Question

The distance of the point (1, -5, 9), from the planar $\left( {\hat i - \hat j + \hat k} \right) = 5$ measured along the line $r = \hat i + \hat j + \hat k$ is

Answer 1

Hint: To solve this question, we will use the concept of finding the equation of a line parallel to a given line and passing through a given point and also, we will use the distance formula.

Complete step-by-step answer:
Given that,
Equation of plane = $\left( {\hat i - \hat j + \hat k} \right) = 5$
Equation of line = $r = \hat i + \hat j + \hat k$
Given passing point = (1, -5, 9).
First, we will convert these vectors form into Cartesian forms,
So,
Equation of plane in cartesian form will be,
$ \Rightarrow x - y + z = 5$ ……. (i)
Equation of line in cartesian form will be,
$ \Rightarrow x = y = z$ ……… (ii)
According to the question,
The line through which the point (1, -5, 9) is passing, is parallel to the line x = y = z.
As we know that, the equation of a line parallel to a given line and passing through a given point is given by,
$ \Rightarrow \dfrac{{x - {x_1}}}{a} = \dfrac{{y - {y_1}}}{b} = \dfrac{{z - {z_1}}}{c}$,
Where a, b and c are the direction cosines of the given line and $\left( {{x_1},{y_1},{z_1}} \right)$ is the given point.
So, the equation of the line parallel to the line $x = y = z$ and passing through the point (1, -5, 9) will be,
$ \Rightarrow \dfrac{{x - 1}}{1} = \dfrac{{y + 5}}{1} = \dfrac{{z - 9}}{1} = \lambda $ [$\lambda $ be any scalar value]
Solving this, we will get
$ \Rightarrow x = \lambda + 1,y = \lambda - 5,z = \lambda + 9$
As we know that, this line is intersecting the given plane, which means this equation must satisfy the equation of plane.
Put these values of x, y and z in equation (i),
$ \Rightarrow \lambda + 1 - \left( {\lambda - 5} \right) + \lambda + 9 = 5$
$ \Rightarrow \lambda + 1 - \lambda + 5 + \lambda + 9 = 5$
$ \Rightarrow \lambda + 10 = 0$
$ \Rightarrow \lambda = - 10$
Now, putting this value in the values of x, y and z, we will get
$ \Rightarrow x = - 10 + 1 = - 9,$
$ \Rightarrow y = - 10 - 5 = - 15,$
$ \Rightarrow z = - 10 + 9 = - 1$
Thus, the point on the plane is (-9, -15, -1).
Now, we will find the distance between the points (1, -5, 9) and (-9, -15, -1) by using the distance formula,
$ \Rightarrow d = \sqrt {{{\left( { - 9 - 1} \right)}^2} + {{\left( { - 15 + 5} \right)}^2} + {{\left( { - 1 - 9} \right)}^2}} $
$ \Rightarrow d = \sqrt {{{\left( { - 10} \right)}^2} + {{\left( { - 10} \right)}^2} + {{\left( { - 10} \right)}^2}} $
$ \Rightarrow d = \sqrt {100 + 100 + 100} $
$ \Rightarrow d = \sqrt {300} $
$ \Rightarrow d = 10\sqrt 3 $ units.
Hence, the distance of the point (1, -5, 9), from the planar $\left( {\hat i - \hat j + \hat k} \right) = 5$ measured along the line $r = \hat i + \hat j + \hat k$ is $10\sqrt 3 $ units.

Note: Whenever we ask such type of question, we also have to remember that the distance of the point $\left( {{x_1},{y_1},{z_1}} \right)$ from the plane $ax + by + cz + d = 0$ is given by $\dfrac{{\left| {a{x_1} + b{y_1} + c{z_1} + d} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$ and this is also called the foot of the perpendicular from $\left( {{x_1},{y_1},{z_1}} \right)$ to the plane $ax + by + cz + d = 0$