Embibe Experts Solutions for Chapter: Vector Algebra, Exercise 2: Exercise-2

If $3$ non zero vectors$\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are such that $\overrightarrow{a}\times \overrightarrow{b}=2(\overrightarrow{a}\times \overrightarrow{c}),\left|\overrightarrow{a}\right|=\left|\overrightarrow{c}\right|=1,\overrightarrow{b}|=4$ and the angle between $\overrightarrow{b}$ and $\overrightarrow{c}$ is ${\cos }^{-1}\frac{1}{4}$ , then $\overrightarrow{b}=l\overrightarrow{c}+\mu \overrightarrow{a}$ where $\left|l\right|+\left|\mu \right|$ is -

If $\theta$ is the angle between the vectors $\overrightarrow{p}=a\hat{i}+b\hat{j}+c\hat{k}$ and vector $\overrightarrow{q}=b\hat{i}+c\hat{j}+a\hat{k}$ then range of $\theta$ is

If the circumcentre of the tetrahedron $OABC$ is given by $\frac{\overrightarrow{a^2}\left(\overrightarrow{b}\times \overrightarrow{c}\right)+\overrightarrow{b^2}\left(\overrightarrow{c}\times \overrightarrow{a}\right)+\overrightarrow{c^2}\left(\overrightarrow{a}\times \overrightarrow{b}\right)}{\alpha },$ where $\overrightarrow{a},\overrightarrow{b}\&\overrightarrow{c}$ are the position vectors of the points $A,B,C$ respectively relative to the origin $O$ such that $\left[\overrightarrow{a,} \overrightarrow{b}, \overrightarrow{c}\right]=36$, then $\alpha$ is

About the line $\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+3}{1}$, the plane $3x+4y+6z+7=0$ is rotated till the plane passes through the origin. Now, $4x+\alpha y+\beta z=0$ is the equation of plane in new position. The value of ${\alpha }^2+{\beta }^2$ is 

The vertices of a triangle are$A(1,1,2), B(4,3,1)$ and $C(2,3,5) .$A vector representing the bisector of the angle $A$ is:

Let $\overrightarrow{V}=2\hat{i}+\hat{j}-\hat{k} \& \overrightarrow{W}=\hat{i}+3\widehat{k.}$ If $\overrightarrow{U}$ is a unit vector, then the value of the scalar triple product $\left[\overrightarrow{U}\overrightarrow{v}\overrightarrow{w}\right]$ may be

If$\overrightarrow{A}+\overrightarrow{B}=\overrightarrow{a},\overrightarrow{A}·\overrightarrow{a}=1$ and $\overrightarrow{A}\times \overrightarrow{B}=\overrightarrow{b}$, then 

If $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ and $\overrightarrow{d}$ are the position vectors of the points $A,B,C$ and $D$, respectively in three-dimensional space no three of $A,B,C \text{and} D$ are collinear and satisfy the relation $3\overrightarrow{a}-2\overrightarrow{b}+\overrightarrow{c}-2\overrightarrow{d}=\overrightarrow{0},$ then