Question
If ∣ a ⃗ ∣ = 2 |\vec{a}| = 2 ∣ a ∣ = 2 ∣ b ⃗ ∣ = 3 |\vec{b}| = 3 ∣ b ∣ = 3 a ⃗ ⋅ b ⃗ = 4 \vec{a} \cdot \vec{b} = 4 a ⋅ b = 4 ∣ a ⃗ × b ⃗ ∣ |\vec{a} \times \vec{b}| ∣ a × b ∣ a ⃗ \vec{a} a b ⃗ \vec{b} b
(NCERT Class 12 — Vectors)
Solution — Step by Step
a ⃗ ⋅ b ⃗ = ∣ a ⃗ ∣ ∣ b ⃗ ∣ cos θ \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta a ⋅ b = ∣ a ∣∣ b ∣ cos θ 4 = 2 × 3 × cos θ = 6 cos θ 4 = 2 \times 3 \times \cos\theta = 6\cos\theta 4 = 2 × 3 × cos θ = 6 cos θ cos θ = 2 3 \cos\theta = \frac{2}{3} cos θ = 3 2 θ = cos − 1 ( 2 3 ) ≈ 48.19 ° \theta = \cos^{-1}\left(\frac{2}{3}\right) \approx 48.19° θ = cos − 1 ( 3 2 ) ≈ 48.19°
sin 2 θ = 1 − cos 2 θ = 1 − 4 9 = 5 9 \sin^2\theta = 1 - \cos^2\theta = 1 - \frac{4}{9} = \frac{5}{9} sin 2 θ = 1 − cos 2 θ = 1 − 9 4 = 9 5 sin θ = 5 3 ( positive since 0 < θ < π ) \sin\theta = \frac{\sqrt{5}}{3} \quad (\text{positive since } 0 < \theta < \pi) sin θ = 3 5 ( positive since 0 < θ < π )
∣ a ⃗ × b ⃗ ∣ = ∣ a ⃗ ∣ ∣ b ⃗ ∣ sin θ = 2 × 3 × 5 3 = 2 5 |\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta = 2 \times 3 \times \frac{\sqrt{5}}{3} = \mathbf{2\sqrt{5}} ∣ a × b ∣ = ∣ a ∣∣ b ∣ sin θ = 2 × 3 × 3 5 = 2 5
Why This Works
The dot product and cross product capture different geometric information about two vectors:
a ⃗ ⋅ b ⃗ = ∣ a ⃗ ∣ ∣ b ⃗ ∣ cos θ \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta a ⋅ b = ∣ a ∣∣ b ∣ cos θ ∣ a ⃗ × b ⃗ ∣ = ∣ a ⃗ ∣ ∣ b ⃗ ∣ sin θ |\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta ∣ a × b ∣ = ∣ a ∣∣ b ∣ sin θ
Together, they are connected by the identity:
∣ a ⃗ ⋅ b ⃗ ∣ 2 + ∣ a ⃗ × b ⃗ ∣ 2 = ∣ a ⃗ ∣ 2 ∣ b ⃗ ∣ 2 |\vec{a} \cdot \vec{b}|^2 + |\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 ∣ a ⋅ b ∣ 2 + ∣ a × b ∣ 2 = ∣ a ∣ 2 ∣ b ∣ 2 Check: 4 2 + ( 2 5 ) 2 = 16 + 20 = 36 = ( 2 × 3 ) 2 4^2 + (2\sqrt{5})^2 = 16 + 20 = 36 = (2 \times 3)^2 4 2 + ( 2 5 ) 2 = 16 + 20 = 36 = ( 2 × 3 ) 2
Alternative Method
Use the Lagrange identity directly:
∣ a ⃗ × b ⃗ ∣ 2 = ∣ a ⃗ ∣ 2 ∣ b ⃗ ∣ 2 − ( a ⃗ ⋅ b ⃗ ) 2 |\vec{a} \times \vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2 ∣ a × b ∣ 2 = ∣ a ∣ 2 ∣ b ∣ 2 − ( a ⋅ b ) 2 = 4 × 9 − 16 = 36 − 16 = 20 = 4 \times 9 - 16 = 36 - 16 = 20 = 4 × 9 − 16 = 36 − 16 = 20 ∣ a ⃗ × b ⃗ ∣ = 20 = 2 5 |\vec{a} \times \vec{b}| = \sqrt{20} = 2\sqrt{5} ∣ a × b ∣ = 20 = 2 5 This skips the angle calculation entirely and goes straight to the answer.
The Lagrange identity ∣ a ⃗ × b ⃗ ∣ 2 + ( a ⃗ ⋅ b ⃗ ) 2 = ∣ a ⃗ ∣ 2 ∣ b ⃗ ∣ 2 |\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2|\vec{b}|^2 ∣ a × b ∣ 2 + ( a ⋅ b ) 2 = ∣ a ∣ 2 ∣ b ∣ 2
Common Mistake
Students often write ∣ a ⃗ × b ⃗ ∣ = ∣ a ⃗ ∣ ∣ b ⃗ ∣ cos θ |\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\cos\theta ∣ a × b ∣ = ∣ a ∣∣ b ∣ cos θ sin θ \sin\theta sin θ dot = cos, cross = sin . A mnemonic: “cross” and “sin” both have the letter “s” (sort of) — or just remember that the cross product is zero when vectors are parallel (θ = 0 \theta = 0 θ = 0 sin 0 = 0 \sin 0 = 0 sin 0 = 0