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JEE Main 2019 · 8 Apr · Shift II

JEE Main 2019 Mathematics question on Vector Algebra

Let a=3i^+2j^+xk^\mathbf{a} = 3\hat{i} + 2\hat{j} + x\hat{k} and b=i^−j^+k^\mathbf{b} = \hat{i} - \hat{j} + \hat{k}, for some real xx. Then ∣a×b∣=r|\mathbf{a} \times \mathbf{b}| = r is possible if
  1. A.
    0<r≤320 < r \leq \dfrac{\sqrt{3}}{2}
  2. B.
    32<r≤332\dfrac{\sqrt{3}}{2} < r \leq 3\dfrac{\sqrt{3}}{2}
  3. C.
    332<r<5323\sqrt{\dfrac{3}{2}} < r < 5\sqrt{\dfrac{3}{2}}
  4. D.
    r≥532r \geq 5\sqrt{\dfrac{3}{2}}
Show answer and solution

Correct option: D

Solution

Compute a × b where a = (3,2,x) and b = (1,-1,1).

a × b = |i j k; 3 2 x; 1 -1 1|
= i(2·1 - x·(-1)) - j(3·1 - x·1) + k(3·(-1) - 2·1)
= i(2+x) - j(3-x) + k(-5)

So |a × b|² = (2+x)² + (3-x)² + 25

Expand: (2+x)² + (3-x)² = 4 + 4x + x² + 9 - 6x + x² = 2x² - 2x + 13

So |a × b|² = 2x² - 2x + 13 + 25 = 2x² - 2x + 38 = 2(x² - x) + 38

Minimize over x: complete the square → 2(x - 1/2)² - 1/2 + 38 = 2(x-1/2)² + 75/2

Minimum value is 75/2 (at x = 1/2), so minimum of r is √(75/2) = 5√(3/2).

Since r has no upper bound (r → ∞ as x → ±∞), r ≥ 5√(3/2), which is option 4.

[JEE Main 2019 · 8 Apr · Shift II]

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