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Let the position vectors of three vertices of a \triangle be 4vecp+vecq-3vecr, -5vecp+vecq+2vecr and 2vecp-vecq+2vecr If the position vectors of the orthocenter and the circumcenter of the \triangle are fracvecp+vecq+vecr4 and alphavecp+betavecq+gammavecr respectively, then alpha+2beta+5gamma is equal to: [cite: 3266, 3267, 3268, 3269, 3270, 3271, 3272]

Solution & Explanation

### Related Formula 1. Centroid (G) of a \triangle with vertices A, B, C is given by: vecG = fracvecA + vecB + vecC3 2. Euler\'s line property: The orthocenter (O), centroid (G), and circumcenter (C) are collinear, and G divides the segment OC internally in the ratio 2:1. ### Step 1: Compute the Centroid Vector Sum the vectors of the three given vertices [cite: 3266, 3268]: vecA = 4vecp+vecq-3vecr vecB = -5vecp+vecq+2vecr vecC = 2vecp-vecq+2vecr vecG = frac(4 - 5 + 2)vecp + (1 + 1 - 1)vecq + (-3 + 2 + 2)vecr3 = fracvecp + vecq + vecr3 ### Step 2: Apply Euler Line Section Ratio Using the section formula ratio O-G-C as 2:1 [cite: 3931, 3932]:
Euler Line section diagram for Q56 - JEE Main 2025 Evening
Euler Line section diagram for Q56 - JEE Main 2025 Evening
vecG = frac2vecC + vecO3 Rightarrow 3vecG = 2vecC + vecO 2vecC = 3vecG - vecO = 3left(fracvecp + vecq + vecr3right) - fracvecp + vecq + vecr4 2vecC = (vecp + vecq + vecr) - frac14(vecp + vecq + vecr) = frac34(vecp + vecq + vecr) vecC = frac38vecp + frac38vecq + frac38vecr ### Step 3: Coefficient Matching Compare with the given circumcenter format alphavecp + betavecq + gammavecr [cite: 3270, 3939]: alpha = frac38, quad beta = frac38, quad gamma = frac38 Calculate alpha + 2beta + 5gamma [cite: 3272, 3949]: frac38 + 2left(frac38right) + 5left(frac38right) = frac3 + 6 + 158 = frac248 = 3 ### Pattern Recognition Euler line configuration is universally O-G-C in 2:1. Remember the mnemonic 'Oil-Gas-Company' or simply 3G = 2C + O to prevent swapping structural coefficients under exam stress. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra Class 11 Mathematics: Properties of Triangles

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

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