# Find the Vector Equation of the Plane that Contains the Lines → R = ( ˆ I + ˆ J ) + λ ( ˆ I + 2 ˆ J − ˆ K ) and the Point (–1, 3, –4). Also, Find the Length of the Perpendicular - Mathematics

#### Question

Sum

Find the equation of the plane passing through the intersection of the planes vecr . (hati + hatj + hatk) and vecr.(2hati + 3hatj - hatk) + 4 = 0 and parallel to the x-axis. Hence, find the distance of the plane from the x-axis.

#### Solution

We have,

vecr.(hati + hatj + hatk) - 1 = 0         ....(1)

vecr.(2hati + 3hatj - hatk) + 4 = 0     .....(2)

Equation of plane passing through the intersection of the planes (i) and (ii), is given by

[vecr.(hati + hatj + hatk) - 1 ] = λ[vecr. (2hati + 3hatj - hatk) + 4 ] = 0

⇒ vecr.[(1 + 2λ)hati + (1 + 3λ)hatj + (1 - λ)hatk] - 1 + 4λ = 0        ....(iii)

If plane (iii) is parallel to x-axis, then

1 + 2λ = 0 ⇒ λ = -1/2

Therefore, the equation of the required plane is vecr.(-hatj + 3hatk) = 6

Distance of the plane vecr.(-hatj + 3hatk) = 6 from x-axis is given by,

d = ((1 xx 0 - 3 xx 0 + 6)/(sqrt(1^2 + 3^2))) = 6/sqrt10

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