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Question: The distance of the point (1, -5, 9), from the planar \(\left( {\hat i - \hat j + \hat k} \right) = ...

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

Explanation

Solution

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 = (i^−j^+k^)=5\left( {\hat i - \hat j + \hat k} \right) = 5
Equation of line = r=i^+j^+k^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,
⇒x−y+z=5\Rightarrow x - y + z = 5 ……. (i)
Equation of line in cartesian form will be,
⇒x=y=z\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,
⇒x−x1a=y−y1b=z−z1c\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 (x1,y1,z1)\left( {{x_1},{y_1},{z_1}} \right) is the given point.
So, the equation of the line parallel to the line x=y=zx = y = z and passing through the point (1, -5, 9) will be,
⇒x−11=y+51=z−91=λ\Rightarrow \dfrac{{x - 1}}{1} = \dfrac{{y + 5}}{1} = \dfrac{{z - 9}}{1} = \lambda [λ\lambda be any scalar value]
Solving this, we will get
⇒x=λ+1,y=λ−5,z=λ+9\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),
⇒λ+1−(λ−5)+λ+9=5\Rightarrow \lambda + 1 - \left( {\lambda - 5} \right) + \lambda + 9 = 5
⇒λ+1−λ+5+λ+9=5\Rightarrow \lambda + 1 - \lambda + 5 + \lambda + 9 = 5
⇒λ+10=0\Rightarrow \lambda + 10 = 0
⇒λ=−10\Rightarrow \lambda = - 10
Now, putting this value in the values of x, y and z, we will get
⇒x=−10+1=−9,\Rightarrow x = - 10 + 1 = - 9,
⇒y=−10−5=−15,\Rightarrow y = - 10 - 5 = - 15,
⇒z=−10+9=−1\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,
⇒d=(−9−1)2+(−15+5)2+(−1−9)2\Rightarrow d = \sqrt {{{\left( { - 9 - 1} \right)}^2} + {{\left( { - 15 + 5} \right)}^2} + {{\left( { - 1 - 9} \right)}^2}}
⇒d=(−10)2+(−10)2+(−10)2\Rightarrow d = \sqrt {{{\left( { - 10} \right)}^2} + {{\left( { - 10} \right)}^2} + {{\left( { - 10} \right)}^2}}
⇒d=100+100+100\Rightarrow d = \sqrt {100 + 100 + 100}
⇒d=300\Rightarrow d = \sqrt {300}
⇒d=103\Rightarrow d = 10\sqrt 3 units.
Hence, the distance of the point (1, -5, 9), from the planar (i^−j^+k^)=5\left( {\hat i - \hat j + \hat k} \right) = 5 measured along the line r=i^+j^+k^r = \hat i + \hat j + \hat k is 10310\sqrt 3 units.

Note: Whenever we ask such type of question, we also have to remember that the distance of the point (x1,y1,z1)\left( {{x_1},{y_1},{z_1}} \right) from the plane ax+by+cz+d=0ax + by + cz + d = 0 is given by ∣ax1+by1+cz1+d∣a2+b2+c2\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 (x1,y1,z1)\left( {{x_1},{y_1},{z_1}} \right) to the plane ax+by+cz+d=0ax + by + cz + d = 0

           ![](https://www.vedantu.com/question-sets/fe05ffe8-df5b-47df-a9e2-833dbe475e3e3120752638044210679.png)