If direction ratios of two lines are 5, -12, 13 and -3, 4, 5, then the angle between them is

The direction-ratios of the two lines are 5, – 12, 13 and – 3, 4, 5.Let θ be the angle between the lines.

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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 $\cos \phi =\overset{\wedge }{\mathop{a}}\,.\overset{\wedge }{\mathop{b}}\,$

Tardigrade - CET NEET JEE Exam App

Find the acute angle between the lines whose direction ratios are 5, 12, -13 and 3, - 4, 5.

`cos theta=|(a_1a_2+b_1b_2+c_1c_2)/(sqrt(a_1^2+b_1^2+c_1^2)sqrt(a_2^2+b_2^2+c_2^2))|`

`=|(15-48-65)/(sqrt(25+144+169)sqrt(9+16+25))|`

`=|-98/(13sqrt2xx5sqrt2)|`

`=|-98/(13xx5xx2)|`

`=49/65`

`theta=cos^-1 (49/65)`

Concept: Angle Between Two Lines

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