Question

Find the angle between the following pairs of lines:
(i) $\dfrac{{x + 4}}{3} = \dfrac{{y - 1}}{5} = \dfrac{{z + 3}}{4}$ and $\dfrac{{x + 1}}{1} = \dfrac{{y - 4}}{1} = \dfrac{{z - 5}}{2}$.

Answer 1

In our question the equation of line is given in Cartesian form. Let ${L_1}$ and ${L_2}$ be any two straight lines in three dimensional space, in which the lines passing through the origin and with direction ratios: ${a_1},{b_1},{c_1}$ and ${a_2},{b_2},{c_2}$ to the lines ${L_1}$ and ${L_2}$ respectively. In Cartesian form, if $\theta$ is the angle between the lines,
${L_1}:\dfrac{{x - {x_1}}}{{{a_1}}} = \dfrac{{y - {y_1}}}{{{b_1}}} = \dfrac{{z - {z_1}}}{{{c_1}}}$
${L_2}:\dfrac{{x - {x_2}}}{{{a_2}}} = \dfrac{{y - {y_2}}}{{{b_2}}} = \dfrac{{z - {z_2}}}{{{c_2}}}$
Then angle between the lines is: $\cos \theta = \left| {\dfrac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {{a_1}^2 + {b_1}^2 + {c_1}^2} \sqrt {{a_2}^2 + {b_2}^2 + {c_2}^2} }}} \right|$, let us find the angle by using this formula.

Complete answer:
We are given the problem,
(i) $\dfrac{{x + 4}}{3} = \dfrac{{y - 1}}{5} = \dfrac{{z + 3}}{4}$ and $\dfrac{{x + 1}}{1} = \dfrac{{y - 4}}{1} = \dfrac{{z - 5}}{2}$.
The direction ratios of the first line are $3,5,4$ and the direction ratios of the second line are $1,1,2$, that is ${a_1} = 3,{b_1} = 5,{c_1} = - 3$ and ${a_2} = 1,{b_2} = 1,{c_2} = 2$
If θ is the angle between them, then substitute the values in the formula:$\cos \theta = \left| {\dfrac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {{a_1}^2 + {b_1}^2 + {c_1}^2} \sqrt {{a_2}^2 + {b_2}^2 + {c_2}^2} }}} \right|$
$\cos \theta = \left| {\left. {\dfrac{{3.1 + 5.1 + 4.2}}{{\sqrt {{3^2} + {5^2} + {4^2}} \sqrt {{1^2}{{.1}^2}{{.2}^2}} }}} \right|} \right.$
The dot operator represents the multiplication, so by performing multiplication in numerator and by squaring in the denominator,
$= \left| {\dfrac{{3 + 5 + 8}}{{\sqrt {9 + 25 + 16} \sqrt {1 + 1 + 4} }}} \right|$
The values inside the modulus $\left( {{\text{||}}} \right)$ is $+$ve, so removing the modulus.
$= \dfrac{{16}}{{\sqrt {50} \sqrt 6 }}$
The $\sqrt {50}$ can be also written as $\sqrt {25 \times 2}$ because $25 \times 2 = 50$, similarly $\sqrt 6$ is written as $\sqrt {2 \times } 3$,
$= \dfrac{{16}}{{\sqrt {25 \times 2} \sqrt {2 \times 3} }}$
The square root of $25$ is 5$$$$\left( {\sqrt {25} = 5} \right), therefore by taking square root of $25$,
$= \dfrac{{16}}{{5\sqrt 2 \sqrt 2 \sqrt 3 }}$
Now multiply $\sqrt 2 \times \sqrt 2$ we will get $2$i.e., $\sqrt 2 \times \sqrt 2 = 2$
$= \dfrac{{16}}{{5 \times 2 \times \sqrt 3 }}$
Cancelling the numerator $16$ by the denominator $2$, we will get,
$= \dfrac{8}{{5\sqrt 3 }}$
$\cos \theta = \dfrac{8}{{5\sqrt 3 }}$
Bringing the $\cos$ into the R.H.S. it will become inverse, that is, ${\cos ^{ - 1}}$
$\theta = {\cos ^{ - 1}}\dfrac{8}{{5\sqrt 3 }}$.
Therefore the angle between $\dfrac{{x + 4}}{3} = \dfrac{{y - 1}}{5} = \dfrac{{z + 3}}{4}$ and $\dfrac{{x + 1}}{1} = \dfrac{{y - 4}}{1} = \dfrac{{z - 5}}{2}$ is $\theta = {\cos ^{ - 1}}\dfrac{8}{{5\sqrt 3 }}$.

Note:
The angle between the pairs of lines can also be found by using the vector method. In vector method let $\mathop b\limits^ \to$ the vectors parallel to the given line. If $\mathop b\limits^ \to = a\hat i + b\hat j + c\hat k$, then $a,b,c$ are direction ratios of the line with $\hat i,\hat j,\hat k$ are the unit vectors in the direction of $x$-axis, $y$-axis and $z$-axis respectively.
Then the angle $\theta$ between the lines is given by: $\cos \theta = \left| {\dfrac{{\mathop {{b_1}}\limits^ \to .\mathop {{b_2}}\limits^ \to }}{{\left| {\mathop {{b_1}}\limits^ \to } \right|\left| {\mathop {{b_2}}\limits^ \to } \right|}}} \right|$.
In our question the lines are given in Cartesian form, let convert it into vector form to apply this formula.
Equation of the first line is $\dfrac{{x + 4}}{3} = \dfrac{{y - 1}}{5} = \dfrac{{z + 3}}{4} \Rightarrow \mathop {{b_1}}\limits^ \to = 3\hat i + 5\hat j + 4\hat k$ and equation of the second line is $\dfrac{{x + 1}}{1} = \dfrac{{y - 4}}{1} = \dfrac{{z - 5}}{2} \Rightarrow \mathop {{b_2}}\limits^ \to = 1\hat i + 1\hat j + 2\hat k$.
Substituting the values in the formula, $\cos \theta = \left| {\dfrac{{\mathop {{b_1}}\limits^ \to .\mathop {{b_2}}\limits^ \to }}{{\left| {\mathop {{b_1}}\limits^ \to } \right|\left| {\mathop {{b_2}}\limits^ \to } \right|}}} \right|$
The magnitude of a vector $\mathop b\limits^ \to = a\hat i + b\hat j + c\hat k$ is $\left| {\mathop b\limits^ \to } \right| = \sqrt {{a^2} + {b^2} + {c^2}}$, by applying this, $\left| {\mathop {{b_1}}\limits^ \to } \right| = \sqrt {{3^2} + {5^2} + {4^2}}$ and $\left| {\mathop {{b_2}}\limits^ \to } \right| = \sqrt {{1^2} + {1^2} + {2^2}}$.
$\cos \theta = \dfrac{{\left( {3\hat i + 5\hat j + 4\hat k} \right)\left( {1\hat i + 1\hat j + 2\hat k} \right)}}{{\sqrt {{3^2} + {5^2} + {4^2}} \sqrt {{1^2} + {1^2} + {2^2}} }}$
Now in numerator multiply the coefficients of $\hat i$, $\hat j$ and $\hat k$ respectively and in denominator calculate the values of square,
$= \dfrac{{3 + 5 + 8}}{{\sqrt {9 + 25 + 16} \sqrt {1 + 1 + 4} }}$
By doing addition,
$= \dfrac{{16}}{{\sqrt {50} + \sqrt 6 }}$
By simplifying as we do as in the earlier method we will get,
$\cos \theta = \dfrac{8}{{5\sqrt 3 }}$
$\theta = {\cos ^{ - 1}}\dfrac{8}{{5\sqrt 3 }}$.
Here we use two different methods to find the angle between the pairs of lines, one is the Cartesian method and the other is the vector method. In both methods we only use the values in the denominator from a line equation to find the angle and not the numerator values so don’t get confused with this.