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Question: Perpendiculars are drawn from points on the line \[\dfrac{{x + 2}}{2} = \dfrac{{y + 1}}{{ - 1}} = \d...

Perpendiculars are drawn from points on the line x+22=y+1−1=z3\dfrac{{x + 2}}{2} = \dfrac{{y + 1}}{{ - 1}} = \dfrac{z}{3} to the plane x + y + z = 3. The feet of the perpendiculars lie on the line is
(a)x5=y−18=z−2−13\left( a \right)\dfrac{x}{5} = \dfrac{{y - 1}}{8} = \dfrac{{z - 2}}{{ - 13}}
(b)x2=y−13=z−2−5\left( b \right)\dfrac{x}{2} = \dfrac{{y - 1}}{3} = \dfrac{{z - 2}}{{ - 5}}
(c)x4=y−13=z−2−7\left( c \right)\dfrac{x}{4} = \dfrac{{y - 1}}{3} = \dfrac{{z - 2}}{{ - 7}}
(d)x2=y−1−7=z−25\left( d \right)\dfrac{x}{2} = \dfrac{{y - 1}}{{ - 7}} = \dfrac{{z - 2}}{5}

Explanation

Solution

In this particular question use the concept that the points from which the perpendicular are drawn satisfying the equation of given line, given plane and one of the given options so first find out that point by equating the line to constant value and solve for x, y and z and then satisfies these points in the equation of plane so use these concepts to reach the solution of the question.

Complete step-by-step answer :
Given data:
Equation of line, x+22=y+1−1=z3\dfrac{{x + 2}}{2} = \dfrac{{y + 1}}{{ - 1}} = \dfrac{z}{3} and equation of plane, x + y + z = 3.............. (1)
Let, x+22=y+1−1=z3=C\dfrac{{x + 2}}{2} = \dfrac{{y + 1}}{{ - 1}} = \dfrac{z}{3} = C, where C is constant value.
⇒x+22=C\Rightarrow \dfrac{{x + 2}}{2} = C
⇒x=2C−2\Rightarrow x = 2C - 2
And
⇒y+1−1=C\Rightarrow \dfrac{{y + 1}}{{ - 1}} = C
⇒y=−C−1\Rightarrow y = - C - 1
And
⇒z3=C\Rightarrow \dfrac{z}{3} = C
⇒z=3C\Rightarrow z = 3C
So the points satisfies the equation of line is (x, y, z) = (2C – 2, -C – 1, 3C)
Now satisfy the above point in the equation of plane we have,
⇒(2C−2)+(−C−1)+3C=3\Rightarrow \left( {2C - 2} \right) + \left( { - C - 1} \right) + 3C = 3
Now simplify we have,
⇒4C=6\Rightarrow 4C = 6
⇒C=32\Rightarrow C = \dfrac{3}{2}.
So the points becomes,
⇒x=2C−2=2(32)−2=3−2=1\Rightarrow x = 2C - 2 = 2\left( {\dfrac{3}{2}} \right) - 2 = 3 - 2 = 1
⇒y=−C−1=−32−1=−52\Rightarrow y = - C - 1 = - \dfrac{3}{2} - 1 = \dfrac{{ - 5}}{2}
⇒z=3C=3(32)=92\Rightarrow z = 3C = 3\left( {\dfrac{3}{2}} \right) = \dfrac{9}{2}
Therefore, (x, y, z) = (1,−52,92)\left( {1,\dfrac{{ - 5}}{2},\dfrac{9}{2}} \right)
Now these points satisfies one of the given options
Option (a)x5=y−18=z−2−13\left( a \right)\dfrac{x}{5} = \dfrac{{y - 1}}{8} = \dfrac{{z - 2}}{{ - 13}}
Now satisfies the points we have,
(a)15=−52−18=92−2−13\left( a \right)\dfrac{1}{5} = \dfrac{{\dfrac{{ - 5}}{2} - 1}}{8} = \dfrac{{\dfrac{9}{2} - 2}}{{ - 13}}
(a)15=−716=−526\left( a \right)\dfrac{1}{5} = \dfrac{{ - 7}}{{16}} = \dfrac{{ - 5}}{{26}}
So option (a) is not correct.
Option (b)x2=y−13=z−2−5\left( b \right)\dfrac{x}{2} = \dfrac{{y - 1}}{3} = \dfrac{{z - 2}}{{ - 5}}
Now satisfies the points we have,
⇒12=−52−13=92−2−5\Rightarrow \dfrac{1}{2} = \dfrac{{\dfrac{{ - 5}}{2} - 1}}{3} = \dfrac{{\dfrac{9}{2} - 2}}{{ - 5}}
⇒12=−76=−11\Rightarrow \dfrac{1}{2} = \dfrac{{ - 7}}{6} = \dfrac{{ - 1}}{1}
So option (b) is also not correct.
Option (c)x4=y−13=z−2−7\left( c \right)\dfrac{x}{4} = \dfrac{{y - 1}}{3} = \dfrac{{z - 2}}{{ - 7}}
Now satisfies the points we have,
⇒14=−52−13=92−2−7\Rightarrow \dfrac{1}{4} = \dfrac{{\dfrac{{ - 5}}{2} - 1}}{3} = \dfrac{{\dfrac{9}{2} - 2}}{{ - 7}}
⇒14=−76=−57\Rightarrow \dfrac{1}{4} = \dfrac{{ - 7}}{6} = \dfrac{{ - 5}}{7}
So option (c) is also not correct.
Option (d)x2=y−1−7=z−25\left( d \right)\dfrac{x}{2} = \dfrac{{y - 1}}{{ - 7}} = \dfrac{{z - 2}}{5}
Now satisfies the points we have,
⇒12=−52−1−7=92−25\Rightarrow \dfrac{1}{2} = \dfrac{{\dfrac{{ - 5}}{2} - 1}}{{ - 7}} = \dfrac{{\dfrac{9}{2} - 2}}{5}
⇒12=12=12\Rightarrow \dfrac{1}{2} = \dfrac{1}{2} = \dfrac{1}{2}
Hence option (d) is the correct answer.
So option (d) is the required feet of the perpendiculars lie on the line.

Note : Whenever we face such types of questions the key concept we have to remember is that always recall that the point of intersection of the given two planes satisfies the equation of perpendiculars from which the perpendiculars are drawn so first find out the intersection point as above and then satisfies the points in the options one by one we will get the required answer.