Question

परिमाण 6 का एक सदिश ज्ञात कीजिए जो दोनों ही सदिशों  ${\displaystyle 2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}}$ और ${\displaystyle 4\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}}$ पर लंब है।

Answer 1

मान लीजिए कि ${\displaystyle 2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}}$ और ${\displaystyle 4\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}}$

हम जानते हैं कि ${\displaystyle \overrightarrow{\text{a}}}$ और ${\displaystyle \overrightarrow{\text{b}}=\frac{(\overrightarrow{\text{a}}\times \overrightarrow{\text{b}})}{|\overrightarrow{\text{a}}\times \overrightarrow{\text{b}}|}}$ के लंबवत मात्रक सदिश

 ${\displaystyle \overrightarrow{\text{a}}\times \overrightarrow{\text{b}}=\left|\begin{array}{ccc}\hat{\text{i}}& \hat{\text{j}}& \hat{\text{k}}\\ 2& -1& 2\\ 4& -1& 3\end{array}\right|}$

= ${\displaystyle \hat{\text{i}}(-3+2)-\hat{\text{j}}(6-8)+\hat{\text{k}}(-2+4)}$

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

∴ ${\displaystyle |\overrightarrow{\text{a}}\times \overrightarrow{\text{b}}|=\sqrt{{(-1)}^2+{\left(2\right)}^2+{\left(2\right)}^2}}$

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

= ${\displaystyle \sqrt{9}}$

= 3

तो, ${\displaystyle \frac{(\overrightarrow{\text{a}}\times \overrightarrow{\text{b}})}{|\overrightarrow{\text{a}}\times \overrightarrow{\text{b}}|}=\frac{-\text{i}+2\hat{\text{j}}+2\hat{\text{k}}}{3}}$

= ${\displaystyle \frac{1}{3}(-\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}})}$

अब परिमाण का सदिश 6 = ${\displaystyle \frac{1}{3}(-\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}})\cdot 6}$

= ${\displaystyle 2(-\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}})}$

= ${\displaystyle -2\hat{\text{i}}+4\hat{\text{j}}+4\hat{\text{k}}}$

अत: अभीष्ट सदिश ${\displaystyle -2\hat{\text{i}}+4\hat{\text{j}}+4\hat{\text{k}}}$ है।