Determine the vector equation of the plane passing through the intersection of the planes and , and the point (1, 1, 1)?

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AAI ATC Junior Executive 25 March 2021 Official Paper (Shift 2)
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Answer (Detailed Solution Below)

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Detailed Solution

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Concept:

As we know that, if a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 represents two different planes, then equation of plane passing through the intersection of these planes is given by:

(a1x + b1y + c1z + d1) + λ × (a2x + b2y + c2z + d2) = 0.

Calculation:

Given:

 and 

Convert these planes in the cartesian form.

Put 

So 

⇒ x + y + z = 6      .........(1)

Similarly,

⇒ 2x + 3y + 4z = -5      .........(2)

So, the plane passing through the intersection of two given planes is:

⇒ (x + y + z – 6) + λ × (2x +3y + 4z + 5) = 0 

∵ it is given that the plane passing through the intersection of two given planes also passes through the point (1, 1, 1)

⇒ The point (3, 2, 1) will satisfy equation (1)

⇒ (1 + 1 + 1 - 6) + λ × (2 + 3 + 4 + 5) = 0 ⇒ λ = 3/14.

So, by substituting the value of λ in equation (1), we get

⇒ (x + y + z – 6) + (3/14) × (2x +3y + 4z + 5) = 0 

20x +23y + 26z = 69

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