Let vecc and vecd be vectors such that |vecc+vecd|=sqrt29 and vecctimes(2hati+3hatj+4hatk)=(2hati+3hatj+4hatk)timesvecd . If lambda_1,lambda_2(lambda_1>lambda_2) are the possible values of (vecc+vecd).(-7hati+2hatj+3hatk) , then the equation K^2x^2+(K^2-5K+lambda_1)xy+left(3K+fraclambda_22right)y^2-8x+12y+lambda_2=0 represents a circle, for k equal to:

Solution & Explanation

### Related Formula vecu times vecv = -(vecv times vecu) General equation of a circle requires coefficient of x^2 = coefficient of y^2 and coefficient of xy = 0. ### Core Logic Given vecc times (2hati + 3hatj + 4hatk) = (2hati + 3hatj + 4hatk) times vecd Rearranging: vecc times (2hati + 3hatj + 4hatk) + vecd times (2hati + 3hatj + 4hatk) = vec0 (vecc + vecd) times (2hati + 3hatj + 4hatk) = vec0 This implies vecc + vecd is parallel to 2hati + 3hatj + 4hatk. Let vecc + vecd = lambda (2hati + 3hatj + 4hatk). ### Step 1: Find the scaling factor Given |vecc + vecd| = sqrt29 |lambda| sqrt2^2 + 3^2 + 4^2 = sqrt29 |lambda| sqrt4 + 9 + 16 = sqrt29 |lambda| sqrt29 = sqrt29 Rightarrow lambda = pm 1 Therefore, vecc + vecd = pm(2hati + 3hatj + 4hatk). ### Step 2: Evaluate the dot product limits We need values of (vecc + vecd) cdot (-7hati + 2hatj + 3hatk). Dot product = lambda(2(-7) + 3(2) + 4(3)) = lambda(-14 + 6 + 12) = 4lambda. Since lambda = pm 1, the possible values are 4 and -4. Given lambda_1 > lambda_2, we have lambda_1 = 4 and lambda_2 = -4. ### Step 3: Apply circle constraints Substitute lambda_1 and lambda_2 into the conic equation: K^2 x^2 + (K^2 - 5K + 4)xy + left(3K - 2right)y^2 - 8x + 12y - 4 = 0 For this to represent a circle: 1) Coefficient of xy must be 0: K^2 - 5K + 4 = 0 Rightarrow K = 1, 4 2) Coefficient of x^2 must equal coefficient of y^2: K^2 = 3K - 2 Rightarrow K^2 - 3K + 2 = 0 Rightarrow K = 1, 2 The common value satisfying both conditions is K = 1. ### Pattern Recognition Cross-product equations mapping to X times A = -Y times A perfectly factor out to (X+Y) times A = 0, guaranteeing (X+Y) is a scalar multiple of A. This immediately unlocks vector magnitudes. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Vector Algebra Class 11 Maths: Conic Sections

Reference Study Guides

Vector Algebra

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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