If $a = 4i + 6j$ and $b = 3j + 4k$,what is the vector component of $a$ along the direction of $b$?

$\vec{b}$ and $\vec{c}$ are non-collinear vectors and $(\vec{c} \cdot \vec{c}) \vec{a} = \vec{c}$. If $(\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} + (\vec{a} \cdot \vec{b}) \vec{b} = (4 - 2 \beta - \sin \alpha) \vec{b} + (\beta^2 - 1) \vec{c}$,then $\sin (\alpha + \beta) =$

Let $\bar{a}$ and $\bar{b}$ be two vectors such that $|\bar{a}|=|\bar{b}|$ and $|\bar{a}+2 \bar{b}|=|2 \bar{a}-\bar{b}|$. If $\bar{c}$ is a vector parallel to $\bar{a}$,then the angle between $\bar{b}$ and $\bar{c}$ is (in $^{\circ}$)

The angle between unit vectors $\bar{a}$ and $\bar{b}$ in $\mathbb{R}^3$ is $\theta$. Then,the value of $\left|\frac{\bar{a} \cdot \bar{a}}{\bar{a} \cdot \bar{b}} \cdot \frac{\bar{b} \cdot \bar{a}}{\bar{b} \cdot \bar{b}}\right| + |\bar{a} \times \bar{b}|^2$ is:

$AB=a$ 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

The magnitude of vectors $\vec{a}$ and $\vec{b}$ are $1$ and $2$ respectively,and $\vec{a} \cdot \vec{b} = 1$. Then,the angle between the two vectors $\vec{a}$ and $\vec{b}$ is . . . . . . .