Let A (2, 3, 5) and C(-3, 4, -2) be opposite vertices of a parallelogram ABCD if the diagonal overrightarrowBD = hati + 2hatj + 3hatk then the area of the parallelogram is equal to

Solution & Explanation

### Related Formula textArea of parallelogram = frac12 |overrightarrowd_1 times overrightarrowd_2| ### Core Logic The diagonals of the parallelogram are overrightarrowAC and overrightarrowBD. First, calculate the diagonal vector overrightarrowAC: overrightarrowAC = textPosition vector of C - textPosition vector of A overrightarrowAC = (-3 - 2)hati + (4 - 3)hatj + (-2 - 5)hatk = -5hati + 1hatj - 7hatk Alternatively, taking overrightarrowCA = 5hati - hatj + 7hatk. Let's use overrightarrowCA or overrightarrowAC, the magnitude of the cross product will be the same. The second diagonal is given: overrightarrowBD = hati + 2hatj + 3hatk ### Step 1: Finding Cross Product overrightarrowArea = frac12 |overrightarrowAC times overrightarrowBD| overrightarrowAC times overrightarrowBD = beginvmatrix hati & hatj & hatk \\ -5 & 1 & -7 \\ 1 & 2 & 3 endvmatrix = hati(3 - (-14)) - hatj(-15 - (-7)) + hatk(-10 - 1) = hati(17) - hatj(-8) + hatk(-11) = 17hati + 8hatj - 11hatk ### Step 2: Calculating Magnitude textArea = frac12 |17hati + 8hatj - 11hatk| = frac12 sqrt17^2 + 8^2 + (-11)^2 = frac12 sqrt289 + 64 + 121 = frac12 sqrt474 ### Pattern Recognition When opposite vertices and one full diagonal vector are provided, immediately calculate the second diagonal vector via displacement and evaluate half the magnitude of their cross product. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Vector Algebra

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More Vector Algebra Previous-Year Questions — Page 7

Q26 jee_main_2024_31_jan_evening Vector Triple Product
Let veca = 3hati + 2hatj + hatk, vecb = 2hati - hatj + 3hatk and vecc be a vector such that (veca + vecb) times vecc = 2(veca times vecb) + 24hatj - 6hatk and (veca - vecb + hati) cdot vecc = -3. Then |vecc|^2 is equal to
Numerical Answer. Answer: 38 to 38

Solution

### Core Logic Evaluate base vectors: veca + vecb = (5, 1, 4) veca times vecb = beginvmatrix hati & hatj & hatk \\ 3 & 2 & 1 \\ 2 & -1 & 3 endvmatrix = (7, -7, -7) Substitute into given cross product equation, letting vecc = (x, y, z): (5hati + hatj + 4hatk) times (xhati + yhatj + zhatk) = 2(7hati - 7hatj - 7hatk) + 24hatj - 6hatk beginvmatrix hati & hatj & hatk \\ 5 & 1 & 4 \\ x & y & z endvmatrix = (14, -14, -14) + (0, 24, -6) (z-4y)hati - (5z-4x)hatj + (5y-x)hatk = (14, 10, -20) Equating components: z - 4y = 14 implies z = 4y + 14 4x - 5z = 10 5y - x = -20 implies x = 5y + 20 Now use the dot product constraint: veca - vecb + hati = (3-2+1)hati + (2+1)hatj + (1-3)hatk = (2, 3, -2). (veca - vecb + hati) cdot vecc = -3 2x + 3y - 2z = -3 Substitute x = 5y + 20 and z = 4y + 14: 2(5y + 20) + 3y - 2(4y + 14) = -3 10y + 40 + 3y - 8y - 28 = -3 5y + 12 = -3 implies 5y = -15 implies y = -3 Calculate x and z: x = 5(-3) + 20 = 5 z = 4(-3) + 14 = 2 So, vecc = (5, -3, 2). |vecc|^2 = 5^2 + (-3)^2 + 2^2 = 25 + 9 + 4 = 38 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Vector Algebra
Q12 jee_main_2024_31_jan_morning Cross and Dot Product Operations
Let veca = 3hati + hatj - 2hatk, vecb = 4hati + hatj + 7hatk and vecc = hati - 3hatj + 4hatk be three vectors. If a vector vecp satisfies vecp times vecb = vecc times vecb and vecp cdot veca = 0, then vecp cdot (hati - hatj - hatk) is equal to
  • A. 24
  • B. 36
  • C. 28
  • D. 32

Solution

### Core Logic Given vecp times vecb = vecc times vecb. (vecp - vecc) times vecb = vec0 Thus, vecp - vecc = lambda vecb implies vecp = vecc + lambda vecb. ### Step 1: Utilize Dot Product Condition Given vecp cdot veca = 0. (vecc + lambda vecb) cdot veca = 0 vecc cdot veca + lambda (vecb cdot veca) = 0 Calculate vecc cdot veca: (1)(3) + (-3)(1) + (4)(-2) = 3 - 3 - 8 = -8. Calculate vecb cdot veca: (4)(3) + (1)(1) + (7)(-2) = 12 + 1 - 14 = -1. -8 + lambda(-1) = 0 implies lambda = -8 ### Step 2: Substitute and Solve Substitute lambda back to find vecp: vecp = vecc - 8vecb = (hati - 3hatj + 4hatk) - 8(4hati + hatj + 7hatk) vecp = -31hati - 11hatj - 52hatk Now compute vecp cdot (hati - hatj - hatk): = (-31)(1) + (-11)(-1) + (-52)(-1) = -31 + 11 + 52 = 32 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Vector Algebra
Q28 jee_main_2024_31_jan_morning Vector Triple Product
Let veca and vecb be two vectors such that |veca| = 1, |vecb| = 4 and veca cdot vecb = 2. If vecc = (2veca times vecb) - 3vecb and the angle between vecb and vecc is alpha, then 192sin^2alpha is equal to
Numerical Answer. Answer: 48 to 48

Solution

### Core Logic vecb cdot vecc = vecb cdot ((2veca times vecb) - 3vecb) |b||c|cosalpha = 2(vecb cdot (veca times vecb)) - 3|b|^2 Since vecb cdot (veca times vecb) = 0, we have |b||c|cosalpha = -3|b|^2. |c|cosalpha = -3|b| = -12 implies |c|^2 cos^2 alpha = 144 ### Step 1: Compute Modulus of c |c|^2 = |2veca times vecb - 3vecb|^2 = 4|veca times vecb|^2 + 9|vecb|^2 - 12((veca times vecb) cdot vecb) = 4|veca times vecb|^2 + 9|vecb|^2 Given veca cdot vecb = 2 implies |a||b|costheta = 2 implies 1 cdot 4 costheta = 2 implies theta = fracpi3. |veca times vecb|^2 = |a|^2|b|^2sin^2theta = 1 cdot 16 cdot frac34 = 12 |c|^2 = 4(12) + 9(16) = 48 + 144 = 192 ### Step 2: Final Calculation We know |c|^2 cos^2 alpha = 144. 192 cos^2 alpha = 144 192(1 - sin^2 alpha) = 144 192sin^2 alpha = 192 - 144 = 48 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Vector Algebra

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