Last updated at May 29, 2018 by Teachoo

Transcript

Example 23 Find a unit vector perpendicular to each of the vectors 𝑎 + 𝑏 and 𝑎 − 𝑏 where 𝑎 = 𝑖 + 𝑗 + 𝑘, b = 𝑖 + 2 𝑗 + 3 𝑘 . 𝑎 = 𝑖 + 𝑗 + 𝑘 = 1 𝑖 + 1 𝑗 + 1 𝑘 𝑏 = 𝑖 + 2 𝑗 + 3 𝑘 = 1 𝑖 + 2 𝑗 + 3 𝑘 ( 𝑎 + 𝑏) = (1 + 1) 𝑖 + (1 + 2) 𝑗 + (1 + 3) 𝑘 = 2 𝑖 + 3 𝑗 + 4 𝑘 ( 𝑎 − 𝑏) = (1 − 1) 𝑖 + (1 − 2) 𝑗 + (1 − 3) 𝑘 = 0 𝑖 − 1 𝑗 − 2 𝑘 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 ( 𝑎 + 𝑏) × ( 𝑎 − 𝑏) = 𝑐 𝑐 = 𝑖 𝑗 𝑘2340−1−2 = 𝑖 3×−2−(−1×4) − 𝑗 2×−2−(0×4) + 𝑘 2×−1−(0×3) = 𝑖 −6− −4 − 𝑗 −4−0 + 𝑘 −2−0 = 𝑖 (−6 + 4) − 𝑗 (−4) + 𝑘(−2) = -2 𝑖 + 4 𝑗 − 2 𝑘 ∴ 𝒄 = -2 𝒊 + 4 𝒋 − 2 𝒌 Now, Unit vector of 𝑐 = 1𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝑐 × 𝑐 Magnitude of 𝑐 = −22+42+ −22 𝑐 = 4+16+4 = 24 = 2×2×6 = 2 6 Unit vector of 𝑐 = 1 𝑐 × 𝑐 = 12 6 −2 𝑖+4 𝑗−2 𝑘 = 12 6 × 2 − 𝑖+2 𝑗− 𝑘 = −1 6 𝑖 + 2 6 𝑗 − 1 6 𝑘 Therefore, required the unit vector is = −𝟏 𝟔 𝒊 + 𝟐 𝟔 𝒋 − 𝟏 𝟔 𝒌 Note: There are always two perpendicular vectors So, another vector would be = − −1 6 𝑖 + 2 6 𝑗 − 1 6 𝑘 = 𝟏 𝟔 𝒊 − 𝟐 𝟔 𝒋 + 𝟏 𝟔 𝒌 Hence, the perpendicular vectors are −1 6 𝑖 + 2 6 𝑗 − 1 6 𝑘 & 1 6 𝑖 − 2 6 𝑗 + 1 6 𝑘

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Example 23 Important You are here

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Example 26 (Supplementary NCERT)

Example 27 (Supplementary NCERT)

Example 28 (Supplementary NCERT)

Example 29 (Supplementary NCERT)

Example 30 (Supplementary NCERT)

Example 31 (Supplementary NCERT)

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.