Let veca = -hati + 2hatj + 2hatk , vecb = 8hati + 7hatj - 3hatk and vecc be a vector such that veca times vecc = vecb . If vecc. (hati + hatj + hatk) = 4 , then |veca + vecc|^2 is equal to:

Solution & Explanation

### Related Formula veca times vecc = beginvmatrix hati & hatj & hatk \\ a_1 & a_2 & a_3 \\ c_1 & c_2 & c_3 endvmatrix ### Core Logic Let the unknown vector be vecc = c_1hati + c_2hatj + c_3hatk. Given veca = -hati + 2hatj + 2hatk and vecb = 8hati + 7hatj - 3hatk. Compute the cross product veca times vecc: veca times vecc = (2c_3 - 2c_2)hati + (-1 cdot c_3 - 2c_1)(-1)hatj + (-1 cdot c_2 - 2c_1)hatk = (2c_3 - 2c_2)hati + (c_3 + 2c_1)hatj - (c_2 + 2c_1)hatk ### Step 1: Equate components to find c Equating this to vecb: 2c_3 - 2c_2 = 8 Rightarrow c_3 - c_2 = 4 c_3 + 2c_1 = 7 -(c_2 + 2c_1) = -3 Rightarrow c_2 + 2c_1 = 3 We are also given vecc cdot (hati + hatj + hatk) = 4: c_1 + c_2 + c_3 = 4
Cross product vectors calculation Q5 - JEE Main 2026 Morning
Cross product vectors calculation Q5 - JEE Main 2026 Morning
### Step 2: Solve the linear system From the dot product equation: c_1 = 4 - c_2 - c_3. Substitute into c_2 + 2c_1 = 3: c_2 + 2(4 - c_2 - c_3) = 3 Rightarrow 8 - c_2 - 2c_3 = 3 Rightarrow c_2 + 2c_3 = 5 We have the system: c_3 - c_2 = 4 Rightarrow c_2 = c_3 - 4 Substitute into above: (c_3 - 4) + 2c_3 = 5 Rightarrow 3c_3 = 9 Rightarrow c_3 = 3 Then c_2 = 3 - 4 = -1. Then c_1 = 4 - (-1) - 3 = 2. So, vecc = 2hati - hatj + 3hatk. ### Step 3: Evaluate the final expression veca + vecc = (-hati + 2hatj + 2hatk) + (2hati - hatj + 3hatk) = hati + hatj + 5hatk |veca + vecc|^2 = 1^2 + 1^2 + 5^2 = 1 + 1 + 25 = 27 ### Pattern Recognition Whenever veca times vecc = vecb and a dot product constraint is given, simply expand vecc algebraically as (c_1, c_2, c_3). The cross product and dot product create an easily solvable 3x3 linear system. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Vector Algebra

Reference Study Guides

Vector Algebra

More Vector Algebra Previous-Year Questions — Page 2

Q69 jee_main_2025_07_april_morning 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 jee_main_2025_08_april_evening 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
Q69 jee_main_2025_29_jan_evening Vector Products and Angles
Let hatmathbfa be a unit vector perpendicular to the vectors vecmathsfb = hatmathsfi -2hatmathsfj +3hatmathsfk and vecmathbfc = 2hatmathbfi +3hatmathbfj -hatmathbfk, and makes an angle of cos^-1left(-frac13right) with the vector hatmathrmi +hatmathrmj +hatmathrmk. If hatmathbfa makes an angle of fracpi3 with the vector hatmathrmi +alpha hatmathrmj +hatmathrmk, then the value of alpha is :
  • A. -sqrt3
  • B. sqrt6
  • C. -sqrt6
  • D. sqrt3

Solution

### Related Formula Cross product for vector perpendicular direction alignment: vecu = vecb times vecc Angle projection formula: costheta = fracveca cdot vecv|veca||vecv| ### Core Logic Compute cross product of vecb and vecc: vecb times vecc = beginvmatrix hati & hatj & hatk \\ 1 & -2 & 3 \\ 2 & 3 & -1 endvmatrix = -7hati + 7hatj + 7hatk = -7(hati - hatj - hatk) Hence, unit vector hata matches form: hata = pm frachati - hatj - hatksqrt3 ### Step 1: Isolate Core Angle Direction Check conditions against vector vecv = hati + hatj + hatk: Using hata = frachati - hatj - hatksqrt3: costheta = frac1 - 1 - 1sqrt3sqrt3 = -frac13 This confirms the direction for hata. ### Step 2: Solve for Unknown Scalar Variable Now compute angle with vector hati + alphahatj + hatk for theta = fracpi3: cosfracpi3 = frac1sqrt3 cdot frac1 - alpha - 1sqrt2 + alpha^2 frac12 = frac-alphasqrt3sqrtalpha^2 + 2 Since left hand side is positive, alpha must be strictly negative. Squaring both sides: frac14 = fracalpha^23(alpha^2 + 2) implies 3alpha^2 + 6 = 4alpha^2 implies alpha^2 = 6 Since alpha < 0, alpha = -sqrt6. ### Pattern Recognition Keep strict track of signs when dealing with algebra containing square roots. Checking value constraints early on allows you to drop phantom positive/negative branches seamlessly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra
Q73 jee_main_2025_28_jan_morning Vector Dot and Cross Products
Let veca = hati +hatj +hatk,quad vecb = 2hati +2hatj +hatk and vecd = vecatimes vecb. If vecc is a vector such that veca.vecc = |vecc |, |vecc -2veca|^2 = 8 and the angle between vecd and vecc is fracpi4, then |10 - 3vecb.vecc| + |vecdtimes vecc|^2 is equal to ....
Numerical Answer. Answer: 6 to 6

Solution

### Related Formula Vector magnitude expansion identity: |vecu - vecv|^2 = |vecu|^2 + |vecv|^2 - 2(vecu cdot vecv) ### Core Logic Calculate the reference vector values: veca = hati + hatj + hatk implies |veca|^2 = 3 vecb = 2hati + 2hatj + hatk vecd = veca times vecb = -hati + hatj implies |vecd|^2 = 2 Expanding |vecc - 2veca|^2 = 8: |vecc|^2 + 4|veca|^2 - 4(veca cdot vecc) = 8 Substituting |veca|^2 = 3 and veca cdot vecc = |vecc|: |vecc|^2 - 4|vecc| + 4 = 0 implies (|vecc| - 2)^2 = 0 implies |vecc| = 2 ### Step 1: Evaluating the Cross Product Vector Component Using the given angle fracpi4 between vecd and vecc: |vecd times vecc|^2 = (|vecd||vecc|sinfracpi4)^2 = left(sqrt2 cdot 2 cdot frac1sqrt2right)^2 = 4 ### Step 2: Solving for the Vector Dot Product Expanding using standard vector identities yields vecb cdot vecc = frac83. Substituting this back into the target expression: textValue = left|10 - 3left(frac83 ight)right| + 4 = |10 - 8| + 4 = 6 ### Pattern Recognition Recognizing a perfect square trinomial (x^2 - 4x + 4 = 0) instantly isolates the unknown vector magnitude. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Vectors
Q73 jee_main_2025_03_april_morning Vector Triple Products and Vector Equations
Let veca = hati + hatj + hatk, vecb = 3hati + 2hatj - hatk, vecc = lambda hatj + mu hatk and hatd be a unit vector such that veca times hatd = vecb times hatd and vecc cdot hatd = 1[cite: 694]. If vecc is perpendicular to veca[cite: 694], then |3lambda hatd + mu vecc|^2 is equal to:
Numerical Answer. Answer: 5 to 5

Solution

### Related Formula Vector cross distribution property: (veca - vecb) times hatd = 0 implies hatd parallel (veca - vecb) ### Core Logic Gather cross products to solve for collineation lines [cite: 1472, 1473]: (veca - vecb) times hatd = 0 implies hatd = t(veca - vecb) [cite: 1473, 1474] Compute the baseline difference vector [cite: 1475]: veca - vecb = (1-3)hati + (1-2)hatj + (1 - (-1))hatk = -2hati - hatj + 2hatk [cite: 1475] hatd = t(-2hati - hatj + 2hatk) [cite: 1475] Since hatd is a unit vector[cite: 1476]: |t| cdot sqrt(-2)^2 + (-1)^2 + 2^2 = 1 implies 3|t| = 1 implies |t| = frac13 [cite: 1476, 1479] ### Step 1: Applying orthogonality constraints Using orthogonal information given for vectors vecc and veca [cite: 1482]: vecc cdot veca = 0 implies (0)(1) + lambda(1) + mu(1) = 0 implies mu = -lambda [cite: 1483, 1484] vecc = lambda(hatj - hatk) implies |vecc|^2 = 2lambda^2 [cite: 1485] Use product condition vecc cdot hatd = 1 to isolate scalar values [cite: 1486]: t(-2hati - hatj + 2hatk) cdot lambda(hatj - hatk) = 1 [cite: 1487] tlambda(-1 - 2) = 1 implies -3tlambda = 1 implies tlambda = -frac13 [cite: 1488] Since |t|^2 = frac19, squaring components yields [cite: 1488]: lambda^2 = 1 [cite: 1488] ### Step 2: Vector magnitude resolution Expand target expression using standard inner dot product expansions [cite: 1488]: |3lambda hatd + mu vecc|^2 = 9lambda^2|hatd|^2 + mu^2|vecc|^2 + 6lambdamu(hatd cdot vecc) [cite: 1488] Substitute values evaluated throughout sections [cite: 1488, 1489]: = 9(1)(1) + (lambda)^2(2lambda^2) + 6lambda(-lambda)(1) = 9 + 2lambda^4 - 6lambda^2 = 9 + 2(1) - 6(1) = 5 [cite: 1489, 1490] ### Pattern Recognition Translating vector cross equalities directly into linear scale parameter multipliers prevents manual determinant expansions, leaving clean system variables behind. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra

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