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Question: A perpendicular is drawn from a point \[P(1,6,3)\] on the line \(\dfrac{x}{1} = \dfrac{{y - 1}}{2} =...

A perpendicular is drawn from a point P(1,6,3)P(1,6,3) on the line x1=y−12=z−23\dfrac{x}{1} = \dfrac{{y - 1}}{2} = \dfrac{{z - 2}}{3}. Find the coordinates of the foot and length of perpendicular?

Explanation

Solution

The equation of the given line which is given to us is its Cartesian equation. Let us consider O to be the foot of the perpendicular. Let x1=y−12=z−23=k\dfrac{x}{1} = \dfrac{{y - 1}}{2} = \dfrac{{z - 2}}{3} = k. Use this particular equation to compute the direction ratio of the perpendicular which is drawn.
To solve this question, use the direction ratios of the given line and the fact that its dot product with the direction ratio of the perpendicular will give the value as 0 to find the coordinates of O. Then, use the distance formula.
d=(x2−x1)2+(y2−y1)2+(z2−z1)2d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2} + {{({z_2} - {z_1})}^2}} to find the perpendicular distance between the point O and P(1,6,3)P(1,6,3).

Complete step by step answer:
Given an equation of a linex1=y−12=z−23\dfrac{x}{1} = \dfrac{{y - 1}}{2} = \dfrac{{z - 2}}{3} and a pointP(1,6,3)P(1,6,3)
Let us look at the diagrammatic representation of this problem.
Let P(1,6,3)P(1,6,3) and AB be the line which is represented by the equation x1=y−12=z−23\dfrac{x}{1} = \dfrac{{y - 1}}{2} = \dfrac{{z - 2}}{3}
Also, let the point O be the foot of the perpendicular which is drawn from point P to the line AB.
Thus, the figure we get is:

There are two parts which we need to find in the given question.

  1. The Coordinates of foot of the perpendicular and
  2. The perpendicular distance
    The equation given to us is x1=y−12=z−23\dfrac{x}{1} = \dfrac{{y - 1}}{2} = \dfrac{{z - 2}}{3}. This is basically the Cartesian equation of the line passing through the point (0,1,2)(0,1,2) and the direction ratios as 1,2,31,2,3.
    Let us consider O≡(x,y,z)O \equiv (x,y,z).
    We will use the above Cartesian equation in order to find out the coordinates of the point O.
    Let x1=y−12=z−23=k\dfrac{x}{1} = \dfrac{{y - 1}}{2} = \dfrac{{z - 2}}{3} = k
    Hence, for any point (x,y,z)(x,y,z) on the line AB:
    x=kx = k
    y=2k+1y = 2k + 1
    z=3k+2z = 3k + 2
    Therefore, we have O(k,2k+1,3k+2)O(k,2k + 1,3k + 2).
    Now, let us consider the direction ratios of the two points O(k,2k+1,3k+2)O(k,2k + 1,3k + 2) and P(1,6,3)P(1,6,3). The direction ratios of these two points are OP‾≡(k−1,2k−5,3k−1)\overline {OP} \equiv (k - 1,2k - 5,3k - 1)
    Also, we are given that the direction ratios of x1=y−12=z−23\dfrac{x}{1} = \dfrac{{y - 1}}{2} = \dfrac{{z - 2}}{3} is given by.
    As OP is perpendicular to the line AB, we know that OP‾⋅AB‾=0\overline {OP} \cdot \overline {AB} = 0
    Therefore, (k−1,2k−5,3k−1)⋅(1,2,3)=0(k - 1,2k - 5,3k - 1) \cdot (1,2,3) = 0
    ⇒(k−1)1+(2k−5)2+(3k−1)3=0\Rightarrow (k - 1)1 + (2k - 5)2 + (3k - 1)3 = 0
    ⇒k−1+4k−10+9k−3=0\Rightarrow k - 1 + 4k - 10 + 9k - 3 = 0
    14k=1414k = 14
    k=1k = 1
    On putting the value in O, we get,
    O(1,3,5)O(1,3,5)
    Therefore, O(1,3,5)O(1,3,5) are the coordinates of the foot of the perpendicular.
    To answer the second part, we will only need the distance formula for two points.
    Let the length of OP be equal to ‘d’. So,
    d=(1−1)2+(3−6)2+(5−3)2d = \sqrt {{{(1 - 1)}^2} + {{(3 - 6)}^2} + {{(5 - 3)}^2}}
    d=(0)2+(−3)2+(2)2d = \sqrt {{{(0)}^2} + {{( - 3)}^2} + {{(2)}^2}}
    d=9+4d = \sqrt {9 + 4}
    d=13d = \sqrt {13}
    Therefore, the length of the perpendicular is d=13d = \sqrt {13} .

Note:
Students often tend to make substitution errors and in the end they get wrong answers. In this case, the point O has 3 coordinates, therefore, we have modified the distance formula in a way that we are able to compute the distance between 2 points which have 3 coordinates.