Vector Algebra

$\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]=0$

$\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 6\hfill & \hfill 3\hfill \\ \hfill 3\hfill & \hfill 2\hfill & \hfill 1\hfill \\ \hfill \alpha +1\hfill & \hfill \beta -1\hfill & \hfill 1\hfill \end{array}\right|=0$

$8\beta -24=0\Rightarrow \beta =3$

$\left|\overrightarrow{c}\right|=\sqrt[]{6}$

${\left(\alpha +1\right)}^2+{\left(\beta -1\right)}^2+1=6\Rightarrow {\left(\alpha +1\right)}^2=1$

a = 1 = 1, 1 $\Rightarrow \alpha =-2,0$ $$

+ = 1, 3

$\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}+\overrightarrow{d}\right|=\sum {\left|\overrightarrow{a}\right|}^2+2\sum \overrightarrow{a}.\overrightarrow{b}$  

$=4+2\overrightarrow{d}.\left(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right)$  (a, b, c are mutually ^ r)

Let   $\overrightarrow{d}=\lambda \overrightarrow{a}+\mu \overrightarrow{b}+v\overrightarrow{c}$

Also   ${\lambda }^2+{\mu }^2+v^2=1=3co{s}^2\theta orcos\theta =\frac{1}{\sqrt[]{3}}$

$\therefore {\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}+\overrightarrow{d}\right|}^2=4\pm 2.\frac{3}{\sqrt[]{3}}$        

$=4\pm 2\frac{3}{\sqrt[]{3}}$       

  $\left|\overline{a}*\left(\overline{a}*\overline{c}\right)\right|=\left|3\overline{b}\right|=3\left|\overline{b}\right|$

$\Rightarrow 3=3.2sin\theta \Rightarrow sin\theta =\frac{1}{2}\Rightarrow co{s}^2\theta =\frac{3}{4}$  

You can use the position vector to find the real displacement of a point from a reference point.

Two vectors which are parallel to each other, irrespective of direction, are called parallel vectors.

Types are given below

• Collinear vectors
• Parallel vectors
• Antiparallel vectors

A vector with zero magnitude is considered a zero vector. Since it has no magnitude, its direction is considered arbitrary. This means the zero vector has no specific direction.

|3a+4b|² = 9|a|² + 16|b|² + 24a·b
But a·b = 0, |a|=|b|=k
|3a+4b| = 5k
|4a-3b|
10k = 20? k = 2 = |a| = |b|

  $\overrightarrow{c}=\alpha \overrightarrow{a}+\beta \overrightarrow{b}.....\left(i\right)$

  $\overrightarrow{a}.\overrightarrow{c}=7\overrightarrow{b}.\overrightarrow{c}=0$         

$\overrightarrow{a}=-\widehat{i}+\widehat{j}+\widehat{k}\Rightarrow \left|\overrightarrow{a}\right|=\sqrt[]{3}$           

$\overrightarrow{b}=2\widehat{i}+\widehat{k}\Rightarrow \left|\overrightarrow{b}\right|=\sqrt[]{5}\overrightarrow{a}.\overrightarrow{b}=-2+1=-1$           

From   $\left(i\right)\overrightarrow{a}.\overrightarrow{c}=\alpha {\left|\overrightarrow{a}\right|}^2-\beta$

$3\alpha -\beta =7..........\left(ii\right)$           

$\overline{b}.\overline{c}=\alpha \overline{b}.\overline{a}+\beta {\left|\overrightarrow{b}\right|}^2\Rightarrow -\alpha +5\beta =0.......\left(iii\right)$           

Solving $\alpha =\frac{5}{2}and\beta =\frac{1}{2}$

$\Rightarrow 2{\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|}^2=75$            

${\left|2\left(\overrightarrow{a}*\overrightarrow{b}\right)\right|}^2+4{\left(\overrightarrow{a}.\overrightarrow{b}\right)}^2$

= $4{\left|\overrightarrow{a}\right|}^2{\left|\overrightarrow{b}\right|}^2$

=4 * 16 * 9 = 576

$\overrightarrow{a}=\left(t,t,t\right)$

$\frac{3t+4t}{5}=7\Rightarrow t=5$

$\overrightarrow{a}=\left(5,5,5\right)$

$\overrightarrow{b}=\mathcal{l}\overrightarrow{a}+m\widehat{i}$

$=\left(5\mathcal{l}+m,5\mathcal{l},5\mathcal{l}\right)$

$\frac{\overrightarrow{b}.\left(3,4,0\right)}{5}=\frac{\pm \frac{5}{\sqrt[]{2}}\left(-6+4+0\right)}{5}=\pm \left(\sqrt[]{2}\right)$