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Question
If `veca = hati + hatj + hatk, veca.vecb` = 1 and `veca xx vecb = hatj - hatk`, then find `|vecb|`.
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Solution
Given, `veca = hati + hatj + hatk`
`veca.vecb` = 1
And `veca xx vecb = hatj - hatk`
Let `vecb = ahati + bhatj + chatk`
Now, `veca.vecb` = 1
⇒ `(hati + hatj + hatk)(b_1hati + b_2hatj + b_3hatk)` = 1
⇒ `b_1 + b_2 + b_3` = 1 ...(i)
And `veca xx vecb = hatj - hatk`
⇒ `|(hati, hatj, hatk),(1, 1, 1),(b_1, b_2, b_3)| = hatj - hatk`
⇒ `hati(b_3 - b_2) - hatj(b_3 - b_1) + hatk(b_2 - b_1) = hatj - hatk`
On comparing both sides, we get
–(b3 – b1) = 1 and b2 – b1 = –1
⇒ b3 – b1 = –1 and b2 – b1 = –1
⇒ b3 = –1 + a and b2 = –1 + b1 ...(ii)
Now from equation (i), we get
b1 + (–1 + b1) + (–1 + b1) = 1
⇒ 3b1 = 3
⇒ b1 = 1
From equation (ii), we get
b2 = 0 and b3 = 0
∴ `vecb = hati`
Therefore, `|vecb|` = 1
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