Misc 12 - Chapter 10 Class 12 Vector Algebra

Transcript

Misc 12 Let 𝑎﷯ = 𝑖﷯ + 4 𝑗﷯ + 2 𝑘﷯, 𝑏﷯ = 3 𝑖﷯ − 2 𝑗﷯ + 7 𝑘﷯ and 𝑐﷯ = 2 𝑖﷯ − 𝑗﷯ + 4 𝑘﷯ . Find a vector 𝑑﷯ which is perpendicular to both 𝑎﷯ and 𝑏﷯ and 𝑐﷯ ⋅ 𝑑﷯ = 15 . 𝑎﷯ = 𝑖﷯ + 4 𝑗﷯ + 2 𝑘﷯ = 1 𝑖﷯ + 4 𝑗﷯ + 2 𝑘﷯ 𝑏﷯ = 3 𝑖﷯ + 2 𝑗﷯ + 7 𝑘﷯ 𝑐﷯ = 2 𝑖﷯ + 𝑗﷯ + 4 𝑘﷯ = 2 𝑖﷯ − 1 𝑗﷯ + 4 𝑘﷯ Let 𝑑﷯ = x 𝑖﷯ + y 𝑗﷯ + z 𝑘﷯ Since 𝑑﷯ is perpendicular to 𝑎﷯ and 𝑏﷯ 𝑑﷯ . 𝑎﷯ = 0 & 𝑑﷯ . 𝑏﷯ = 0 Also, 𝑐﷯ . 𝑑﷯ = 15 (2 𝑖﷯ – 1 𝑗﷯ + 4 𝑘﷯). (x 𝑖﷯ + y 𝑗﷯ + z 𝑘﷯) = 15 (2 × x) + (−1 × y) + (4 × 2) = 15 2x − y + 4z = 15 Now, we need to solve equations x + 4y + 2z = 0 …(1) 3x − 2y + 7z = 0 …(2) & 2x − y + 4z = 15 …(3) Solving x + 4y + 2z = 0 …(1) 3x – 2y + 7z = 0 …(2) 𝑥﷮28 − (−4) ﷯ = 𝑦﷮6 − 7 ﷯ = 𝑧﷮−2 − 12 ﷯ 𝑥﷮32 ﷯ = 𝑦﷮−1 ﷯ = 𝑧﷮−14 ﷯ Writing x & y in terms of z ∴ x = 32𝑧﷮−14﷯ = −16𝑧﷮7﷯ & y = −1𝑧﷮−14﷯ = 𝑧﷮14﷯ Putting values of x and y in (3) 2x – y + 4z = 15 2 −16𝑧﷮7﷯ ﷯ − 𝑧﷮14﷯ ﷯ + 4z = 5 −32﷮7﷯ z − 1﷮14﷯ z + 4﷮1﷯ z = 15 (−64 − 1 + 56)﷮14﷯ z = 15 −9﷮14﷯ z = 15 z = 15 × 14﷮−9﷯ z = −𝟕𝟎﷮𝟑﷯ Putting value of z in x & y, x = −16𝑧﷮7﷯ = −16﷮7﷯ × −70﷮3﷯ = 𝟏𝟔𝟎﷮𝟑﷯ y = 𝑧﷮14﷯ = 1﷮14﷯ × −70﷮3﷯ = −𝟓﷮𝟑﷯ Therefore, the required vector 𝑑﷯ = x 𝑖﷯ + y 𝑗﷯ + z 𝑘﷯ = 160﷮3﷯ 𝑖﷯ − 5﷮3﷯ 𝑗﷯ – 70﷮3﷯ 𝑘﷯ = 𝟏﷮𝟑﷯ (160 𝒊﷯ − 5 𝒋﷯ – 70 𝒌﷯)