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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`

  Is there an error in this question or solution?
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