Question

Show that the vectors ${\displaystyle -\hat{\text{i}}-2\hat{\text{j}}-6\hat{\text{k}},2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}}$ and find ${\displaystyle -\hat{\text{i}}+3\hat{\text{j}}+5\hat{\text{k}}}$ form a right angled triangle

Answer 1

Let ${\displaystyle \overrightarrow{\text{AB}}\mid =-\hat{\text{i}}-2\hat{\text{j}}-6\hat{\text{k}}}$

${\displaystyle \overrightarrow{\text{BC}}=2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}}$

and ${\displaystyle \overrightarrow{\text{CA}}=-\hat{\text{i}}-2\hat{\text{j}}-5\hat{\text{k}}}$

${\displaystyle \left|\overrightarrow{\text{AB}}\right|=|-\hat{\text{i}}-2\hat{\text{j}}-6\hat{\text{k}}|}$

= ${\displaystyle \sqrt{{(-1)}^2+{(-2)}^2+{(-6)}^2}}$

AB = ${\displaystyle \sqrt{1+4+36}}$

= ${\displaystyle \sqrt{41}}$

${\displaystyle \left|\overrightarrow{\text{BC}}\right|=|2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}|}$

= ${\displaystyle \sqrt{2^2+{(-1)}^2+1^2}}$

BC = ${\displaystyle \sqrt{4+1+1}}$

= ${\displaystyle \sqrt{6}}$

${\displaystyle \left|\overrightarrow{\text{CA}}\right|=|-\hat{\text{i}}+3\hat{\text{j}}+5\hat{\text{k}}|}$

= ${\displaystyle \sqrt{{(-1)}^2+3^2+5^2}}$

CA = ${\displaystyle \sqrt{1+9+25}}$

= ${\displaystyle \sqrt{35}}$

AB ≠ BC + CA

∴ The given vectors form a triangle, Also

AB² = 41, BC² = 6, CA² = 35

AB² = BC² + CA²

∴ ∆ABC is a right angled triangle.