk and 3hatmathbfi - mathbf4hatmathbfj - mathbf4hatmathbfk . Let G be the centroid of the triangle ABC. Then 6leftleft|overlineAGright|^2 + left|overlineBGright|^2 + left|overlineCGright|^2right) is equal to

Numerical Answer Type:
Enter a numerical value Answer: 164 to 164 +4 marks

Solution & Explanation

### Core Logic Let the vertices of the triangle be A, B, and C. The vectors representing the side paths are: overlineAB = 2hati - hatj + hatk overlineCA = hati - 3hatj - 5hatk overlineCB = 3hati - 4hatj - 4hatk Notice that overlineAB + overlineCA = (2+1)hati + (-1-3)hatj + (1-5)hatk = 3hati - 4hatj - 4hatk = overlineCB. This structurally validates vector addition rules.
Vector algebra diagram for Q75 - JEE Main 2025 Evening
Vector algebra diagram for Q75 - JEE Main 2025 Evening
### Step 1: Finding Position Vectors relative to A Let's set vertex A as the origin origin point (Position vector vecA = vec0): - Position vector of B: vecB = 2hati - hatj + hatk - Position vector of C: Since overlineCA = vecA - vecC = -vecC implies vecC = -hati + 3hatj + 5hatk Now, calculate the position vector of the centroid G: vecG = fracvecA + vecB + vecC3 = fracvec0 + (2hati - hatj + hatk) + (-hati + 3hatj + 5hatk)3 = frac13left(hati + 2hatj + 6hatkright) ### Step 2: Calculating Squared Lengths to the Centroid Let's find each individual vector distance block: - overlineAG = vecG - vecA = frac13(hati + 2hatj + 6hatk) implies |overlineAG|^2 = frac19(1^2 + 2^2 + 6^2) = frac419 - overlineBG = vecG - vecB = left(frac13-2right)hati + left(frac23+1right)hatj + left(2-1right)hatk = -frac53hati + frac53hatj + 1hatk |overlineBG|^2 = left(-frac53right)^2 + left(frac53right)^2 + 1^2 = frac259 + frac259 + 1 = frac599 - overlineCG = vecG - vecC = left(frac13+1right)hati + left(frac23-3right)hatj + left(2-5right)hatk = frac43hati - frac73hatj - 3hatk |overlineCG|^2 = left(frac43right)^2 + left(-frac73right)^2 + (-3)^2 = frac169 + frac499 + 9 = frac1469 ### Step 3: Final Targeted Evaluation Summing the squared values and multiplying by 6: textValue = 6 left[ |overlineAG|^2 + |overlineBG|^2 + |overlineCG|^2 right] = 6 left[ frac419 + frac599 + frac1469 right] textValue = 6 times frac2469 = 2 times frac2463 = 2 times 82 = 164 ### Pattern Recognition Setting one vector node as the origin point (vecA = vec0) heavily dampens intermediate coordinate math steps, avoiding dealing with an absolute baseline origin orientation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra

Reference Study Guides

Vector Algebra

More Vector Algebra Previous-Year Questions

Q68 2025 Properties of Vectors
Let veca = 2hati - 3hatj + hatk, vecb = 3hati + 2hatj + 5hatk and a vector vecc be such that (veca - vecc) times vecb = -18hati - 3hatj + 12hatk and veca cdot vecc = 3. If vecb times vecc = vecd, then |veca cdot vecd| is equal to:
  • A. 18
  • B. 12
  • C. 9
  • D. 15

Solution

### Related Formula textVector Cross product distributes over subtraction: (veca - vecc) times vecb = veca times vecb - vecc times vecb textScalar Triple Product cyclic identity: veca cdot (vecb times vecc) = (veca times vecb) cdot vecc textAntisymmetry: vecc times vecb = - vecb times vecc ### Core Logic Instead of solving for the individual coordinates of vector vecc, we apply vector algebraic identities to compute the target scalar triple product directly. ### Step 1: Expand and rewrite the cross product Given (veca - vecc) times vecb = -18hatmathrmi - 3hatmathrmj + 12hatmathrmk: veca times vecb - vecc times vecb = -18hatmathrmi - 3hatmathrmj + 12hatmathrmk veca times vecb + vecb times vecc = -18hatmathrmi - 3hatmathrmj + 12hatmathrmk vecb times vecc = (-18hatmathrmi - 3hatmathrmj + 12hatmathrmk) - (veca times vecb) quad text--- (1) ### Step 2: Calculate a x b Evaluate the cross product: veca times vecb = beginvmatrix hatmathrmi & hatmathrmj & hatmathrmk \\ 2 & -3 & 1 \\ 3 & 2 & 5 endvmatrix veca times vecb = hatmathrmi(-15 - 2) - hatmathrmj(10 - 3) + hatmathrmk(4 - (-9)) = -17hatmathrmi - 7hatmathrmj + 13hatmathrmk ### Step 3: Solve for the vector d Substitute veca times vecb back into equation (1): vecd = vecb times vecc = (-18hatmathrmi - 3hatmathrmj + 12hatmathrmk) - (-17hatmathrmi - 7hatmathrmj + 13hatmathrmk) vecd = -hatmathrmi + 4hatmathrmj - hatmathrmk ### Step 4: Compute the final dot product Now compute the requested dot product: veca cdot vecd = (2hatmathrmi - 3hatmathrmj + hatmathrmk) cdot (-hatmathrmi + 4hatmathrmj - hatmathrmk) veca cdot vecd = 2(-1) + (-3)(4) + 1(-1) = -2 - 12 - 1 = -15 left| veca cdot vecd right| = 15 ### Pattern Recognition Scalar triple product shortcut: Recognizing that veca cdot vecd = veca cdot (vecb times vecc) = [ veca \, vecb \, vecc ] allows you to find the scalar value through simple determinants and linear equations instead of solving for the vector components. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra
Q 2025 Vector Projections
If veca is a nonzero vector such that its projections on the vectors 2hatmathbfi - hatmathbfj + 2hatmathbfk, hatmathbfi + 2hatmathbfj - 2hatmathbfk and hatmathbfk are equal, then a unit vector along veca is:
  • A. frac1sqrt155left(-7hatmathrmi + 9hatmathrmj + 5hatmathrmkright)
  • B. frac1sqrt155left(-7hatmathrmi + 9hatmathrmj - 5hatmathrmkright)
  • C. frac1sqrt155left(7hatmathrmi + 9hatmathrmj + 5hatmathrmkright)
  • D. frac1sqrt155left(7hatmathrmi + 9hatmathrmj - 5hatmathrmkright)

Solution

### Related Formula The projection of vector veca onto vector vecb is: textProj_vecbveca = fracveca cdot vecb|vecb| ### Core Logic Let veca = a_1hatmathbfi + a_2hatmathbfj + a_3hatmathbfk. Set up equations equating the three projections to determine the ratios of components. ### Step 1: Set up Projection Ratios Let vecb = 2hatmathbfi-hatmathbfj+2hatmathbfk implies |vecb| = 3 Let vecc = hatmathbfi+2hatmathbfj-2hatmathbfk implies |vecc| = 3 Let vecd = hatmathbfk implies |vecd| = 1 Equating projections: frac2a_1 - a_2 + 2a_33 = fraca_1 + 2a_2 - 2a_33 = fraca_31 ### Step 2: Solve System of Linear Equations From the last equality: 2a_1 - a_2 + 2a_3 = 3a_3 implies 2a_1 - a_2 = a_3 quad dots (1) a_1 + 2a_2 - 2a_3 = 3a_3 implies a_1 + 2a_2 = 5a_3 quad dots (2) Multiply (1) by 2 and add to (2): 4a_1 - 2a_2 + a_1 + 2a_2 = 2a_3 + 5a_3 implies 5a_1 = 7a_3 implies a_1 = frac75a_3 Substitute back to find a_2: a_2 = 2a_1 - a_3 = 2left(frac75a_3right) - a_3 = frac95a_3 ### Step 3: Normalize to Unit Vector The vector components are in ratio a_1 : a_2 : a_3 = frac75 : frac95 : 1 = 7 : 9 : 5. textMagnitude factor = sqrt7^2 + 9^2 + 5^2 = sqrt49 + 81 + 25 = sqrt155 Thus, the unit vector is: hata = frac1sqrt155left(7hatmathbfi + 9hatmathbfj + 5hatmathbfkright) ### Pattern Recognition Looking at the options, only option (3) provides components with positive signs for all three unit basis elements matching our proportional derivation of 7:9:5 directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Vector Algebra
Q74 2025 Vector Products
Let veca = hati + 2hatj + hatk, vecb = 3hati - 3hatj + 3hatk, vecc = 2hati - hatj + 2hatk and vecd be a vector such that vecb times vecd = vecc times vecd and veca cdot vecd = 4. Then |(veca times vecd)|^2 is equal to
Numerical Answer. Answer: 128 to 128

Solution

### Related Formula Lagrange's Identity: |vecu times vecv|^2 = |vecu|^2 |vecv|^2 - (vecu cdot vecv)^2 Also, if vecu times vecw = vecv times vecw implies (vecu - vecv) times vecw = 0 implies vecw is parallel/collinear to vecu - vecv. ### Core Logic Rearranging the cross product: vecb times vecd - vecc times vecd = vec0 implies (vecb - vecc) times vecd = vec0 Thus, vecd = lambda (vecb - vecc). ### Step 1: Finding vector vecd Calculate (vecb - vecc): vecb - vecc = (3hati - 3hatj + 3hatk) - (2hati - hatj + 2hatk) = hati - 2hatj + hatk Therefore: vecd = lambda(hati - 2hatj + hatk) Substitute into veca cdot vecd = 4: lambda(hati + 2hatj + hatk) cdot (hati - 2hatj + hatk) = 4 lambda(1 - 4 + 1) = 4 implies -2lambda = 4 implies lambda = -2 Thus: vecd = -2(hati - 2hatj + hatk) = -2hati + 4hatj - 2hatk |vecd|^2 = (-2)^2 + 4^2 + (-2)^2 = 4 + 16 + 4 = 24 ### Step 2: Calculating |veca times vecd|^2 using Lagrange's Identity Given |veca|^2 = 1^2 + 2^2 + 1^2 = 6: |veca times vecd|^2 = |veca|^2 |vecd|^2 - (veca cdot vecd)^2 |veca times vecd|^2 = 6 times 24 - 4^2 = 144 - 16 = 128 ### Pattern Recognition Applying Lagrange's Identity avoids computing the actual cross product determinants vector-by-vector. This is extremely efficient and reduces chances of sign errors. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra
Q69 2025 Dot and Cross Products
Let the angle theta, 0 < theta < fracpi2 between two unit vectors hata and hatb be sin^-1left(fracsqrt659right) . If the vector vecc = 3hata + 6hatb + 9(hata times hatb) , then the value of 9(vecc cdot hata) - 3(vecc cdot hatb) is
  • A. 31
  • B. 27
  • C. 29
  • D. 24

Solution

### Related Formula Properties of dot and cross vector setups: hata cdot hata = |hata|^2 = 1 hata cdot hatb = |hata||hatb|costheta = costheta hata cdot (hata times hatb) = 0, quad hatb cdot (hata times hatb) = 0 ### Core Logic Given sintheta = fracsqrt659. Since theta lies in the first quadrant: costheta = sqrt1 - sin^2theta = sqrt1 - frac6581 = sqrtfrac1681 = frac49 Now, let's take individual dot product equations using vecc = 3hata + 6hatb + 9(hata times hatb): 1. Find vecc cdot hata: vecc cdot hata = 3(hata cdot hata) + 6(hatb cdot hata) + 9((hata times hatb) cdot hata) vecc cdot hata = 3(1) + 6costheta + 0 = 3 + 6left(frac49right) = 3 + frac249 = frac519 ### Step 1: Compute Second Dot Product Term 2. Find vecc cdot hatb: vecc cdot hatb = 3(hata cdot hatb) + 6(hatb cdot hatb) + 9((hata times hatb) cdot hatb) vecc cdot hatb = 3costheta + 6(1) + 0 = 3left(frac49right) + 6 = frac129 + 6 = frac12 + 549 = frac669 = frac223 ### Step 2: Substitute and Finalize Result Evaluate the target expression layout: textValue = 9(vecc cdot hata) - 3(vecc cdot hatb) textValue = 9left(frac519right) - 3left(frac223right) = 51 - 22 = 29 ### Pattern Recognition Remember that a cross product vector (hata times hatb) is orthogonal to both constituent vectors. Thus, taking their dot product evaluates to 0 immediately, allowing you to ignore that entire component during computation steps. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra
Q59 2025 Coplanar and Perpendicular Vectors
Let veca = hati + 2hatj + hatk and vecb = 2hati + hatj - hatk. Let hatc be a unit vector in the plane of the vectors veca and vecb and be perpendicular to veca. Then such a vector hatc is:
  • A. frac1sqrt5 (hatmathrmj - 2hatmathrmk)
  • B. frac1sqrt3left(-hatmathrmi + hatmathrmj - hatmathrmkright)
  • C. frac1sqrt3left(hatmathrmi - hatmathrmj + hatmathrmkright)
  • D. frac1sqrt2left(-hatmathrmi + hatmathrmkright)

Solution

### Related Formula vecp = K(veca + lambdavecb) vecp cdot veca = 0 ### Core Logic Formulate a coplanar parameterization vector, apply the zero dot-product geometric orthogonality constraint to pin down the linear parameter, and then normalize. ### Step 1: Define Coplanar Structural Form Let the targeting vector path be: vecp = K(veca + lambdavecb) = Kleft( (1+2lambda)hati + (2+lambda)hatj + (1-lambda)hatk right) ### Step 2: Force Orthogonality Constraint Impose vecp cdot veca = 0: 1(1+2lambda) + 2(2+lambda) + 1(1-lambda) = 0 1 + 2lambda + 4 + 2lambda + 1 - lambda = 0 implies 6 + 3lambda = 0 implies lambda = -2 ### Step 3: Substitute and Normalize Substitute lambda = -2 back into the base formulation: vecp = K(-3hati + 3hatk) Normalizing to turn this vector into a proper unit scale form: hatc = pm frac-hati + hatksqrt2 ### Pattern Recognition Finding coplanar vectors orthogonal to one base component matches taking cross expansions like (veca times vecb) times veca up to scalar metrics. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra

More Vector Algebra Questions — jee_main_2025_04_april_evening

Practice all Vector Algebra previous-year questions →

YOUR FIRST PREP STEP STARTS HERE

We Map Every Repeating Question in Competitive Exams.

Say goodbye to generic mock test fatigue. RankBit uses smart analysis to group past exam questions into their foundational Repeating Question Types. Find chapter weightage, track repeating questions, and score higher with targeted practice.

Select Your Target Exam

Choose an exam track below to find formulas per chapter and patterns.

Syncing Exam Intelligence

Mapping formulas and patterns across all tracks…

PATH A — FULL LENGTH PRACTICE

Full Mock Test Hub

Simulate real NTA exam conditions with fully tracked mocks. Time yourself against past papers.

Under Development
PATH B — TARGETED PRACTICE

Topic-wise Practice Hub

Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.

Loading Questions... Browse Topics
Latest from the Blog
View all →

Loading articles...