Ex 10.4, 8 If either 𝑎 ⃗ = 0 ⃗ or 𝑏 ⃗ = 0 ⃗, then 𝑎 ⃗ × 𝑏 ⃗ = 0 ⃗ . Is the converse true? Justify your answer with an example. Converse : If 𝑎 ⃗ × 𝑏 ⃗ = 0 ⃗, then either 𝑎 ⃗ = 0 ⃗ or 𝑏 ⃗ = 0 ⃗
𝑎 ⃗ × 𝑏 ⃗ = |𝑎 ⃗ ||𝑏 ⃗ | sin θ 𝑛 ̂
where, θ = angle between 𝑎 ⃗ and 𝑏 ⃗
𝑛 ̂ = unit vector perpendicular to 𝑎 ⃗ 𝜀 𝑏 ⃗
Let 𝑎 ⃗ = 1𝑖 ̂ + 1𝑗 ̂ + 1𝑘 ̂
& 𝑏 ⃗ = 2𝑖 ̂ + 2𝑗 ̂ + 2𝑘 ̂
𝑎 ⃗ × 𝑏 ⃗ = |■8(𝑖 ̂&𝑗 ̂&𝑘 ̂@1&1&1@2&2&2)|
= 𝑖 ̂ (1 × 2 − 2 × 1) − 𝑗 ̂ (1 × 2 −2 × 1) + 𝑘 ̂ (1 × 2 − 2 × 1) + 𝑘 ̂(1 × 2 − 2 × 1)
= 𝑖 ̂ (2 − 2) − 𝑗 ̂ (2 −2) + 𝑘 ̂ (2 − 2)
= 0𝑖 ̂ − 0𝑗 ̂ + 0𝑘 ̂
= 0 ⃗
Here, 𝑎 ⃗ ≠ 0 ⃗ & 𝑏 ⃗≠ 0 ⃗
But 𝑎 ⃗ × 𝑏 ⃗ = 0 ⃗
Therefore, converse is not true.

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.