Embibe Experts Solutions for Chapter: Vector Algebra, Exercise 1: TS EAMCET 2018 (04-May Shift-1)

Let $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}, \overrightarrow{d}$ be four vectors such that $\overrightarrow{a}$ is perpendicular only to$\overrightarrow{c}$. The vector $\overrightarrow{b}$ is parallel to$(\overrightarrow{c}-\overrightarrow{d})$ then$\overrightarrow{c}=$

Consider the vectors $\overline{a}=3\overline{i}+5\overline{j}+2\overline{k}, \overline{b}=2\overline{i}-3\overline{j}-5\overline{k}$ and $\overline{c}=-5\overline{i}-2\overline{j}+3\overline{k}.$ If $l, m$ and $n$ are length of projections of $\overline{a}$ on $\overline{b}, \overline{b}$ on $\overline{c}$ and $\overline{c}$ on $\overline{a}$ respectively, then

Let $(x, y)\in (R\times R)$ and $\overrightarrow{a}=x\hat{i}+2\hat{j}-\hat{k}, \overrightarrow{b}=6\hat{i}-y\hat{j}+2\hat{k}$ be two vectors. If $|\overrightarrow{a}\times \overrightarrow{b}{|}^2+|\overrightarrow{a}.\overrightarrow{b}{|}^2=f\left(x\right)g\left(y\right)$ then $f\left(x\right)+g\left(y\right)-46=0$ represents

If $\overrightarrow{a}=2\hat{i}+2\hat{j}+\hat{k},\overrightarrow{\left|b\right|}=6$ and the angle between $\overrightarrow{a}$ and$\overrightarrow{b}$  is $\frac{\pi }{6},$ then the area of the triangle (in square units) with $\overline{a}$ and $\overline{b}$ as two of its sides is

The four points $A\left(2,-1,3\right), B\left(4,-2,1\right), C\left(4,5,-7\right)$ and $D\left(2,6,-5\right)$ forms a

Let $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ be unit vectors such that $2\overrightarrow{a}+3\overrightarrow{b}+4\overrightarrow{c}=\overrightarrow{0}.$ Then $|\overrightarrow{b}\times \overrightarrow{c}|=$

If $\overrightarrow{r}$ is a vector perpendicular to both the vectors $2\hat{i}+3\hat{j}-4\hat{k}$ and $3\hat{i}-\hat{j}+\hat{k}$ and satisfy $\overrightarrow{r}.\left(3\hat{i}-3\hat{j}+4\hat{k}\right)=5$, then $\left|\overrightarrow{r}\right|=$

If $\overrightarrow{b}=2\hat{i}-\hat{j}-\hat{k}, \overrightarrow{a}=3\hat{i}+4\hat{j}-5\hat{k}$ and $\overrightarrow{b}\times (\overrightarrow{a}\times \overrightarrow{b})=\frac{\overrightarrow{a}-k\overrightarrow{b}}{l}$ then $\frac{k}{l\left|\overrightarrow{b}\right|}$ is