Scalar Product of Two Vectors

Let the unit vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ be perpendicular to each other and the unit vector $\overrightarrow{c}$ be inclined at an angle $\theta$ to both $\overrightarrow{a}$ and $\overrightarrow{b}$. If $\overrightarrow{c}=x\overrightarrow{a}+y\overrightarrow{b}+z\left(\overrightarrow{a}\times \overrightarrow{b}\right)$, then

The moment about the point $\hat{i}+2\hat{j}+3\hat{k}$ of a force represented by $\hat{i}+\hat{j}+\hat{k}$ acting through the point $2\hat{i}+3\hat{j}+\hat{k}$ is 

The resultant vector of $\overrightarrow{\mathit{P}}$ and $\overrightarrow{Q}$ is $\overrightarrow{R}$. On reversing the direction of $\overrightarrow{Q}$ the resultant vector becomes $\overrightarrow{S}$. Find the value of ${\left|\overrightarrow{R}\right|}^2+{\left|\overrightarrow{S}\right|}^2$

A vector of magnitude $3$, bisecting the angle between the vectors $\overrightarrow{a}=2\hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{b}=\hat{i}-2\hat{j}+\hat{k}$ and making an obtuse angle with $\overrightarrow{b}$ is

Let $\overrightarrow{A}=2\hat{i}+\hat{k}, \overrightarrow{B}=\hat{i}+\hat{j}+\hat{k}$ and $\overrightarrow{C}=4\hat{i}-3\hat{j}+7\hat{k}$. The vector $\overrightarrow{R}$ which satisfies the equations $\overrightarrow{R}\times \overrightarrow{B}=\overrightarrow{C}\times \overrightarrow{\mathit{B}}$ and $\overrightarrow{R}·\overrightarrow{A}=0$ is given by

The acute angle between the medians drawn through the acute angle of an isosceles right angled triangle is

The vector $\overrightarrow{b}=3\hat{i}+4\hat{\mathrm{k}}$ is to be written as the sum of a vector $\overrightarrow{b_1}$ parallel to $\overrightarrow{a}=\hat{i}+\hat{j}$ and a vector $\overrightarrow{b_2}$ perpendicular to $\overrightarrow{a}$. Then $b_1$ is equal to

If $\overrightarrow{a}·\overrightarrow{b}=\overrightarrow{a}·\overrightarrow{c}$ and $\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{a}\times \overrightarrow{c}$, then correct statement is

Forces $3\hat{i}+2\hat{j}+5\hat{k}$ and $2\hat{i}+\hat{j}+3\hat{k}$ are acting at a particle which is displaced from point $2\hat{i}-\hat{j}-3\hat{k}$ to the point $4\hat{i}-3\hat{j}+\hat{k}$. The work done by forces is

$X$-component of $\overrightarrow{a}$ is twice its $Y$-component. If the magnitude of the vector is $5\sqrt{2}$ and it makes an angle of $135°$ with $z$-axis then the vector is

If $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}, \overrightarrow{b}=\hat{i}+3j+5\hat{k}$ and $\overrightarrow{c}=7\hat{i}+9\hat{j}+11\hat{k}$, then the area of parallelogram having diagonals $\overrightarrow{a}+\overrightarrow{b}$ and $\overrightarrow{b}+\overrightarrow{c}$ is

If $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$ are unit vectors, then $|\overrightarrow{a}-\overrightarrow{b}{|}^2+|\overrightarrow{b}-\overrightarrow{c}{|}^2+|\overrightarrow{c}-\overrightarrow{a}{|}^2$ does not exceed

Force $\hat{i}+2\hat{j}-3\hat{k}, 2\hat{i}+3\hat{j}+4\hat{k}$ and $-\hat{i}-\hat{j}+\hat{k}$ are acting at the point $P\left(0,1,2\right)$. The moment of these forces about the point $A\left(1,-2,0\right)$ is

Three non-coplanar vectors $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$ are drawn from a common initial point. The angle between the plane passing through the terminal points of these vectors and the vector $\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{a}$ is

If $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are three non-zero vectors such that $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0$ and $m=\overrightarrow{a}·\overrightarrow{b}+\overrightarrow{b}·\overrightarrow{c}+\overrightarrow{c}·\overrightarrow{a}$, then

Find the area of the parallelogram whose diagonals are represented by $3\hat{i}+\hat{j}+\hat{k}$ and $\hat{i}-\hat{j}-\hat{k}$

If $A, B, C$ and $D$ are four points in space satisfying is $\overrightarrow{AB}·\overrightarrow{CD}=\stackrel{}{k}[{\left|\overrightarrow{AD}\right|}^2+{\left|\overrightarrow{BC}\right|}^2-{\left|\overrightarrow{AC}\right|}^2-{\left|\overrightarrow{BD}\right|}^2]$ then the value of $k$ is

Let $\overrightarrow{a}=2\hat{i}+\hat{j}-2\hat{k}$ and $\overrightarrow{b}=\hat{i}+\hat{j}$. If $\overrightarrow{c}$ is a vector such that $\overrightarrow{a}.\overrightarrow{c}=\left|\overrightarrow{c}\right|, \left|\overrightarrow{c}-\overrightarrow{a}\right|=2\sqrt{2}$ and the angle between $\overrightarrow{a}\times \overrightarrow{b}$ and $\overrightarrow{c}$ is ${30}^{°}$, then $\left|\left(\overrightarrow{a}\times \overrightarrow{b}\right)\times \overrightarrow{c}\right|=$

If $\left|\overrightarrow{\mathrm{a}}\right|=3, \left|\overrightarrow{\mathrm{b}}\right|=2, \left|\overrightarrow{\mathrm{c}}\right|=1$, then the value of $|\overrightarrow{\mathrm{a}}·\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}}·\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}}·\overrightarrow{\mathrm{a}}|$ is (given that $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=0$)

If $\overrightarrow{a}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+2\hat{\mathrm{k}}$ and $\overrightarrow{b}=2\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$, then the angle $\theta$ between $\overrightarrow{a}$ and $\overrightarrow{b}$ is given by