In a trapezium,the vector $\overrightarrow{BC} = \lambda \overrightarrow{AD}$. We will then find that $p = \overrightarrow{AC} + \overrightarrow{BD}$ is collinear with $\overrightarrow{AD}$. If $p = \mu \overrightarrow{AD}$,then

Let $ABC$ be a triangle. Let a point $P$ divide $AB$ in the ratio $1:2$ internally and a point $Q$ divide $BC$ in the ratio $1:2$ internally. Let $D$ be the point of intersection of $AQ$ and $CP$. If the area of the triangle $ABC$ is $k$ square units,then the area of the triangle $BCD$ in square units is:

Let $\vec{a}=2 \hat{i}-\hat{j}+\hat{k}$ be the position vector of a point $A$. Let $\vec{b}=\hat{i}+2 \hat{j}-\hat{k}$ and $\vec{c}=\hat{i}+\hat{j}-2 \hat{k}$ be two vectors and $\vec{r}$ be a vector passing through the point $A$ with position vector $\vec{a}$ and parallel to the vector $\vec{b}$. If the projection of $\vec{r}$ on $\vec{c}$ is $\frac{9}{\sqrt{6}}$,then find $|\vec{r}|$.

The magnitude of the projection of the vector $\vec{a} = 4\hat{i} - 3\hat{j} + 2\hat{k}$ on the line which makes equal angles with the coordinate axes is

The value of $\hat{i} \cdot (\hat{j} \times \hat{k}) + \hat{j} \cdot (\hat{k} \times \hat{i}) + \hat{k} \cdot (\hat{i} \times \hat{j})$ is . . . . . . .

Let $\vec{a}=\hat{i}-\hat{j}+2 \hat{k}$ and $\vec{b}$ be a vector such that $\vec{a} \times \vec{b}=2 \hat{i}-\hat{k}$ and $\vec{a} \cdot \vec{b}=3$. Then the projection of $\vec{b}$ on the vector $\vec{a}-\vec{b}$ is :-