Understanding Vector Properties: Perpendicular Unit Vectors and Angle

Let's analyze the given statements regarding the vectors \(\vec{a}=(0,1,1)\) and \(\vec{b}=(1,0,1)\).

Analysis of Statement 1: Number of Unit Vectors Perpendicular to Both Vectors

A vector that is perpendicular to both \(\vec{a}\) and \(\vec{b}\) is parallel to their cross product, \(\vec{a} \times \vec{b}\). Let's calculate the cross product:

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

This vector \((1, 1, -1)\) is perpendicular to both \(\vec{a}\) and \(\vec{b}\).

To find the unit vectors perpendicular to both \(\vec{a}\) and \(\vec{b}\), we need to normalize this cross product vector. First, we find its magnitude:

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

The unit vector in the direction of \(\vec{a} \times \vec{b}\) is:

\[ \hat{n}_1 = \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} = \frac{(1, 1, -1)}{\sqrt{3}} = \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}\right) \]

There is also a unit vector in the opposite direction, which is also perpendicular to both \(\vec{a}\) and \(\vec{b}\):

\[ \hat{n}_2 = -\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} = -\frac{(1, 1, -1)}{\sqrt{3}} = \left(-\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) \]

Thus, there are two unit vectors perpendicular to both \(\vec{a}\) and \(\vec{b}\). Statement 1 claims there is only one, which is incorrect.

Analysis of Statement 2: Angle Between the Vectors

The angle \(\theta\) between two non-zero vectors \(\vec{a}\) and \(\vec{b}\) can be found using the dot product formula:

\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \]

First, calculate the dot product \(\vec{a} \cdot \vec{b}\):

\[ \vec{a} \cdot \vec{b} = (0)(1) + (1)(0) + (1)(1) = 0 + 0 + 1 = 1 \]

Next, calculate the magnitudes of vectors \(\vec{a}\) and \(\vec{b}\):

\[ |\vec{a}| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{0 + 1 + 1} = \sqrt{2} \] \[ |\vec{b}| = \sqrt{1^2 + 0^2 + 1^2} = \sqrt{1 + 0 + 1} = \sqrt{2} \]

Now, substitute these values into the dot product formula to find \(\cos \theta\):

\[ \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{1}{\sqrt{2} \times \sqrt{2}} = \frac{1}{2} \]

The angle \(\theta\) such that \(\cos \theta = \frac{1}{2}\) in the range \(0 \le \theta \le \pi\) is \(\theta = \frac{\pi}{3}\).

Statement 2 claims the angle between the vectors is \(\frac{\pi}{3}\), which is correct.

Conclusion on Statements

• Statement 1 is incorrect because there are two unit vectors perpendicular to both given vectors.
• Statement 2 is correct because the angle between the given vectors is \(\frac{\pi}{3}\).

Therefore, only Statement 2 is correct.

Revision Table: Key Vector Concepts

Additional Information: Vector Operations

Vector operations like the dot product and cross product are fundamental in vector algebra. The dot product gives a scalar and is useful for finding angles and checking orthogonality (\(\vec{a} \cdot \vec{b} = 0\) if non-zero \(\vec{a}, \vec{b}\) are perpendicular). The cross product gives a vector that is orthogonal (perpendicular) to the plane containing the two original vectors, following the right-hand rule. Its magnitude is related to the area of the parallelogram formed by the vectors.

Unit vectors are often used to indicate direction without magnitude. The concept of unit vectors perpendicular to two given vectors is crucial in understanding directions orthogonal to a plane defined by those vectors.

Always remember that there are two opposite directions perpendicular to a plane, hence two unit vectors.