Ex 10.4, 2 Find a unit vector perpendicular to each of the vector 𝑎 ⃗ + 𝑏 ⃗ and 𝑎 ⃗ − 𝑏 ⃗, where 𝑎 ⃗ = 3𝑖 ̂ + 2𝑗 ̂ + 2𝑘 ̂ and 𝑏 ⃗ = 𝑖 ̂ + 2𝑗 ̂ − 2𝑘 ̂ .𝑎 ⃗ = 3𝑖 ̂ + 2𝑗 ̂ + 2𝑘 ̂
𝑏 ⃗ = 1𝑖 ̂ + 2𝑗 ̂ − 2𝑘 ̂
(𝑎 ⃗ + 𝑏 ⃗) = (3 + 1) 𝑖 ̂ + (2 + 2) 𝑗 ̂ + (2 − 2) 𝑘 ̂
= 4𝑖 ̂ + 4𝑗 ̂ + 0𝑘 ̂
(𝑎 ⃗ − 𝑏 ⃗) = (3 − 1) 𝑖 ̂ + (2 − 2) 𝑗 ̂ + (2 − (−2)) 𝑘 ̂
= 2𝑖 ̂ + 0𝑗 ̂ + 4𝑘 ̂
Now, we need to find a vector perpendicular to both 𝑎 ⃗ + 𝑏 ⃗ and 𝑎 ⃗ − 𝑏 ⃗,
We know that
(𝑎 ⃗ × 𝑏 ⃗) is perpendicular to 𝑎 ⃗ and 𝑏 ⃗
Replacing 𝑎 ⃗ by (𝑎 ⃗ + 𝑏 ⃗) & 𝑏 ⃗ by (𝑎 ⃗ − 𝑏 ⃗)
(𝒂 ⃗ + 𝒃 ⃗) × (𝒂 ⃗ − 𝒃 ⃗) will be perpendicular to (𝒂 ⃗ + 𝒃 ⃗) and (𝒂 ⃗ − 𝒃 ⃗)
Let 𝑐 ⃗ = (𝑎 ⃗ + 𝑏 ⃗) × (𝑎 ⃗ − 𝑏 ⃗)
∴ 𝑐 ⃗ = |■8(𝑖 ̂&𝑗 ̂&𝑘 ̂@█(4@2)&█(4@0)&█(0@4))|
= 𝑖 ̂ [(4×4)−(0×0)] − 𝑗 ̂ [(4×4)−(2×0)] + 𝑘 ̂ [(4×0)−(2×4)]
= 𝑖 ̂ (16 − 0) − 𝑗 ̂ (16 − 0) + 𝑘 ̂ (0 − 8)
= 16 𝑖 ̂ − 16𝑗 ̂ − 8𝑘 ̂
∴ 𝑐 ⃗ = 16 𝑖 ̂ − 16𝑗 ̂ − 8𝑘 ̂
Now,
Unit vector of 𝑐 ⃗ = 1/(𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒 𝑜𝑓𝑐 ⃗ ) × 𝑐 ⃗
Magnitude of 𝑐 ⃗ = √(162+(−16)2+(−8)2)
|𝑐 ⃗ | = √(256+256+64) = √576 = 24
Unit vector of 𝑐 ⃗ = 1/|𝑐 ⃗ | × 𝑐 ⃗
= 1/24 × ["16" 𝑖 ̂" − 16" 𝑗 ̂" − 8" 𝑘 ̂ ]
= 𝟐/𝟑 𝒊 ̂ − 𝟐/𝟑 𝒋 ̂ − 𝟏/𝟑 𝒌 ̂ .
Therefore the required unit vector is 2/3 𝑖 ̂ − 2/3 𝑗 ̂ − 1/3 𝑘 ̂ .
Note: There are always two perpendicular vectors
So, another vector would be
= −(𝟐/𝟑 " " 𝒊 ̂" − " 𝟐/𝟑 " " 𝒋 ̂" − " 𝟏/𝟑 " " 𝒌 ̂ )
= (−𝟐)/𝟑 𝒊 ̂ + 𝟐/𝟑 𝒋 ̂ + 𝟏/𝟑 𝒌 ̂
Hence, the perpendicular vectors are 2/3 " " 𝑖 ̂" − " 2/3 " " 𝑗 ̂" − " 1/3 " " 𝑘 ̂ & (−2)/3 𝑖 ̂ + 2/3 𝑗 ̂ + 1/3 𝑘 ̂

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo

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