Example 23 - Chapter 10 Class 12 Vector Algebra

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﷯﷮ 𝑘﷯﷮2﷮3﷮4﷮0﷮−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﷯ = 1﷮𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝑐﷯﷯ × 𝑐﷯ Magnitude of 𝑐﷯ = ﷮ −2﷯2+42+ −2﷯2﷯ 𝑐﷯﷯ = ﷮4+16+4﷯ = ﷮24﷯ = ﷮2×2×6﷯ = 2 ﷮6﷯ Unit vector of 𝑐﷯ = 1﷮ 𝑐﷯﷯﷯ × 𝑐﷯ = 1﷮2 ﷮6﷯﷯ −2 𝑖﷯+4 𝑗﷯−2 𝑘﷯﷯ = 1﷮2 ﷮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﷯