Question

Examine whether the victors $a=2\hat{i}+3\hat{j}+2k,b=1-\hat{j}+2k$ and $c=3\hat{j}+2j-4k$ form a left handed or right handed system

Answer 1

Consider the given vectors.

Let ,

Position vector points A, B and C are

$\overrightarrow{a}=2\hat{i}+3\hat{j}+2\hat{k},\overrightarrow{b}=\hat{i}-\hat{j}+2k,\overrightarrow{c}=3\hat{i}+2\hat{j}-4\hat{k}$

Direction ratios of line AB are ,

$1-2,-1-3,2-2i.e.,-1,-4,0$ .

Direction ratios of line BC are ,

$3-1,2+1,-4-2i.e.2,3,-6.$

Direction ratios of line CA are ,

$2-3,3-2,2+4i.e.,-1,1,6$ .

Now,

Direction cosines of AB are ,

$\Rightarrow \frac{-1}{\sqrt{{(-1)}^2+{(-4)}^2+0^2}},\frac{-4}{\sqrt{{(-1)}^2+{(-4)}^2+0^2}},\frac{0}{\sqrt{{(-1)}^2+{(-4)}^2+0^2}}$

$\Rightarrow \frac{-1}{\sqrt{17}},\frac{-4}{\sqrt{17}},0$

Direction cosines of line BC are,

$\Rightarrow \frac{2}{\sqrt{2^2+3^2+{(-6)}^2}},\frac{3}{\sqrt{2^2+3^2+{(-6)}^2}},\frac{-6}{\sqrt{2^2+3^2+{(-6)}^2}}$

$\Rightarrow \frac{2}{7},\frac{3}{7},\frac{-6}{7}$

Direction cosines of line CA are,

$\Rightarrow \frac{-1}{\sqrt{{(-1)}^2+1^2+6^2}},\frac{1}{\sqrt{{(-1)}^2+1^2+6^2}},\frac{6}{\sqrt{{(-1)}^2+1^2+6^2}}$

$\Rightarrow \frac{-1}{\sqrt{37}},\frac{1}{\sqrt{37}},\frac{6}{\sqrt{37}}$

Hence , this is the answer .