Question

Find the distance of the point $\left( {3,4,5} \right)$ from the plane $x + y + z = 2$ measured parallel to the line $2x = y = z$

Answer 1

Here, we will find the point of intersection of the line in variable by using the given point and the direction ratio in the Cartesian equation. Then we will substitute the point of intersection in place of a variable in the equation of the plane to find the point of intersection. We will then use the distance formula to find the distance of the given point and point of intersection.

Formula Used:
We will use the following formula:

1. The Cartesian equation of line which passes through a point and parallel to the line is given by$\dfrac{{x - {x_1}}}{a} = \dfrac{{y - {y_1}}}{b} = \dfrac{{z - {z_1}}}{c} = \lambda$
2. Distance between the points is given by the formula $d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}}$

Complete step by step solution:
We are given that the plane $x + y + z = 2$ measured parallel to the line $2x = y = z$ .
We are given the equation of line as $2x = y = z$ , now rewriting the equation in the form of Cartesian equation as $\dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c}$ , so, we get
$\Rightarrow$ $\dfrac{x}{{\dfrac{1}{2}}} = \dfrac{y}{1} = \dfrac{z}{1}$
Thus, the direction ratio of the given line is $\left( {\dfrac{1}{2},1,1} \right)$ .
The Cartesian equation of line which passes through a point and parallel to the line is given by $\dfrac{{x - {x_1}}}{a} = \dfrac{{y - {y_1}}}{b} = \dfrac{{z - {z_1}}}{c} = \lambda$
So, equation of Line which passes through the point $\left( {3,4,5} \right)$ and parallel to the line $2x = y = z$ is
$\dfrac{{x - 3}}{{\dfrac{1}{2}}} = \dfrac{{y - 4}}{1} = \dfrac{{z - 5}}{1} = \lambda$
Equation first expression to the last expression, we get
$\dfrac{{x - 3}}{{\dfrac{1}{2}}} = \lambda$
Multiplying both sides by $\dfrac{1}{2}$, we get
$\Rightarrow x - 3 = \dfrac{\lambda }{2}$
Adding 3 on both sides, we get
$\Rightarrow x = \dfrac{\lambda }{2} + 3$
Equating second expression to the last expression, we get
$\dfrac{{y - 4}}{1} = \lambda$
$\Rightarrow y - 4 = \lambda$
Adding 4 on both sides, we get
$\Rightarrow y = \lambda + 4$
Equating third expression to the last expression, we get
$\dfrac{{z - 5}}{1} = \lambda$
$\Rightarrow z - 5 = \lambda$
Adding 4 on both sides, we get
$\Rightarrow z = \lambda + 5$
Thus, the point of intersection is of the form $\left( {\dfrac{\lambda }{2} + 3,\lambda + 4,\lambda + 5} \right)$
We will find the value of $\lambda$.
Now, to , substituting $x = \dfrac{\lambda }{2} + 3$, $y = \lambda + 4$ and $z = \lambda + 5$ in the equation of the plane $x + y + z = 2$, we get
$\dfrac{\lambda }{2} + 3 + \lambda + 4 + \lambda + 5 = 2$
Adding the like terms by taking LCM, we get
$\Rightarrow \dfrac{{\lambda + 2\lambda + 2\lambda }}{2} + 12 = 2$
Subtracting 12 from both the sides, we get
$\Rightarrow \dfrac{{\lambda + 2\lambda + 2\lambda }}{2} = 2 - 12$
$\Rightarrow \dfrac{{5\lambda }}{2} = - 10$
Multiplying both side by 2, we get
$\Rightarrow 5\lambda = - 10 \times 2$
$\Rightarrow 5\lambda = - 20$
Dividing both side by 5, we get
$\Rightarrow \lambda = \dfrac{{ - 20}}{5}$
$\Rightarrow \lambda = - 4$
By substituting $\lambda = - 4$in $\left( {\dfrac{\lambda }{2} + 3,\lambda + 4,\lambda + 5} \right)$ , we get
$\left( {\dfrac{\lambda }{2} + 3,\lambda + 4,\lambda + 5} \right) = \left( {\dfrac{{ - 4}}{2} + 3, - 4 + 4, - 4 + 5} \right)$
$\Rightarrow \left( {\dfrac{\lambda }{2} + 3,\lambda + 4,\lambda + 5} \right) = \left( { - 2 + 3, - 4 + 4, - 4 + 5} \right)$
Adding the terms, we get
$\Rightarrow \left( {\dfrac{\lambda }{2} + 3,\lambda + 4,\lambda + 5} \right) = \left( {1,0,1} \right)$
Thus the point of intersection is $\left( {1,0,1} \right)$ .
Now, we will find the distance between the point of intersection $\left( {1,0,1} \right)$ and the point $\left( {3,4,5} \right)$
Substituting ${x_1} = 1,{x_2} = 3,{y_1} = 0,{y_2} = 4,{z_1} = 1$ and ${z_2} = 5$ in the formula $d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}}$, we get
$\Rightarrow d = \sqrt {{{\left( {3 - 1} \right)}^2} + {{\left( {4 - 0} \right)}^2} + {{\left( {5 - 1} \right)}^2}}$
Subtracting the terms, we get
$\Rightarrow d = \sqrt {{2^2} + {4^2} + {4^2}}$
Applying exponent to the terms, we get
$\Rightarrow d = \sqrt {4 + 16 + 16}$
By adding the numbers, we get
$\Rightarrow d = \sqrt {36}$
$\Rightarrow d = 6units$

Therefore, the distance of the point $\left( {3,4,5} \right)$ from the plane $x + y + z = 2$ measured parallel to the line $2x = y = z$ is $6$ units.

Note:
We know that if the equation of line is parallel to the equation of plane then it can be a vector equation. A vector is an object which has both a magnitude and a direction. The vector equation of a line is used to identify the position vector of every point along the line. Vector equation can be uniquely determined if it passes through a particular point in a specific direction or if it passes through two points. The Cartesian equation is an equation which is represented in three dimensional coordinates. If a line is parallel to the plane, it will be perpendicular to the plane’s normal vector.