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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:

Solution & Explanation

### 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

Reference Study Guides

More Vector Algebra Previous-Year Questions — Page 4

Q55 2025 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

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