Question

The number of unit vectors perpendicular to \(\vec{a} = \hat{i}+\hat{j}\) and \(\vec{b} = \hat{j}+\hat{k}\) is:

• A. infinite
• B. one
• C. two
• D. three

Hint:

The cross product gives a direction perpendicular to both vectors. Normalization yields two unit vectors (opposite directions).

Answer 1

The Correct Option is C

To find the number of unit vectors perpendicular to both \(\vec{a} = \hat{i} + \hat{j}\) and \(\vec{b} = \hat{j} + \hat{k}\), we need to determine the direction of the vector perpendicular to both given vectors. This direction can be found using the cross product of the two vectors, which will give a vector perpendicular to both.

Let's compute the cross product \(\vec{a} \times \vec{b}\):

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

Expanding the determinant, we get:

\[ \vec{a} \times \vec{b} = \hat{i}(1 \cdot 1 - 0 \cdot 1) - \hat{j}(1 \cdot 1 - 0 \cdot 0) + \hat{k}(1 \cdot 1 - 1 \cdot 0) \]

\[ = \hat{i}(1) - \hat{j}(1) + \hat{k}(1) = \hat{i} - \hat{j} + \hat{k} \]

Thus, the vector perpendicular to both \(\vec{a}\) and \(\vec{b}\) is \(\vec{n} = \hat{i} - \hat{j} + \hat{k}\).

Next, we need to find unit vectors in the direction of \(\vec{n}\). The magnitude of \(\vec{n}\) is:

\[ |\vec{n}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3} \]

A unit vector in the direction of \(\vec{n}\) can be calculated by dividing \(\vec{n}\) by its magnitude. Thus, the unit vector \(\hat{u}\) in the direction of \(\vec{n}\) is:

\[ \hat{u} = \frac{\vec{n}}{\sqrt{3}} = \frac{1}{\sqrt{3}} (\hat{i} - \hat{j} + \hat{k}) \]

Since vectors in opposite directions are also perpendicular, the opposite unit vector is:

\[ -\hat{u} = \frac{-1}{\sqrt{3}} (\hat{i} - \hat{j} + \hat{k}) \]

Therefore, there are two unit vectors perpendicular to both \(\vec{a}\) and \(\vec{b}\): \(\hat{u}\) and \(-\hat{u}\).

The correct answer is: two