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# Find the image of the line $\dfrac{{x - 1}}{3} = \dfrac{{y - 3}}{1} = \dfrac{{z - 4}}{{ - 1}}$ in the plane $2x - y + z + 3 = 0$.

Last updated date: 20th Jun 2024
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Hint: Here, we will compare the given equation of the line by $\lambda$ and find the values of $x,y,z$ respectively. Then we will substitute these values in the equation of the plane as the line passes through it. This will give us the value of $\lambda$ and hence, we will be able to find the required image of the given line.

Given equation of the line is $\dfrac{{x - 1}}{3} = \dfrac{{y - 3}}{1} = \dfrac{{z - 4}}{{ - 1}}$
And, equation of the given plane is $2x - y + z + 3 = 0$
Now, first of all, we will equate the equation of the line to $\lambda$, hence we get,
$\dfrac{{x - 1}}{2} = \dfrac{{y - 3}}{{ - 1}} = \dfrac{{z - 4}}{1} = \lambda$
Now, we will cross multiply to find the values of $x,y,z$ respectively
Hence,
$x = 2\lambda + 1$
$y = 3 - \lambda$
$z = \lambda + 4$

Now, since the line is passing through the given plane.
Thus, the points $\left( {2\lambda + 1,3 - \lambda ,\lambda + 4} \right)$ represent any point on the plane.
Now, substituting these points in the equation of the plane, we get,
$\Rightarrow 2\left( {2\lambda + 1} \right) - \left( {3 - \lambda } \right) + \left( {\lambda + 4} \right) + 3 = 0$
$\Rightarrow 4\lambda + 2 - 3 + \lambda + \lambda + 4 + 3 = 0$
Solving further, we get,
$\Rightarrow 6\lambda + 6 = 0$
Dividing both sides by 6, we get
$\Rightarrow \lambda + 1 = 0$
Subtracting 1 from both the sides, we get
$\Rightarrow \lambda = - 1$
Hence, substituting these values to find the values of $x,y,z$ respectively, we get,
$x = 2\left( { - 1} \right) + 1 = - 1$
$y = 3 - \left( { - 1} \right) = 4$
$z = - 1 + 4 = 3$
Hence, $x = - 1$, $y = 4$ and $z = 3$

Therefore, the required image of the given line is:
$\dfrac{{x + 1}}{3} = \dfrac{{y - 4}}{1} = \dfrac{{z - 3}}{{ - 1}}$
Hence, this is the required answer.

Note:
A line is an indefinite straight one-dimensional figure which is extending infinitely in both the directions whereas; a line segment is definite which means that it has two endpoints. Also, a plane is a two-dimensional figure consisting of two linear independent vectors. In geometry, the intersection of a line and a plane in three-dimensional space can be the empty set, a point or a line. If the line is embedded in the plane, it is the entire line and if the line is parallel to the plane, it is an empty set and if it cuts the plane then it is a single point.