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Question

Let a plane Pcontain two lines r=i^+λ(i^+j^), λR and r=j^+μ(j^k^), μR. If Q(α,β,γ) is the foot of the perpendicular drawn from the point M(1,0,1) to P, then 3(α+β+γ) equals:


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Solution

Step 1: Finding the unit vector perpendicular to the plane

Given that the plane containing the vectors

r=i^+λ(i^+j^)

And r=j^+μ(j^k^)

Considering the unit vector be perpendicular to the plane be n^

n=i^j^k^110011

n=-i^+j^+k^

Therefore, the equation of plane containing the point M(1,0,1) and normal n^

1(x-1)+1(y-0)+1(z-0)=0x-y-z-1=0...(i)

Step 2: Finding the value of 3(α+β+γ)

Foot of the perpendicular drawn from the point M(1,0,1) to the plane (i)

(x-1)1=(y-0)-1=(z-1)-1=-(1-0-1-1)3(x-1)=13,y-1=13,(z-1)-1=13x=43,y=-13,z=23

As given Q(α,β,γ)lies on foot of perpendicular

α=43,β=-13,γ=23

α+β+γ=43-13+23α+β+γ=533(α+β+γ)=5

Hence, the 3(α+β+γ) is equal to 5


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