Let $\vec a =\hat i+\hat j,\vec b=\hat i-\hat j$ and $\vec c =\hat i+\hat j +\hat k$. If $\hat m$ is a unit vector perpendicular to both $\vec a$ and $\vec b$, then $|\vec c.\hat m|$ is equal to

1

The correct answer is Option (4) → 1

$\vec a = \hat i + \hat j$

$\vec b = \hat i - \hat j$

$\vec c = \hat i + \hat j + \hat k$

A unit vector $\hat m$ perpendicular to both $\vec a$ and $\vec b$ must be parallel to $\vec a \times \vec b$.

Compute $\vec a \times \vec b$:

$\vec a \times \vec b = \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 1 & 0 \\ 1 & -1 & 0 \end{vmatrix}$

$= \hat k (1 \cdot -1 - 1 \cdot 1) = \hat k(-1 - 1) = -2\hat k$

Thus a direction perpendicular to both is along $\pm \hat k$.

Unit vector:

$\hat m = \pm \hat k$

Now compute $|\vec c \cdot \hat m|$:

$\vec c \cdot \hat m = (\hat i + \hat j + \hat k)\cdot(\pm \hat k) = \pm 1$

Absolute value:

$|\vec c \cdot \hat m| = 1$

Final Answer: 1