Perpendiculars are drawn from points on the line (x + 2)/2 = (y + 1)/(- 1) = z/3 to the plane x+y+z=3. The feet of perpendiculars lie on the line

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Correct answer: (D) x/2 = (y - 1)/(- 7) = (z - 2)/5. The equation od given line is (x + 2)/2 = (y + 1)/(- 1) = z/3 = λ Now any two points on this line are A-2,-1,0,B0,-2,3 forλ=0, 1 Let foot of perpendicular from A(-2,-1,0) on plane is α , β , γ ⇒(α+2)/1=(β+1)/1=(γ-0)/1=μsay Also the point lies on the plane, hence α + β + γ = 3 ⇒μ-2+μ-1+μ=3⇒μ=2 Hence the coordinates of M are ⇒M0, 1, 2 Similarly foot of perpendicular from B(0, -2, 3) on plane is N 2/3 , (- 4)/3 , (1 1)/3 So the required equation of MN is (x-0)/23=(y-1)/(-73)=(z-2)/53 ⇒(x-0)/2=(y-1)/(-7)=(z-2)/5

Perpendiculars are drawn from points on the line $\frac{x + 2}{2} = \frac{y + 1}{- 1} = \frac{z}{3}$ to the plane $x + y + z = 3$ . The feet of perpendiculars lie on the line

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Relevant Questions

Perpendiculars are drawn from points on the line x + 2 2 = y + 1 - 1 = z 3 to the plane x+y+z=3. The feet of perpendiculars lie on the line

Solution

Relevant Questions

Perpendiculars are drawn from points on the line $\frac{x + 2}{2} = \frac{y + 1}{- 1} = \frac{z}{3}$ to the plane $x + y + z = 3$ . The feet of perpendiculars lie on the line