Angle between lines whose direction ratios are 1,2,1 and (2 3,4 is)

Answer

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Hint: Determine the line vectors using the direction ratios, we know for the vector $\overrightarrow{r}=a\overset{\hat{\ }}{\mathop{i}}\,+b\overset{\wedge }{\mathop{j}}\,+c\overset{\wedge }{\mathop{k}}\,$, the direction ratios are (a, b, c) and the proportionality given in the question, then find their scalar product.

Complete step-by-step answer:

Let us assume the line vector with direction ratios 4, -3, 5 to be $\overrightarrow{a}$.Then, $\overrightarrow{a}=4\overset{\hat{\ }}{\mathop{i}}\,-3\overset{\wedge }{\mathop{j}}\,+5\overset{\wedge }{\mathop{k}}\,$.Let us assume the line vector with direction ratios 3, 4, 5 to be $\overrightarrow{b}$.Therefore, the x component is 3.Therefore, the y component is 4.Therefore, the z component is 5.Then, $\overrightarrow{b}=3\overset{\hat{\ }}{\mathop{i}}\,+4\overset{\wedge }{\mathop{j}}\,+5\overset{\wedge }{\mathop{k}}\,$.We know that the scalar product of $\overrightarrow{x}\ and\ \overrightarrow{y}$is$\overrightarrow{x}.\overrightarrow{y}=\left| \overrightarrow{x} \right|\left| \overrightarrow{y} \right|\cos \theta $,Where $\overrightarrow{x}\ and\ \overrightarrow{y}$ are the magnitudes of $\overrightarrow{x}\ and\ \overrightarrow{y}$ respectively and $\theta $ is the angle between $\overrightarrow{x}\ and\ \overrightarrow{y}$.Let us assume the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ i.e. the two given line to be $\phi $, then$\left( \overrightarrow{a}.\overrightarrow{b} \right)=\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\cos \phi $This can also be written as,$\cos \phi =\dfrac{\overrightarrow{a}.\overrightarrow{b}}{\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|}..........(i)$Now we will find the values separately,$\begin{align} & \left| \overrightarrow{a} \right|=\sqrt{{{4}^{2}}+{{\left( -3 \right)}^{2}}+{{5}^{2}}}=\sqrt{50}=5\sqrt{2} \\ & \left| \overrightarrow{b} \right|=\sqrt{{{3}^{2}}+{{4}^{2}}+{{5}^{2}}}=\sqrt{50}=5\sqrt{2} \\ \end{align}$Substituting these values in equation (i), we get$\begin{align} & \cos \phi =\dfrac{\overrightarrow{a}.\overrightarrow{b}}{\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|} \\ & \Rightarrow \cos \phi =\dfrac{\left( 4\overset{\hat{\ }}{\mathop{i}}\,-3\overset{\wedge }{\mathop{j}}\,+5\overset{\wedge }{\mathop{k}}\, \right)\left( 3\overset{\hat{\ }}{\mathop{i}}\,+4\overset{\wedge }{\mathop{j}}\,+5\overset{\wedge }{\mathop{k}}\, \right)}{5\sqrt{2}\times 5\sqrt{2}} \\ & \Rightarrow \cos \phi =\dfrac{4\times 3-3\times 4+5\times 5}{50} \\ & \Rightarrow \cos \phi =\dfrac{25}{50} \\ & \Rightarrow \cos \phi =\dfrac{1}{2} \\ \end{align}$Therefore,$\phi ={{\cos }^{-1}}\dfrac{1}{2}$$\phi =\dfrac{\pi }{3}$Therefore, the angle between the two lines is $\dfrac{\pi }{3}$.

Note:Another approach to find the angle between two lines is by first finding the unit vectors of the given lines by using the formula $\overset{\wedge }{\mathop{a}}\,=\dfrac{\overrightarrow{a}}{\left| \overrightarrow{a} \right|}$. And the angle between two unit vector is given by $\cos \phi =\overset{\wedge }{\mathop{a}}\,.\overset{\wedge }{\mathop{b}}\,$

\[ \text{ The direction ratios of the first line are 1, - 2, 1 and the direction ratios of the second line are 4, 3, 2 } . \]

\[ \text{ Let } \theta \text{ be the angle between these two lines } . \]

\[\text{ Now }, \]

\[\cos \theta = \left| \frac{1\left( 4 \right) + \left( - 2 \right)\left( 3 \right) + 1\left( 2 \right)}{\sqrt{\left( 1 \right)^2 + \left( - 2 \right)^2 + \left( 1 \right)^2} \sqrt{\left( 4 \right)^2 + \left( 3 \right)^2 + \left( 2 \right)^2}} \right|\]

\[ = \left| \frac{4 - 6 + 2}{\sqrt{1 + 4 + 1}\sqrt{16 + 9 + 4}} \right|\]

\[ = \frac{0}{\sqrt{6}\sqrt{29}}\]

\[ = 0 \]

\[ \Rightarrow \theta = \frac{\pi}{2}\]

\[\text { Hence, the required angle is } \frac{\pi}{2} .\]

Text Solution

Solution : The direction ratios of the first line are `1,-2,1` and the direction ratios of the second line are `4,3,2`. Let `theta` be the angle between these two lines. Now , `` `cos theta=|frac{1(4)+(-2)(3)+1(2)}{sqrt{(1)^{2}+(-2)^{2}+(1)^{2}} sqrt{(4)^{2}+(3)^{2}+(2)^{2}}}|` `=|frac{4-6+2}{sqrt{1+4+1} sqrt{16+9+4}}|` `=frac{0}{sqrt{6} sqrt{29}}` `=0` `Rightarrow theta=frac{pi}{2}` `` Hence, the required angle is `frac{pi}{2}`.

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