সৃজনশীলঃ ‘A→=2i^+3j^−k^,B→=i^+2j^−k^,C→=i^+bj^+3k^…’ দিনাজপুর বোর্ড ২০১৭ – ভেক্টর, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে, $\overrightarrow{A}=2\hat{i}+3\hat{j}-\hat{k},\overrightarrow{B}=\hat{i}+2\hat{j}-\hat{k}$

এবং $\overrightarrow{C}=\hat{i}+b\hat{j}+3\hat{k}$

$\therefore |\overrightarrow{A}|=\sqrt{2^2+3^2+(-1{)}^2}=\sqrt{4+9+1}=\sqrt{14}$

$\overrightarrow{A}\cdot \overrightarrow{B}=(2\hat{i}+3\hat{j}-\hat{k})\cdot (\hat{i}+2\hat{j}-\hat{k})=2+6+1=9$

$\overrightarrow{A}$ ভেক্টর বরাবর $\overrightarrow{B}$ ভেক্টরে উপাংশ = $\frac{\overrightarrow{A}\cdot \overrightarrow{B}}{|\overrightarrow{A}{|}^2}\cdot \overrightarrow{A}$

$=\frac{9}{(\sqrt{14}{)}^2}\cdot (2\hat{i}+3\hat{j}-\hat{k})$

$=\frac{9}{14}(2\hat{i}+3\hat{j}-\hat{k})$

$=\frac{9}{7}\hat{i}+\frac{27}{14}\hat{j}-\frac{9}{14}\hat{k}$

$\overrightarrow{A}$ ভেক্টর বরাবর $\overrightarrow{B}$ ভেক্টরের উপাংশ $\overrightarrow{C}$ ভেক্টরের সাথে লম্ব হলে,

$(\frac{9}{7}\hat{i}+\frac{27}{14}\hat{j}-\frac{9}{14}\hat{k})\cdot \overrightarrow{C}=0$

বা, $(\frac{9}{7}\hat{i}+\frac{27}{14}\hat{j}-\frac{9}{14}\hat{k})\cdot (\hat{i}+b\hat{j}+3\hat{k})=0$

বা, $\frac{9}{7}+\frac{27}{14}b-\frac{27}{14}=0$

বা, $\frac{27}{14}b-\frac{9}{14}=0$

বা, $\frac{27}{14}b=\frac{9}{14}$

বা, $b=\frac{9\times 14}{14\times 27}=\frac{1}{3}$

নির্ণেয় নাম : $b=\frac{1}{3}$.

দেওয়া আছে, $\overrightarrow{A}=2\hat{i}+3\hat{j}-\hat{k}$ এবং $\overrightarrow{B}=\hat{i}+2\hat{j}-\hat{k}$

$\therefore \overrightarrow{A}+\overrightarrow{B}=2\hat{i}+3\hat{j}-\hat{k}+\hat{i}+2\hat{j}-\hat{k}=3\hat{i}+5\hat{j}-2\hat{k}$

$|\overrightarrow{A}+\overrightarrow{B}|=\sqrt{3^2+5^2+(-2{)}^2}=\sqrt{9+25+4}=\sqrt{38}$

$\overrightarrow{A}\times \overrightarrow{B}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& -1\\ 1& 2& -1\end{array}\right|$

$=\hat{i}(-3+2)+\hat{j}(-1+2)+\hat{k}(4-3)$

$=-\hat{i}+\hat{j}+\hat{k}$

$|\overrightarrow{A}\times \overrightarrow{B}|=\sqrt{(-1{)}^2+1^2+1^2}=\sqrt{1+1+1}=\sqrt{3}$

$\overrightarrow{A}+\overrightarrow{B}$ এবং $\overrightarrow{A}\times \overrightarrow{B}$ এর মধ্যবর্তী কোণ $\theta$ হলে,

$\cos \theta =\frac{(\overrightarrow{A}+\overrightarrow{B})\cdot (\overrightarrow{A}\times \overrightarrow{B})}{|\overrightarrow{A}+\overrightarrow{B}|\cdot |\overrightarrow{A}\times \overrightarrow{B}|}$

$=\frac{(3\hat{i}+5\hat{j}-2\hat{k})\cdot (-\hat{i}+\hat{j}+\hat{k})}{\sqrt{38}\cdot \sqrt{3}}$

$=\frac{-3+5-2}{\sqrt{114}}=\frac{5-5}{\sqrt{1\cdot 14}}=\frac{0}{\sqrt{114}}=0$

বা, $\theta ={\cos }^{-1}0$

$\therefore \theta ={90}^{\circ }$

নির্ণেয় কোণ ${90}^{\circ }$.