If \(\vec a= \hat i + \hat j + \hat k, \,\vec a .\,\vec b = 1 \) and

1.	If \(\vec a= \hat i + \hat j + \hat k, \,\vec a .\,\vec b = 1 \) and \(\vec a \times \vec b = \hat j - \hat k\), then find \(\|\vec b\|\).
Answer» \(\vec a = \hat i + \hat j + \hat k\) Let \(\vec b = b_1\hat i + b_2\hat j + b_3 \hat k\) \(\therefore \vec a . \vec b = 1\) ⇒ \((\hat i + \hat j + \hat k). (b_1 \hat i + b_2 \hat j + b_3 \hat k)=1\) ⇒ \(b_1 + b_2 + b_3 = 1\) .....(1) \(\vec a \times \vec b = \hat j - \hat k\) ⇒ \(\begin{vmatrix}\hat i &\hat j&\hat k\\1&1&1\\b_1&b_2&b_3\end{vmatrix} = \hat j - \hat k\) ⇒ \(\hat i (b_3 - b_2) + j(b _1 - b_3 )+ \hat k (b_2 - b_1) = \hat j - \hat k\) \(\therefore b_3 - b_2 = 0 \) ⇒ \(b_3 = b_2\) .....(2) \(b_1 - b_3 = 1\) ⇒ \(b_1 = 1 + b_3\) .....(3) \(b_2- b_1 = -1\) .....(4) ⇒ \(b_2 - (1 + b_3 ) = -1\) ⇒ \(b_2 - 1- b_3 = -1\) ⇒ \(b_2 = b_3\) From (1) & (2), \(b_1 + 2 b_2 = 1\) ......(5) From (2), (3), (5), we get \(1 + b_3 + 2b_3 = 1\) ⇒ \(3b_3 = 0\) ⇒ \(b_3 = 0\) \(\therefore b_2 = 0\) (From(2)) \(\therefore b_1= 1\) (From(1)) Hence, \(b_1 = 1 , \, b_2 = 0, \, b_3 = 0\) \(\therefore \vec b= \hat i\) & \(\|\vec b = 1\|\)

If \(\vec a= \hat i + \hat j + \hat k, \,\vec a .\,\vec b = 1 \) and \(\vec a \times \vec b = \hat j - \hat k\), then find \(|\vec b|\).

Answer»

\(\vec a = \hat i + \hat j + \hat k\)

Let \(\vec b = b_1\hat i + b_2\hat j + b_3 \hat k\)

\(\therefore \vec a . \vec b = 1\)

⇒ \((\hat i + \hat j + \hat k). (b_1 \hat i + b_2 \hat j + b_3 \hat k)=1\)

⇒ \(b_1 + b_2 + b_3 = 1\) .....(1)

\(\vec a \times \vec b = \hat j - \hat k\)

⇒ \(\begin{vmatrix}\hat i &\hat j&\hat k\\1&1&1\\b_1&b_2&b_3\end{vmatrix} = \hat j - \hat k\)

⇒ \(\hat i (b_3 - b_2) + j(b _1 - b_3 )+ \hat k (b_2 - b_1) = \hat j - \hat k\)

\(\therefore b_3 - b_2 = 0 \)

⇒ \(b_3 = b_2\) .....(2)

\(b_1 - b_3 = 1\)

⇒ \(b_1 = 1 + b_3\) .....(3)

\(b_2- b_1 = -1\) .....(4)

⇒ \(b_2 - (1 + b_3 ) = -1\)

⇒ \(b_2 - 1- b_3 = -1\)

⇒ \(b_2 = b_3\)

From (1) & (2),

\(b_1 + 2 b_2 = 1\) ......(5)

From (2), (3), (5), we get

\(1 + b_3 + 2b_3 = 1\)

⇒ \(3b_3 = 0\)

⇒ \(b_3 = 0\)

\(\therefore b_2 = 0\) (From(2))

\(\therefore b_1= 1\) (From(1))

Hence,

\(b_1 = 1 , \, b_2 = 0, \, b_3 = 0\)

\(\therefore \vec b= \hat i\) & \(|\vec b = 1|\)