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