Scalar Triple Product

If the vertices of any tetrahedron be $\overrightarrow{a}=\hat{j}+2\hat{k}, \overrightarrow{b}=3\hat{i}+\hat{k}, \overrightarrow{c}=4\hat{i}+3\hat{j}+6\hat{k}$ and $\overrightarrow{d}=2\hat{i}+3\hat{j}+2\hat{k}$ then find its volume

If $\left|\begin{array}{lll}a& a^2& 1+a^3\\ b& b^2& 1+b^3\\ c& c^2& 1+c^3\end{array}\right|=0$ and vectors $\left(1,a,a^2\right),\left(1,b,b^2\right)$ and $\left(1,c,c^2\right)$ are non-coplanar, then the product $a b c$ equals

If $\overrightarrow{\mathit{b}}$ and $\overrightarrow{\mathit{c}}$ are any two non-collinear unit vectors and $\overrightarrow{\mathit{a}}$ is any vector, then $\left(\overrightarrow{a}·\overrightarrow{b}\right)\overrightarrow{b}+\left(\overrightarrow{a}·\overrightarrow{c}\right)\overrightarrow{c}+\frac{\overrightarrow{a}·\left(\overrightarrow{b}\times \overrightarrow{c}\right)}{{\left|\overrightarrow{b}\times \overrightarrow{c}\right|}^2}\left(\overrightarrow{b}\times \overrightarrow{c}\right)$ is equal to

Consider the parallelepiped with side $\overrightarrow{a}=3\hat{i}+2\hat{j}+\hat{k}, \overrightarrow{b}=\hat{i}+\hat{j}+2\hat{k}$ and $\overrightarrow{c}=\hat{i}+3\hat{j}+3\hat{k}$ then the angle between $\overrightarrow{\mathit{a}}$ and the plane containing the face determined by $\overrightarrow{b}$ and $\overrightarrow{\mathit{c}}$ is

If $\overrightarrow{a}=2\hat{i}-3\hat{j}, \overrightarrow{b}=\hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{c}=3\hat{i}-\hat{k}$ represent three coterminous edges of a parallelepiped, then the volume of that parallelepiped is

Let $\overrightarrow{v}=2\hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{w}=\hat{i}+3\hat{k}$. If $\overrightarrow{u}$ is a unit vector, then for the maximum value of the scalar triple product $\left[\overrightarrow{u} \overrightarrow{v} \overrightarrow{w}\right], \overrightarrow{u}=$

If $\overrightarrow{r}=\alpha \left(\overrightarrow{b}\times \overrightarrow{c}\right)+\beta \left(\overrightarrow{c}\times \overrightarrow{a}\right)+\gamma \left(\overrightarrow{a}\times \overrightarrow{b}\right)$ and $\left[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}\right]=2$, then $a+\beta +\gamma$ is equal to

If $a$ is a non-zero real number, then the vectors
$\overrightarrow{\mathrm{\alpha }}=a\hat{\mathrm{i}}+2\mathrm{a}\hat{\mathrm{j}}-3a\hat{\mathrm{k}}, \overrightarrow{\mathrm{\beta }}=\left(2a+1\right)\hat{\mathrm{i}}+\left(2a+3\right)\hat{\mathrm{j}}+\left(a+1\right)\hat{\mathrm{k}}$, $\overrightarrow{\mathrm{\gamma }}=\left(3a+5\right)\hat{\mathrm{i}}+\left(a+5\right)\hat{\mathrm{j}}+\left(a+2\right)\hat{\mathrm{k}}$ are

Let $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$ be three non-zero vectors such that no two of these are collinear. If the vector $\overrightarrow{a}+2\overrightarrow{b}$ is collinear with $\overrightarrow{c}$ and $\overrightarrow{b}+3\overrightarrow{c}$ is collinear with $\overrightarrow{a}$($\lambda$ being some non-zero scalar), then $\overrightarrow{a}+2\overrightarrow{b}+6\overrightarrow{c}$ equals

If vectors $\overrightarrow{a}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overrightarrow{b}=\hat{\mathrm{i}}+2\hat{\mathrm{j}}-\hat{\mathrm{k}}$ and $\overrightarrow{c}=3\hat{\mathrm{i}}+p\hat{\mathrm{j}}+5\hat{\mathrm{k}}$ are coplanar, then the value of $p$ is

The angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ is $\theta$. The value of the triple product $\overrightarrow{A}·\left(\overrightarrow{B}\times \overrightarrow{A}\right)$ is