Question

Let `veca` , `vecb` and `vecc` be three vectors such that `|veca| = 1,|vecb| = 2, |vecc| = 3.` If the projection of `vecb` along `veca` is equal to the projection of `vecc` along `veca`; and `vecb` , `vecc` are perpendicular to each other, then find `|3veca - 2vecb + 2vecc|`.

Answer 1

Given,

`|veca|= 1,|vecb|= 2,|vecc|= 3`

the projection of `vecb    "along"    veca = (vecb·veca)/|veca|`

the projection of `vecc     "along"    veca = (vecc·veca)/|veca|`

According to the question,

Projection of `vecb    "along"  veca = "Projection of"  vecc   "along"  veca`

⇒ `(vecb·veca)/|veca| = (vecc·veca)/|veca|`

⇒ `vecb·veca = vecc·veca`                     ......(i)

since `vecb and vecc` are perpendicular to each other, we have

`vecb. vecc = 0`                                      ......(ii)

`(3veca - 2vecb + 2vecc)·(3veca - 2vecb + 2vecc) = 9|veca|^2 -6veca·vecb + 6veca·vecc - 6vecb·veca + 4 |vecb|^2 -4vecb·vecc + 6vecc·veca - 4vecc·vecb + 4|vecc|^2`

`|3veca - 2vecb + 2vecc|^2 = 9|veca|^2 + 4|vecb|^2 + 4|vecc|^2 -12veca·vecb + 12veca·vecc - 8vecb·vecc`       ....(iiii)

From (i), (ii) and (iii)

`|3veca - 2vecb + 2vecc|^2 = 9|veca|^2 + 4|vecb|^2 + 4|vecc|^2`

⇒ `|3veca - 2vecb + 2vecc|^2` = 9 x 1 + 4 x 4 + 4 x 9 = 61

⇒ `|3veca - 2vecb + 2vecc| = sqrt(61)`