Question

If `overlinea = hati + hatj + hatk, overlinea . overlineb = 1` and `overlinea xx overlineb = hatj - hatk,` then `overlineb` = ______ 

पर्याय

• `hati`

• `hati - hatj + hatk`

• `2hatj - hatk`

• `2hati`

Answer 1

If `overlinea = hati + hatj + hatk, overlinea . overlineb = 1` and `overlinea xx overlineb = hatj - hatk,` then `overlineb` = `underlinehati`.

Explanation:

Let `overlineb = b_1hati + b_2hatj + b_3hatk`

Now, `hatj - hatk = overlinea xx overlineb = |(hati, hatj, hatk), (1, 1, 1), (b_2, b_2, b_3)|`

⇒ `b_3 - b_2 = 0, b_1 - b_3 = 1, b_2 - b_1 = -1`

⇒ `b_3 = b_2, b_1 = b_2 + 1`

Also, `overlinea.overlineb = 1`

⇒ `b_1 + b_2 + b_3 = 1`

⇒ `3b_2 + 1 = 1`

⇒ b₂ = 0

⇒ `b_1 = 1, b_3 = 0`

Thus, `overlineb = hati`