# Let →A,→B and →C Be Three Vectors Such that | → a | = 1 ,| ∣ → B ∣ ∣ = 2 and |→C|=3 .If the Projection of →B Along → a is Equal to the Projection of → C Along → a ; and → B, → C - Mathematics

Sum

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|.

#### Solution

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)

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