Obtain the scalar product of unit vectors in Cartesian co-ordinate system.
Since $\hat{i}, \hat{j}$ and $\hat{k}$ are the unit vectors in the direction of $\mathrm{X}, \mathrm{Y}$ and $\mathrm{Z}$ in Cartesian co-ordinate system.
$(i)$ $\hat{i} \cdot \hat{i}=(1)(1) \cos 0^{\circ} \quad[\because|\hat{i}|=1$, and $\hat{i} \| \hat{i}]$
$\therefore \hat{i} \cdot \hat{i}=1 \quad\left[\because \cos 0^{\circ}=1\right]$
similarly $\hat{j} \cdot \hat{j}=1$ and $\hat{k} \cdot \hat{k}=1$
$(ii)$ $\hat{i} \cdot \hat{j}=(1)(1) \cos 90^{\circ} \quad[\because|\hat{i}|=1,|\hat{j}|=1$ and $\hat{i} \perp \hat{j}]$
$\therefore \hat{i} \cdot \hat{j}=\hat{j} \cdot \hat{i}=0 \quad\left[\because \cos 90^{\circ}=0\right]$
similarly, $\quad \hat{j} \cdot \hat{k}=\hat{k} \cdot \hat{j}=0$
$\hat{k} \cdot \hat{i}=\hat{i} \cdot \hat{k}=0$
The area of the triangle formed by $2\hat i + \hat j - \hat k$ and $\hat i + \hat j + \hat k$ is
If $\overrightarrow{ F }=2 \hat{ i }+\hat{ j }-\hat{ k }$ and $\overrightarrow{ r }=3 \hat{ i }+2 \hat{ j }-2 \hat{ k }$, then the scalar and vector products of $\overrightarrow{ F }$ and $\overrightarrow{ r }$ have the magnitudes respectively as
Two vector $A$ and $B$ have equal magnitudes. Then the vector $\mathop A\limits^ \to + \mathop B\limits^ \to $ is perpendicular to