Question

Let `veca = αhati + 3hatj - hatk, vecb = 3hati - βhatj + 4hatk` and `vecc = hati + 2hatj - 2hatk` where α, β ∈ R, be three vectors. If the projection of a `veca` on `vecc` is `10/3` and `vecb xx vecc = -6hati + 10hatj + 7hatk`, then the value of α + β is equal to ______.

Options

• 3

• 4

• 5

• 6

Answer 1

Let `veca = αhati + 3hatj - hatk, vecb = 3hati - βhatj + 4hatk` and `vecc = hati + 2hatj - 2hatk` where α, β ∈ R, be three vectors. If the projection of a `veca` on `vecc` is `10/3` and `vecb xx vecc = -6hati + 10hatj + 7hatk`, then the value of α + β is equal to 3.

Explanation:

Given: `veca = αhati + 3hatj - hatk`

`vecb = 3hati - βvecj + 4hatk`

`vecc = hati + 2hatj - 2hatk`

The projection of `veca` on `vecc` is `10/3`

As we know, the projection of `vecx` on `vecy` is given by `(vecx.vecy)/|vecy|`

⇒ `(veca.vecb)/|vecc| = 10/3`

⇒ `((αhati + 3hatj - hatk)(hati + 2hatj - 2hatk))/|hati + 2hatj - 2hatk| = 10/3`

⇒ `(α + 6 + 2)/sqrt(1 + 4 + 4) = 10/3`

⇒ α + 8 = 10

⇒ α = 2

Also, given that `vecb xx vecc = -6hati + 10hatj + 7hatk`

⇒ `(3hati - βhatj + 4hatk) xx (hati + 2hatj - 2hatk) = -6hati + 10hatj + 7hatk`

⇒ `|(hati, hatj, hatk),(3, -β, 4),(1, 2, -2)| = -6hati + 10hatj + 7hatk`

⇒ `(2β - 8)hati - hatj(-6 - 4) + hatk(6 + β)hatk = -6hati + 10hatj + 7hatk`

⇒ 2β – 8 = –6 or 6 + β = 7

⇒ β = 1

∴ α + β = 2 + 1 = 3