Find the angle between the vectors i-2j+3k and 3i-2j+k. - Vidyadip

Question

Find the angle between the vectors $\hat{i}-2\hat{j}+3\hat{k}\ \text{and}\ 3\hat{i}-2\hat{j}+\hat{k}.$

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Answer

$\text{Given:}\ \ \ \text{Let}\ \ \ \ \vec{a}=\hat{i}-2\hat{j}+3\hat{k}\ \text{and}\ \vec{b}=3\hat{i}-2\hat{j}+\hat{k}$ $\Rightarrow\ \ \big|\vec{a}\big|=\sqrt{1+4+9}=\sqrt{14}\ \text{and}\ \Big|\vec{b}\Big|=\sqrt{9+4+1}=\sqrt{14}$ $\Big[\because\ \text{x}\hat{i}+\text{y}\hat{j}+\text{z}\hat{k}=\sqrt{\text{x}^2+\text{y}^2+\text{z}^2}\ \Big]$ $\text{Also}\ \ \ \ \vec{a}.\vec{b}$= Product of coefficients of $\hat{i}$ + Product of coefficients of $\hat{j}$ + Product of coefficients $\hat{k}$
= 1(3) + (-2)(-2) + 3(1) = 3 + 4 + 3 = 10
Let $\theta$ be the angle between the vector $\vec{a}\ \text{and}\ \vec{b}.$ We know that $\text{cos}\ \theta=\frac{\vec{a}.\vec{b}}{\big|\vec{a}\big|.\big| \vec{b}\big|}$$\Rightarrow\ \ \text{cos}\ \theta=\frac{10}{\sqrt{14}\ .\sqrt{14}}=\frac{10}{14}=\frac{5}{7}$ $ \Rightarrow\ \text{cos}\ \theta=\frac{5}{7}$
$\Rightarrow\ \ \theta=\cos^{-1}\frac{5}{7}$

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