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 4

Q51 jee_main_2025_24_jan_morning Coplanar Vectors and Vector Products
Let vecmathbfa = hatmathbfi + 2hatmathbfj + 3hatmathbfk , vecmathbfb = 3hatmathbfi + hatmathbfj - hatmathbfk and \|\vec{\mathbf{c}}\| be three vectors such that vecmathbfc is coplanar with vecmathbfa and vecmathbfb . If the vector vecmathbfc is perpendicular to vecmathbfb and vecmathbfa cdot vecmathbfc = 5 , then \|vecmathbfc\| is equal to :
  • A. frac13sqrt2
  • B. 18
  • C. 16
  • D. sqrtfrac116

Solution

### Related Formula A vector vecmathbfc coplanar with vecmathbfa and vecmathbfb and perpendicular to vecmathbfb can be expressed using the vector triple product command template: vecmathbfc = lambda (vecmathbfb times (vecmathbfa times vecmathbfb)) = lambda [(vecmathbfb cdot vecmathbfb)vecmathbfa - (vecmathbfa cdot vecmathbfb)vecmathbfb] ### Core Logic First, find the dot products of the given vectors: vecmathbfb cdot vecmathbfb = 3^2 + 1^2 + (-1)^2 = 9 + 1 + 1 = 11 vecmathbfa cdot vecmathbfb = (1)(3) + (2)(1) + (3)(-1) = 3 + 2 - 3 = 2 Substituting these values into the expression for vecmathbfc: vecmathbfc = lambda [11vecmathbfa - 2vecmathbfb] vecmathbfc = lambda [11(hatmathbfi + 2hatmathbfj + 3hatmathbfk) - 2(3hatmathbfi + hatmathbfj - hatmathbfk)] vecmathbfc = lambda (5hatmathbfi + 20hatmathbfj + 35hatmathbfk) = 5lambda (hatmathbfi + 4hatmathbfj + 7hatmathbfk) ### Step 1: Determine Lambda Using the given condition vecmathbfa cdot vecmathbfc = 5: vecmathbfa cdot [5lambda (hatmathbfi + 4hatmathbfj + 7hatmathbfk)] = 5 5lambda (1 cdot 1 + 2 cdot 4 + 3 cdot 7) = 5 5lambda (1 + 8 + 21) = 5 implies 30lambda = 1 implies lambda = frac130 Thus, the vector vecmathbfc is: vecmathbfc = frac530(hatmathbfi + 4hatmathbfj + 7hatmathbfk) = frac16(hatmathbfi + 4hatmathbfj + 7hatmathbfk) ### Step 2: Calculate Magnitude The magnitude of vecmathbfc is evaluated as: \|vecmathbfc\| = fracsqrt1^2 + 4^2 + 7^26 = fracsqrt1 + 16 + 496 = fracsqrt666 = sqrtfrac6636 = sqrtfrac116 ### Pattern Recognition Whenever a vector is specified to be coplanar with vecmathbfa, vecmathbfb and perpendicular to vecmathbfb, direct setup with the cross-product template vecmathbfb times (vecmathbfa times vecmathbfb) circumvents solving cumbersome linear systems of scalar variables. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra
Q52 jee_main_2025_28_jan_evening Vector Operations and Components
Let A, B, C be three points in xy-plane, whose position vector are given by sqrt3hati+hatj, hati+sqrt3hatj and ahati+(1-a)hatj respectively with respect to the origin O. If the distance of the point C from the line bisecting the angle between the vectors overlineOA and overlineOB is frac9sqrt2 then the sum of all the possible values of a is:
  • A. 1
  • B. 9/2
  • C. 0
  • D. 2

Solution

### Related Formula The line bisecting the angle between two symmetric vectors passing through the origin in the first quadrant is given by y = x or x - y = 0. Distance from point (x_1, y_1) to line Ax + By + C = 0 is: d = frac|Ax_1 + By_1 + C|sqrtA^2 + B^2 ### Core Logic Vectors overlineOA = sqrt3hati+hatj and overlineOB = hati+sqrt3hatj are symmetric about the line y = x. Therefore, the angle bisector of overlineOA and overlineOB is the line x - y = 0. Point C has coordinates (a, 1 - a). ### Step 1: Calculate Distance and Solve for a The perpendicular distance from C(a, 1 - a) to x - y = 0 is: d = frac|a - (1 - a)|sqrt1^2 + (-1)^2 = frac|2a - 1|sqrt2 Given that this distance is frac9sqrt2: frac|2a - 1|sqrt2 = frac9sqrt2 implies |2a - 1| = 9 This gives two solutions: 1) 2a - 1 = 9 implies 2a = 10 implies a = 5 2) 2a - 1 = -9 implies 2a = -8 implies a = -4 ### Step 2: Find the Sum of Values Sum of all possible values of a: textSum = 5 + (-4) = 1 ### Pattern Recognition Notice that overlineOA and overlineOB have swapped coordinates, meaning they are symmetric with respect to y=x. Thus, the angle bisector equation is immediate (x-y=0), simplifying the problem to a standard point-to-line distance calculation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Straight Lines Class 12 Mathematics: Vector Algebra
Q53 jee_main_2025_28_jan_evening Components of Vectors
If the components of veca=alphahati+betahatj+gammahatk along and perpendicular to vecb=3hati+hatj-hatk respectively, are frac1611(3hati+hatj-hatk) and frac111(-4hati-5hatj-17hatk), then alpha^2+beta^2+gamma^2 is equal to:
  • A. 23
  • B. 18
  • C. 16
  • D. 26

Solution

### Related Formula Any vector veca can be written as the sum of its component parallel to vecb (along vecb) and its component perpendicular to vecb: veca = veca_parallel + veca_perp Magnitude squared: |veca|^2 = alpha^2 + beta^2 + gamma^2 ### Core Logic Given: veca_parallel = frac1611(3hati+hatj-hatk) veca_perp = frac111(-4hati-5hatj-17hatk) ### Step 1: Reconstruct Vector a Add both components to find veca: veca = frac1611(3hati+hatj-hatk) + frac111(-4hati-5hatj-17hatk) veca = frac111 left[ (48 - 4)hati + (16 - 5)hatj + (-16 - 17)hatk right] veca = frac111 left[ 44hati + 11hatj - 33hatk right] = 4hati + hatj - 3hatk Therefore, alpha = 4, beta = 1, gamma = -3. ### Step 2: Calculate Sum of Squares alpha^2 + beta^2 + gamma^2 = 4^2 + 1^2 + (-3)^2 = 16 + 1 + 9 = 26 ### Pattern Recognition Since parallel and perpendicular components are orthogonal vectors, you can also use directly |veca|^2 = |veca_parallel|^2 + |veca_perp|^2 to save algebra step: |veca_parallel|^2 = frac16^211^2(9+1+1) = frac25611, |veca_perp|^2 = frac111^2(16+25+289) = frac330121 = frac3011. Total = frac28611 = 26. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra
Q55 jee_main_2025_29_jan_morning Vector Cross Product and Dot Product
Let veca = 2hati - hatj + 3hatk , vecb = 3hati - 5hatj + hatk and vecc be a vector such that veca times vecc = vecc times vecb and left(vecmathbfa + vecmathbfcright) . left(vecmathbfb + vecmathbfcright) = 168. Then the maximum value of |vecmathbfc|^2 is:
  • A. 77
  • B. 462
  • C. 308
  • D. 154

Solution

### Related Formula vecu times vecv = -vecv times vecu textIf vecu times vecv = 0 implies vecu parallel vecv implies vecu = lambda vecv ### Core Logic Given veca times vecc = vecc times vecb implies veca times vecc + vecb times vecc = 0 (veca + vecb) times vecc = 0 implies vecc = lambda(veca + vecb) ### Step 1: Compute a + b veca + vecb = (2+3)hati + (-1-5)hatj + (3+1)hatk = 5hati - 6hatj + 4hatk vecc = lambda(5hati - 6hatj + 4hatk) |vecc|^2 = lambda^2(25 + 36 + 16) = 77lambda^2 ### Step 2: Expand the Dot Product Condition (veca + vecc) cdot (vecb + vecc) = 168 veca cdot vecb + vecc cdot (veca + vecb) + |vecc|^2 = 168 Evaluate veca cdot vecb = (2)(3) + (-1)(-5) + (3)(1) = 6 + 5 + 3 = 14. Substitute vecc cdot (veca + vecb) = lambda |veca + vecb|^2 = 77lambda: 14 + 77lambda + 77lambda^2 = 168 implies 77lambda^2 + 77lambda - 154 = 0 lambda^2 + lambda - 2 = 0 implies lambda = 1 text or lambda = -2 ### Step 3: Maximize |vecc|^2 Maximum value occurs when lambda = -2: |vecc|^2 = 77(-2)^2 = 77 times 4 = 308 ### Pattern Recognition Recognize the cross-product rule inversion immediately: vecx times vecy = vecy times vecz implies (vecx+vecz) parallel vecy. This linear reduction circumvents solving complex linear systems. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra
Q8 jee_main_2024_01_february_morning Vector Triple Product
Let veca=-5hati+hatj-3hatk, vecb=hati+2hatj-4hatk and vecc=(((vecatimesvecb)timeshati)timeshati)timeshati. Then vecccdot(-hati+hatj+hatk) is equal to:
  • A. -12
  • B. -10
  • C. -13
  • D. -15

Solution

### Related Formula Vector Triple Product Identity: (vecu times vecv) times vecw = (vecu cdot vecw)vecv - (vecv cdot vecw)veca ### Core Logic Given vectors veca and vecb: First, let's find the inner vector triple product (veca times vecb) times hati: (veca times vecb) times hati = (veca cdot hati)vecb - (vecb cdot hati)veca From the given components: - veca cdot hati = -5 - vecb cdot hati = 1 (veca times vecb) times hati = -5vecb - veca ### Step 1: Substitute Vector Coordinates Substitute coordinates of veca and vecb into the expression: -5vecb - veca = -5(hati + 2hatj - 4hatk) - (-5hati + hatj - 3hatk) = -5hati - 10hatj + 20hatk + 5hati - hatj + 3hatk = -11hatj + 23hatk Now, perform successive cross products with hati: textNext Step = (-11hatj + 23hatk) times hati = -11(hatj times hati) + 23(hatk times hati) = 11hatk + 23hatj textFinal Vector vecc = (23hatj + 11hatk) times hati = 23(hatj times hati) + 11(hatk times hati) = -23hatk + 11hatj ### Step 2: Calculate Dot Product Now evaluate vecc cdot (-hati + hatj + hatk): vecc = 11hatj - 23hatk vecc cdot (-hati + hatj + hatk) = (0)( -1) + (11)(1) + (-23)(1) = 11 - 23 = -12 ### Pattern Recognition Sees: Repeated vector cross products with standard unit vectors. Shortcut: Observe that crossing a vector in the y-z plane with hati rotates its components by 90^circ in that plane. Doing it twice returns it to the original plane with flipped coefficients and signs (vecv times hati times hati = -vecv). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra

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