Question

let a= i +j - k   and c = 2i - 3j + 2k then the number of  vector b such b x c = a and |b| belongs to {1,2,3...,10}

A    0

B    1

C    2

D    3

Answer 1

Correct option is (A) 0

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

\(\vec c = 2\hat i - 3\hat j + 2\hat k\)

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

Then 

\(\vec b \times \vec c = \begin{vmatrix} \hat i&\hat j & \hat k\\b_1 &b_2 &b_3\\2 &-3&2\end{vmatrix}\)

\(= \hat i (2 b_2 + 3b_3) - \hat j(2b_1 - 2b_3) + \hat k (-3b_1 - 2b_2)\)

\(= \hat a\)   (Given)

\(\therefore 2b_2 + 3b_3 = 1\)   .....(1)

\(2b_3 - 2b_1 = 1\)

⇒ \(b_3 = \frac12 + b_1\)     .....(2)

\(3b_1 - 2b_2 = -1 \)

⇒ \(3b_1 + 2b_2 = 1\)    ....(3)

\(\therefore \) From (1), we get

\(2b_2 + \frac32 + 3{b_1} = 1\)

⇒ \(2b_2 + 3b_1 = 1 - \frac32 = \frac{-1}2\)

But 

\(2b_2 + 3b_1 = 1\)  (From(3))

\(\therefore\) Such \(\vec b\) is not possible.