Given \(\rm \vec A = 2\hat i - \hat j + 2\hat k\) and \(\rm \vec B = -\hat i - 2\hat j + 2\hat k\) (î, ĵ and k̂ are unit vectors along x, y & z axes respectively). The unit vector in the direction of \(\rm \vec A -\vec B\) is:

Concept:

• The unit vector û in the direction of a vector \(\rm \vec A\) is given by: \(\rm \hat u = \frac{\vec A}{\left|\vec A\right|}\), where \(\rm \left|\vec A \right|\) is the magnitude of the vector \(\rm \vec A\).
• The magnitude of a vector \(\rm \vec A\) is given as: 

\(\rm \left|\vec A\right|=\sqrt{\vec A.\vec A}\).

\(\rm \left|a\hat i +b\hat j + c\hat k\right|=\sqrt{\left(a\hat i +b\hat j + c\hat k\right).\left(a\hat i +b\hat j + c\hat k\right)}\)

\(\rm \left|a\hat i +b\hat j + c\hat k\right|=\sqrt{a^2+b^2+c^2}\)

Calculation:

We have \(\rm \vec A = 2\hat i - \hat j + 2\hat k\) and \(\rm \vec B = -\hat i - 2\hat j + 2\hat k\).

∴ \(\rm \vec A-\vec B = \left(2\hat i - \hat j + 2\hat k\right)-\left(-\hat i - 2\hat j + 2\hat k \right)\)

⇒ \(\rm \vec A-\vec B = 3\hat i + \hat j\)

The magnitude of \(\rm \vec A -\vec B\) will be:

\(\rm \left|3\hat i +1\hat j + 0\hat k\right|=\sqrt{3^2+1^2+0^2}=\sqrt{10}\)

Now, the unit vector along the direction of \(\rm \vec A -\vec B\) will be:

\(\rm \hat u = \frac{\vec A-\vec B}{\left|\vec A-\vec B\right|}\)

⇒ \(\rm \hat u = \frac{3\hat i + \hat j}{\sqrt{10}}\).