Question

The direction ratios of two lines are $1$, $- 3$, $4$ and $- 2$, $0$, $6$. Find the direction cosines of a line perpendicular to both the lines.

Answer 1

We need to find the cross product of the vectors along to the given lines as it is given that the third line is perpendicular to the given two lines. The cross product defines the condition of perpendicularity.

Complete step by step solution:
We know that if $a$, $b$ and $c$ are the direction ratios of a line then the vector along the line is $\vec n = a\hat i + b\hat j + c\hat k$.
Let the vector along the line with direction ratios $1$, $- 3$, $4$ be $\vec u$. So, we get
$\vec u = 1\hat i - 3\hat j + 4\hat k$
Let the vector along the line with direction ratios $- 2$, $0$, $6$ be $\vec v$. So, we get
$\vec u = - 2\hat i + 6\hat k$
To find the direction cosines of the line perpendicular to the given lines, we need to find the cross product of the vectors $\vec u$ and $\vec v$.
\vec u \times \vec v = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\\ 1&{ - 3}&4 \\\ { - 2}&0&6 \end{array}} \right|
$\vec u \times \vec v = \left( { - 18 - 0} \right)\hat i - \left( {6 + 8} \right)\hat j + \left( {0 - 6} \right)\hat k$
$\vec u \times \vec v = - 18\hat i - 14\hat j - 6\hat k$
So, the direction ratios of the line which is perpendicular to the given lines are $- 18$, $- 14$, $- 6$.
The direction cosines of the line are given by $\left( {\dfrac{a}{{\sqrt {{a^2} + {b^2} + {c^2}} }},\dfrac{b}{{\sqrt {{a^2} + {b^2} + {c^2}} }},\dfrac{c}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right)$.
Here, the value of $a$ is $- 18$, $b$ is $- 14$, $c$ is $- 6$.
Now, the direction cosines of the required line can be calculated as:
$\left( {\dfrac{{ - 18}}{{2\sqrt {139} }},\dfrac{{ - 14}}{{2\sqrt {139} }},\dfrac{{ - 6}}{{2\sqrt {139} }}} \right)$

Therefore, the required direction cosines of the line which is perpendicular to the given lines are $\left( {\dfrac{{ - 18}}{{2\sqrt {139} }},\dfrac{{ - 14}}{{2\sqrt {139} }},\dfrac{{ - 6}}{{2\sqrt {139} }}} \right)$.

Note:
The direction of a line can be any three numbers. The vector for the required line can also be written as $n = r\left( {\cos \alpha \hat i + \cos \beta \hat j + \cos \gamma \hat k} \right)$ such that $a$, $b$ and $c$ are direction ratios. The direction cosines are $\cos \alpha$, $\cos \beta$ and $\cos \gamma$.