The horizontal force and the force inclined at an angle $60^\circ$ with the vertical,whose resultant is in the vertical direction with a magnitude of $P \ kg$,are:

If $\overrightarrow{a} = \hat{i} + 2\hat{k}$,$\overrightarrow{b} = \hat{i} + \hat{j} + \hat{k}$,and $\overrightarrow{c} = 7\hat{i} - 3\hat{j} + 4\hat{k}$,such that $\overrightarrow{r} \times \overrightarrow{b} + \overrightarrow{b} \times \overrightarrow{c} = \overrightarrow{0}$ and $\overrightarrow{r} \cdot \overrightarrow{a} = 0$,then $\overrightarrow{r} \cdot \overrightarrow{c}$ is equal to:

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:

$a$ and $b$ are unit vectors such that $a+2b$ is also a unit vector. If $\theta$ is the angle between $a$ and $b$,then $\sin \theta + \cos^3 \theta + \tan^5 \theta$ is equal to

If the components of $\vec{a}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$ along and perpendicular to $\vec{b}=3 \hat{i}+\hat{j}-\hat{k}$ respectively,are $\frac{16}{11}(3 \hat{i}+\hat{j}-\hat{k})$ and $\frac{1}{11}(-4 \hat{i}-5 \hat{j}-17 \hat{k})$,then $\alpha^2+\beta^2+\gamma^2$ is equal to :

$AB=a$ and $AC=b$ are the sides of $\triangle ABC$. $P$ is a point on $AB$ and $Q$ is a point on $BC$ such that $\frac{AP}{PB}=\frac{1}{2}$ and $\frac{BQ}{QC}=\frac{1}{2}$. If the point of intersection of $AQ$ and $CP$ is $D$ and the area of $\triangle BCD$ is $7$ square units,then the area of the $\triangle ABC$ (in the same square units) is