Question

Let `veca = hati + hatj + sqrt(2) hatk, vecb = b_1hati + b_2hatj + sqrt(2)hatk` and `vecc = 5hati + hatj + sqrt(2)hatk` be three vectors such that the projection vector of `vecb` on `veca`. If `veca + vecb` is perpendicular to `vecc`, then `|vecb|` is equal to ______.

पर्याय

• `sqrt(32)`

• 6

• `sqrt(22)`

• 4

Answer 1

Let `veca = hati + hatj + sqrt(2) hatk, vecb = b_1hati + b_2hatj + sqrt(2)hatk` and `vecc = 5hati + hatj + sqrt(2)hatk` be three vectors such that the projection vector of `vecb` on `veca`. If `veca + vecb` is perpendicular to `vecc`, then `|vecb|` is equal to 6.

Explanation:

Projection of `vecb` on `veca = (vecb.veca)/|veca| = (b_1 + b_2 + 2)/4`

According to question `(b_1 + b_2 + 2)/2 = sqrt(1 + 1 + 2)` = 2

`\implies` b₁ + b₂ = 2  ...(i)

Since, `veca + vecb` is is perpendicular to `vecc`.

Hence, `veca.vecc + vecb .vecc` = 0

`\implies` 8 + 5b₁ + b₂ + 2 = 0

From (i) and (ii),

b₁ = –3, b₂ = 5

`\implies vecb = -3.hati + 5hatj + sqrt(2)hatk`

`|vecb| = sqrt(9 + 25 + 2)` = 6