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Question
Let bar"b" = 4hat"i" + 3hat"j" and bar"c" be two vectors perpendicular to each other in the XY-plane. Find the vector in the same plane having projection 1 and 2 along bar"b" and bar"c" respectively.
Solution
`bar"b" = 4hat"i" + 3hat"j"`
∴ `|bar"b"| = sqrt(4^2 + 3^2) = sqrt(16 + 9) = 5`
Let `bar"c" = "m"hat"i" + "n"hat"j"` be perpendicular to `bar"b"`
Then `bar"b".bar"c" = 0`
∴ `(4hat"i" + 3hat"j").("m"hat"i" + "n"hat"j") = 0`
∴ 4m + 3n = 0
∴ n = `- "4m"/3`
∴ `bar"c" = "m"hat"i" - "4m"/3hat"j" = "m"/3(3hat"i" - 4hat"j")`
∴ `bar"c" = "p"(3hat"i" - 4hat"j") .....["p" = "m"/3]`
∴ `|bar"c"| = "p" sqrt(3^2 + (- 4)^2) = "p"sqrt(9 + 16) = 5"p"`
Let `bar"d" = "x"hat"i" + "y"hat"j"` be the vector having projections 1 and 2 along `bar"b" and bar"c"`.
∴ `(bar"b".bar"d")/|bar"b"| = 1`
∴ `((4hat"i" + 3hat"j").("x"hat"i" + "y"hat"j"))/5 = 1`
∴ 4x + 3y = 5 .....(1)
Also, `(bar"c".bar"d")/|bar"c"| = 2`
∴ `((3"p"hat"i" - 4"p"hat"j").("x"hat"i" + "y"hat"j"))/"5p" = 2`
∴ 3px - 4py = 10p
∴ 3x - 4y = 10
From (1), 3y = 5 - 4x
∴ y = `(5 - 4"x")/3`
Substituting for y in (2), we get
`3"x" - 4((5 - "4x")/3) = 10`
∴ 9x - 20 + 16x = 30
∴ 25x = 50
∴ x = 2
y = `(5 - 4"x")/3 = (5 - 4(2))/3 = - 1`
∴ `bar"d" = 2hat"i" - hat"j"`
Hence, the required vector is `2hat"i" - hat"j"`.
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