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JEE Main 2018

JEE Main 2018 Mathematics question on Vector Algebra

Let u⃗\vec{u} be a vector coplanar with the vectors a⃗=2i^+3j^−k^\vec{a} = 2\hat{i} + 3\hat{j} - \hat{k} and b⃗=j^+k^\vec{b} = \hat{j} + \hat{k}. If u⃗\vec{u} is perpendicular to a⃗\vec{a} and u⃗⋅b⃗=24\vec{u} \cdot \vec{b} = 24, then ∣u⃗∣2|\vec{u}|^2 is equal to
  1. A.
    336
  2. B.
    315
  3. C.
    256
  4. D.
    84
Show answer and solution

Correct option: A

Solution

Since u is coplanar with a and b, write u = λa + μb = λ(2i + 3j - k) + μ(j + k) = 2λi + (3λ + μ)j + (-λ + μ)k.

Condition u · a = 0:
u · a = 2λ(2) + (3λ + μ)(3) + (-λ + μ)(-1) = 4λ + 9λ + 3μ + λ - μ = 14λ + 2μ = 0, so μ = -7λ.

Condition u · b = 24 (b = j + k):
(3λ + μ) + (-λ + μ) = 2λ + 2μ = 2λ + 2(-7λ) = -12λ = 24, so λ = -2, μ = 14.

Then u = 2(-2)i + (3(-2) + 14)j + (-(-2) + 14)k = -4i + 8j + 16k.

|u|² = 16 + 64 + 256 = 336.

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