Vector or Cross Product of Two Vectors

Let the vectors $\overrightarrow{PQ}, \overrightarrow{QR}, \overrightarrow{RS}, \overrightarrow{ST}, \overrightarrow{TU}$ and $\overrightarrow{UP}$ represent the sides of a regular hexagon.

Statement I: $\overrightarrow{PQ}\times \left(\overrightarrow{RS}+\overrightarrow{ST}\right)\ne 0$, because

Statement II: $\overrightarrow{PQ}\times \overrightarrow{RS}=0$ and $\overrightarrow{PQ}\times \overrightarrow{ST}\ne 0$.

The vector $\overrightarrow{c}$ is perpendicular to the vectors $\overrightarrow{a}=\left(2,-3,1\right)$, $\overrightarrow{b}=\left(1,-2,3\right)$ and satisfies the condition $\overrightarrow{c}·\left( \hat{\mathrm{i}}+2\hat{\mathrm{j}}-7\hat{\mathrm{k}}\right)=10$.Then, the vector $\overrightarrow{c}$ is equal to

For two particular vectors $\overrightarrow{A}$ and $\overrightarrow{B}$, it is known that  $\overrightarrow{A}\times \overrightarrow{B}=\overrightarrow{B}\times \overrightarrow{A}$. What must be true about the two vectors?

A force $\overrightarrow{F}$= $\mathrm{2}\hat{i}+\hat{j}–\hat{k}$ acts at a point A, whose position vector is $\mathrm{2}\hat{i}–\hat{j}$. The moment of $\overrightarrow{F}$about the origin is

A force of magnitude $6$ acts along the vector $\left(9,6,-2\right)$ and passes through a point $A\left(4,-1,-7\right)$. Then the moment of a force about point $O\left(1,-3,2\right)$ is

The moment of a force represented by $\overrightarrow{F}=\hat{i}+2\hat{j}+3\hat{k}$ about the point $2\hat{i}-\hat{j}+\hat{k}$ is equal to

The moment of the force $\overrightarrow{F}$ acting at a point $\overrightarrow{P}$, about the point $\overrightarrow{C}$ is

If $\overrightarrow{a}$ and $\overrightarrow{b}$ are two vectors, then ${\left(\overrightarrow{a}\times \overrightarrow{b}\right)}^2$

A tetrahedron has vertices at $O\left(0,0,0\right), A\left(1,2,1\right), B\left(2,1,3\right)$ and $C\left(–1,1,2\right)$. Then, the angle between the faces $OAB$ and $ABC$ will be

If $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ and $\overrightarrow{d}$ are the unit vectors such that $\left(\overrightarrow{a}\times \overrightarrow{b}\right)·\left(\overrightarrow{c}\times \overrightarrow{d}\right)=1$ and $\overrightarrow{a}·\overrightarrow{c}=\frac{1}{2}$, then

Let $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ be unit vectors such that $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0$. Which one of the following is correct$?$

For any vector $\overrightarrow{a}$, the value of ${\left(\overrightarrow{a}\times \hat{i}\right)}^2+{\left(\overrightarrow{a}\times \hat{j}\right)}^2+{\left(\overrightarrow{a}\times \hat{k}\right)}^2$ is equal to

If $\overrightarrow{\mathrm{u}}$ and $\overrightarrow{\mathrm{v}}$ are unit vectors and θ is the acute angle between them, then $2\overrightarrow{\mathrm{u}}\times 3\overrightarrow{\mathrm{v}}$ is a unit vector for

The vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ are not perpendicular and $\overrightarrow{c}$ and $\overrightarrow{d}$ are two vectors satisfying $\overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{b}\times \overrightarrow{d}$ and $\overrightarrow{a}·\overrightarrow{d}=0$. Then, the vector $\overrightarrow{d}$ is equal to

Let $\overrightarrow{a}=2\hat{i}+\hat{j}-2\stackrel{⏜}{k}, \overrightarrow{b}=\hat{i}+\hat{j}$ and $\overrightarrow{c}$ are the vectors such that $\left|\overrightarrow{c}-\overrightarrow{a}\right|=3, \left|\left(\overrightarrow{a}\times \overrightarrow{b}\right)\times \overrightarrow{c}\right|=3$ and the angle between $\overrightarrow{c}$ and$\overrightarrow{a}\times \overrightarrow{b}$ is $30°$then $\overrightarrow{a}·\overrightarrow{c}$equals to

The shortest distance between a diagonal of a unit cube and a diagonal of a face skew to it is

A parallelepiped is formed by planes drawn parallel to coordinate axes through the points $A=\left(1,2,3\right)$ and $B=\left(9,8,5\right)$. The volume of that parallelepiped is equal to (in cubic units)

If $\overrightarrow{p},\overrightarrow{q}$ are two non-collinear vectors such that $\left(\overrightarrow{b}-\overrightarrow{c}\right) \overrightarrow{p}\times \overrightarrow{q}+\left(\overrightarrow{c}-\overrightarrow{a}\right)\overrightarrow{p}+\left(\overrightarrow{a}-\overrightarrow{b}\right) \overrightarrow{q}=0$ Where $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are lengths of sides of a triangle, then the triangle is

Let $\overrightarrow{a},\overrightarrow{b}\&\overrightarrow{c}$ be the three vectors having magnitudes $1,5$ and $3$respectively such that the angle between $\overrightarrow{a}\&\overrightarrow{b}$ is $\theta$ and $\overrightarrow{a}\times \left(\overrightarrow{a}\times \overrightarrow{b}\right)=\overrightarrow{c}$, Then $\tan \theta$ is equal to

If $\overrightarrow{a},\overrightarrow{b}$ are unit vectors, then the vector defined as $\overrightarrow{V}=\left(\overrightarrow{a}\times \overrightarrow{b}\right)\times \left(\overrightarrow{a}+\overrightarrow{b}\right)$ is collinear to the vector