
Three lines are drawn from the origin O with direction cosines proportional to $(1, - 1,1)$ , $(2, - 3,0)$ and $(1,0,3)$. The three lines are
A. Not coplanar
B. Coplanar
C. Perpendicular to each other
D. Coincident
Answer
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Hint: In this question, we have given the direction cosines of three lines drawn from the origin O and we have to check whether these lines are coplanar or not. In order to solve this, you have to consider the determinant formed by the direction cosines and then check its value.
Complete step-by-step answer:
Here in the question there are three lines with the direction cosines proportional to $(1, - 1,1)$ , $(2, - 3,0)$ and $(1,0,3)$.
The equation for line I is given by,
$\dfrac{x}{1} = \dfrac{y}{{ - 1}} = \dfrac{z}{1}$
The equation for line II is given by
$\dfrac{x}{2} = \dfrac{y}{{ - 3}} = \dfrac{z}{0}$
And the equation for line III is given by
$\dfrac{x}{1} = \dfrac{y}{0} = \dfrac{z}{3}$
To check the coplanar first consider the direction cosines as the determinant and check its value.
Taking the determinant as:
$\left| {\begin{array}{*{20}{c}}
1&{ - 1}&1 \\
2&{ - 3}&0 \\
1&0&3
\end{array}} \right|$
Now on solving this determinant we get
$\Delta = 1( - 9) + 1(6 - 0) + 1(0 + 3)$
On further solving, we get
$\Delta = 0$
Since, the value for determinant is zero
So, the lines are coplanar in nature.
Therefore, the correct option is (B).
Additional information:
Some real-world examples of coplanar lines are
The lines that are drawn on a notebook are coplanar to each other.
The grid that is drawn on a graphing paper is also coplanar.
The hands of our clocks or watches (the second, minute, and hour hands) are also coplanar to each other.
Note:
Remember that the coplanar lines are the lines that lie on the same plane. Two lines in a three-dimensional plane are said to be coplanar if there is a plane that includes them both. This condition occurs if the lines intersect with each other or if they are parallel to each other. The two lines that are not coplanar are commonly known as skew lines.
Complete step-by-step answer:
Here in the question there are three lines with the direction cosines proportional to $(1, - 1,1)$ , $(2, - 3,0)$ and $(1,0,3)$.
The equation for line I is given by,
$\dfrac{x}{1} = \dfrac{y}{{ - 1}} = \dfrac{z}{1}$
The equation for line II is given by
$\dfrac{x}{2} = \dfrac{y}{{ - 3}} = \dfrac{z}{0}$
And the equation for line III is given by
$\dfrac{x}{1} = \dfrac{y}{0} = \dfrac{z}{3}$
To check the coplanar first consider the direction cosines as the determinant and check its value.
Taking the determinant as:
$\left| {\begin{array}{*{20}{c}}
1&{ - 1}&1 \\
2&{ - 3}&0 \\
1&0&3
\end{array}} \right|$
Now on solving this determinant we get
$\Delta = 1( - 9) + 1(6 - 0) + 1(0 + 3)$
On further solving, we get
$\Delta = 0$
Since, the value for determinant is zero
So, the lines are coplanar in nature.
Therefore, the correct option is (B).
Additional information:
Some real-world examples of coplanar lines are
The lines that are drawn on a notebook are coplanar to each other.
The grid that is drawn on a graphing paper is also coplanar.
The hands of our clocks or watches (the second, minute, and hour hands) are also coplanar to each other.
Note:
Remember that the coplanar lines are the lines that lie on the same plane. Two lines in a three-dimensional plane are said to be coplanar if there is a plane that includes them both. This condition occurs if the lines intersect with each other or if they are parallel to each other. The two lines that are not coplanar are commonly known as skew lines.
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