Scalar Triple Product

If the vectors $p\hat{i}+\hat{j}+\hat{k}, \hat{i}+q\hat{j}+\hat{k}$ and $\hat{i}+\hat{j}+r\hat{k}$ (where $p\ne q\ne r\ne 1)$) are coplanar, then the value of $pqr-(p+q+r)$ is

If $\overrightarrow{V}=2\stackrel{⏜}{i}+\stackrel{⏜}{j}-\stackrel{⏜}{k}$ and $\overrightarrow{W}=\stackrel{⏜}{i}+3\stackrel{⏜}{k}$. If $\overrightarrow{U}$ is a unit vector, then the maximum value of the scalar triple product $\left[\overrightarrow{U} \overrightarrow{V} \overrightarrow{W}\right]$ is

If $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$ are non-coplanar vectors and $\overrightarrow{d}=\lambda \overrightarrow{a}+\mu \overrightarrow{b}+v\overrightarrow{c},$then $\lambda$ is equal to

If $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ are three non-coplanar vectors, then $\frac{\overrightarrow{a}·\left(\overrightarrow{b}\times \overrightarrow{c}\right)}{\left(\overrightarrow{c}\times \overrightarrow{a}\right)·\overrightarrow{b}}+\frac{\overrightarrow{b}·\left(\overrightarrow{a}\times \overrightarrow{c}\right)}{\overrightarrow{c}·\left(\overrightarrow{a}\times \overrightarrow{b}\right)}$ is equal to

If $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ are any three vectors and their inverse are $a^{–1}, b^{–1}, {\mathrm{c}}^{–1}$ and $\left[\overrightarrow{a}\overrightarrow{ b} \overrightarrow{c}\right]\ne 0,$ then $\left[a^{–1} b^{–1} {\mathrm{c}}^{–1}\right]$ will be

If $\overrightarrow{u}$, $\overrightarrow{v}$ and $\overrightarrow{w}$ are three non-coplanar vectors, then $\left(\overrightarrow{u}+\overrightarrow{v}–\overrightarrow{w}\right)·\left[\left(\overrightarrow{u}–\overrightarrow{v}\right)\times \left(\overrightarrow{v}–\overrightarrow{w}\right)\right]$ is equal to

Let $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}, \overrightarrow{b}=\hat{i}-\hat{j}+\hat{k}$ and $\overrightarrow{c}=\hat{i}-\hat{j}-\hat{k}$ be three vectors. A vector $\overrightarrow{V}$ in the plane of $\overrightarrow{a}$ and $\overrightarrow{b}$, whose projection on $\overrightarrow{c}$ is $\frac{1}{\sqrt{3}}$, is given by

The number of distinct real values of $\lambda$, for which the vectors $-{\lambda }^2\hat{i}+\hat{j}+\hat{k}, \hat{i}-{\lambda }^2\hat{k}+\hat{j}$ and $\hat{i}+\hat{j}-{\mathrm{\lambda }}^2\hat{k}$ are coplanar, is

The vector $\overrightarrow{\mathrm{a}}=\alpha \hat{i}+2\hat{j}+\beta \hat{k}$ lies in the plane of the vectors $\overrightarrow{\mathrm{b}}=\hat{i}+\hat{j}$ and $\overrightarrow{\mathrm{c}}=\hat{j}+\hat{k}$ and bisects the angle between $\overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$. Then, which one of the following given possible values of $\alpha$ and $\beta$?

If $\overrightarrow{a}, \overrightarrow{b},\overrightarrow{ c}$ are non-coplanar vectors and $\lambda$ is a real number, then $\left[\lambda \left(\overrightarrow{a}+\overrightarrow{b}\right) {\lambda }^2\overrightarrow{b} \lambda \overrightarrow{c}\right]=\left[\overrightarrow{a}\overrightarrow{b}+\overrightarrow{c}\overrightarrow{b}\right]$ for

If $\overrightarrow{\mathit{u}}$, $\overrightarrow{\mathit{v}}$ and $\overrightarrow{\mathit{w}}$ are non-coplanar vectors and $p \text{and} q$ are real numbers, then the equality $\left[3\overrightarrow{\mathit{u}} p\overrightarrow{\mathit{v}} p\overrightarrow{\mathit{w}}\right]–\left[p\overrightarrow{\mathit{v}}\mathit{ }\overrightarrow{\mathit{w}}\mathit{ }q\overrightarrow{\mathit{u}}\right]–\left[2\overrightarrow{\mathit{w}}\mathit{ }q\overrightarrow{\mathit{v}}\mathit{ }q\overrightarrow{\mathit{u}}\right]=0$ holds for

Let $\overrightarrow{u}=\hat{i}+\hat{j}, \overrightarrow{v}=\hat{i}-\hat{j}$ and $\overrightarrow{\mathrm{w}}=\hat{i}+2\hat{j}+3\hat{k}$ If $\hat{n}$ is a unit vector such that $\overrightarrow{u}·\hat{n}=0$ and$\overrightarrow{v}·\hat{n}=0,$ then $\left|\overrightarrow{w}·\hat{n}\right|$ is equal to

Let $\overrightarrow{a}=\stackrel{⏜}{i}-\stackrel{⏜}{k}, \overrightarrow{b}=x\stackrel{⏜}{i}+\stackrel{⏜}{j}+\left(1-x\right)\stackrel{⏜}{k}$ and $\overrightarrow{c}=y\stackrel{⏜}{i}+x\stackrel{⏜}{j}+\left(1+x-y\right)\stackrel{⏜}{k}$.Then,  $\left[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}\right]$ depends on

Let $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}, \overrightarrow{b}=\hat{i}-\hat{j}+2\hat{k} \& \overrightarrow{c}=x\hat{i}+\left(x-2\right)\hat{j}-\hat{k}$. If the vector $\overrightarrow{c}$ lies in the plane of $\overrightarrow{a}$ and $\overrightarrow{b}$, then $x$ is equal to

The value of $a$, so that the volume of parallelepiped formed by $\stackrel{⏜}{i}+a\stackrel{⏜}{j}+\stackrel{⏜}{k}, \stackrel{⏜}{j}+a\stackrel{⏜}{k}$ and $a\stackrel{⏜}{i}+\stackrel{⏜}{k}$ become minimum, is

The unit vector which is orthogonal to the vector $3\hat{\mathrm{i}}+2\hat{\mathrm{j}}+6\hat{\mathrm{k}}$ and is coplanar with the vectors $2\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}},$ 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 the maximum value of the scalar triple product $\left[\overrightarrow{u} \overrightarrow{v} \overrightarrow{w}\right]$ is

If $\alpha \left(\overrightarrow{a}\times \overrightarrow{b}\right)+\beta \left(\overrightarrow{b}\times \overrightarrow{c}\right)+\gamma \left(\overrightarrow{c}\times \overrightarrow{a}\right)=0$, then

Let $\overrightarrow{b}, \overrightarrow{a}$ and $\overrightarrow{c}$ be three non-coplanar vectors and $\overrightarrow{d}$ be a non-zero vector, which is perpendicular to $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}$. Now, if $\overrightarrow{d}=\left(\sin x\right)\left(\overrightarrow{a}\times \overrightarrow{b}\right)+\left(\cos y\right)\left(\overrightarrow{b}\times \overrightarrow{c}\right)+2\left(\overrightarrow{c}\times \overrightarrow{a}\right),$then the minimum value of $x^2+y^2$ is equal to

If $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$ are any three vectors forming a linearly independent system, then $\left[\overrightarrow{a}\cos \theta +\overrightarrow{b}\sin \theta +\overrightarrow{c}\cos 2\theta \overrightarrow{a}\cos \left(\frac{2\pi }{3}+\theta \right)+\overrightarrow{b}\sin \left(\frac{2\pi }{3}+\theta \right)+\overrightarrow{c}\cos 2\left(\frac{2\pi }{3}+\theta \right) \overrightarrow{a}\cos \left(\theta -\frac{2\pi }{3}\right)+\overrightarrow{b}\sin \left(\theta -\frac{2\pi }{3}\right)+\overrightarrow{c}\cos 2\left(\theta -\frac{2\pi }{3}\right)\right]$   $\forall \theta \in R$ equals