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1982

JEE Main 1982 Mathematics question on Vector Algebra

For non-zero vectors a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c}, ∣(a⃗×b⃗)⋅c⃗∣=∣a⃗∣∣b⃗∣∣c⃗∣|(\vec{a} \times \vec{b}) \cdot \vec{c}|= |\vec{a}||\vec{b}||\vec{c}| holds, if and only if
  1. A.
    a⃗⋅b⃗=0,b⃗⋅c⃗=0\vec{a} \cdot \vec{b} = 0, \vec{b} \cdot \vec{c} = 0
  2. B.
    b⃗⋅c⃗=0,c⃗⋅a⃗=0\vec{b} \cdot \vec{c} = 0, \vec{c} \cdot \vec{a} = 0
  3. C.
    c⃗⋅a⃗=0,a⃗⋅b⃗=0\vec{c} \cdot \vec{a} = 0, \vec{a} \cdot \vec{b} = 0
  4. D.
    a⃗⋅b⃗=b⃗⋅c⃗=c⃗⋅a⃗=0\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = 0
Show answer and solution

Correct option: D

Solution

|(a × b) · c| = |a × b||c||cosα| where α is the angle between (a × b) and c. Also |a × b| = |a||b|sinθ where θ is the angle between a and b.

So |(a × b) · c| = |a||b||c| |sinθ||cosα|.

For this to equal |a||b||c|, we need |sinθ| = 1 and |cosα| = 1, meaning θ = 90° (a ⊥ b) and α = 0° (c parallel to a × b).

Since a × b is perpendicular to both a and b, if c is parallel to a × b then c is perpendicular to both a and b. Thus all three conditions a·b = 0, b·c = 0, c·a = 0 must hold simultaneously.

[1982]

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