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Question: The coordinates of the foot of the perpendicular drawn from the point \( A\left( 1,0,3 \right) \) to...

The coordinates of the foot of the perpendicular drawn from the point A(1,0,3)A\left( 1,0,3 \right) to the join of the points B(4,7,1)B\left( 4,7,1 \right) and C(3,5,3)C\left( 3,5,3 \right) are:
A. (53,73,173)\left( \dfrac{5}{3},\dfrac{7}{3},\dfrac{17}{3} \right)
B. (5,7,17)\left( 5,7,17 \right)
C. (53,−73,173)\left( \dfrac{5}{3},-\dfrac{7}{3},\dfrac{17}{3} \right)
D. (−53,73,−173)\left( -\dfrac{5}{3},\dfrac{7}{3},-\dfrac{17}{3} \right)

Explanation

Solution

Hint : Find the coordinates of the general point of BC. Now find the directional cosine of AD line, BC line. Use the condition of the perpendicular, by this find variable assumed. From this variable find the value of D. This point D is the required result in the question.

Complete step-by-step answer :
Given condition in the question is written as:
Foot of perpendicular from A to line joining points BC,
The values of coordinates of point A are given (1,0,3)\left( 1,0,3 \right) .
The values of coordinates of point B are given (4,7,1)\left( 4,7,1 \right) .
The values of coordinates of point C are given (3,5,3)\left( 3,5,3 \right) .
The coordinates of general point on line through (a,b,c)(d,e,f)\left( a,b,c \right)\left( d,e,f \right) is given by formula x−aa−d=y−bb−d=z−cc−f=k\dfrac{x-a}{a-d}=\dfrac{y-b}{b-d}=\dfrac{z-c}{c-f}=k
By using above formula on the line BC, we get it is:
x−44−3=y−77−5=z−11−3=k\dfrac{x-4}{4-3}=\dfrac{y-7}{7-5}=\dfrac{z-1}{1-3}=k ………………….. (1)
By taking first term of equation (1), we get it as:
x−41=k\dfrac{x-4}{1}=k
By simplifying we write it as following equation
x=4+kx=4+k .
Similarly, we can write the value of y is the form:
y=2k+7y=2k+7
Similarly, we can write the value of z in the form:
z=−2k+1z=-2k+1 .
By this we get the coordinates of D as:
D=((k+4),2k+7,−2k+1)D=\left( \left( k+4 \right),2k+7,-2k+1 \right)

Direction cosines of AD are proportional to:

(k+4,)−1,(2k+7)−0,(−2k+1)−3=k+3,2k+7,−2k−2\left( k+4, \right)-1,\left( 2k+7 \right)-0,\left( -2k+1 \right)-3=k+3,2k+7,-2k-2
Direction cosines of line BC are proportion to:
4−3,7−5,1−3,=1,2,−24-3,7-5,1-3,=1,2,-2
We know condition of perpendicular of 2 lines of direction cosines as (x,y,z)\left( x,y,z \right) , (m,n,p)\left( m,n,p \right) is xm+yn+zp=0xm+yn+zp=0
By using above equation, we can write the equation as:
(k+3).1+(2k+7).2+(−2k−2)(−2)=0\left( k+3 \right).1+\left( 2k+7 \right).2+\left( -2k-2 \right)\left( -2 \right)=0
By using distributive law in each term, we can write them as:
k+3+4k+14+4k+4=0k+3+4k+14+4k+4=0
By grouping terms and subtracting 21 on both sides, we get:
9k=−219k=-21
By dividing with 9 on both sides, we get it in form of:
k=−73k=-\dfrac{7}{3}
By substituting value of k in point D we get as:
D=((−73+4),(−143+7),(+143+1))=(53,73,173)D=\left( \left( -\dfrac{7}{3}+4 \right),\left( -\dfrac{14}{3}+7 \right),\left( +\dfrac{14}{3}+1 \right) \right)=\left( \dfrac{5}{3},\dfrac{7}{3},\dfrac{17}{3} \right)
Therefore, option (a) is the correct answer for this question.

Note : Be careful while calculating point D in terms of K as it is the most crucial step used in the solution. Be careful with +,−+,- sign as options differ only with sign. There may be a lot of confusion. Calculate direction cosine value correctly because value of k depends on it.