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"`.