Question
The unit vector which is orthogonal to the vector $$3\hat i + 2\hat j + 6\hat k$$ and is coplanar with the vectors $$2\hat i + \hat j + \hat k$$ and $$\hat i - \hat j + \hat k$$ is :
A.
$$\frac{{2\hat i - 6\hat j + \hat k}}{{\sqrt {41} }}$$
B.
$$\frac{{2\hat i - 3\hat j}}{{\sqrt {13} }}$$
C.
$$\frac{{3\hat i - \hat k}}{{\sqrt {10} }}$$
D.
$$\frac{{4\hat i + 3\hat j - 3\hat k}}{{\sqrt {34} }}$$
Answer :
$$\frac{{3\hat i - \hat k}}{{\sqrt {10} }}$$
Solution :
Any vector coplanar to $${\vec a}$$ and $${\vec b}$$ can be written as
$$\eqalign{
& \vec r = \vec a + \lambda \vec b \cr
& \vec r = \left( {1 + 2\lambda } \right)\hat i + \left( { - 1 + \lambda } \right)\hat j + \left( {1 + \lambda } \right)\hat k \cr} $$
Since $${\vec r}$$ is orthogonal to $$5\hat i + 2\hat j + 6\hat k$$
$$\eqalign{
& \Rightarrow 5\left( {1 + 2\lambda } \right) + 2\left( { - 1 + \lambda } \right) + 6\left( {1 + \lambda } \right) = 0 \cr
& \Rightarrow 9 + 18\lambda = 0 \cr
& \Rightarrow \lambda = - \frac{1}{2} \cr
& \therefore \,\vec r{\text{ is 3}}\hat j - \hat k \cr} $$
Since $${\hat r}$$ is a unit vector, $$\therefore \,\,\hat r = \frac{{{\text{3}}\hat j - \hat k}}{{\sqrt {10} }}$$