Find the vector equation of the plane passing through the intersection of the planes and through the point (2, 1, 3)

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Solution

It is given that the planes r → ⋅( 2 i ^ +2 j ^ −3 k ^ )=7 and r → ⋅( 2 i ^ +5 j ^ +3 k ^ )=9 passes through the point ( 2,1,3 ).

The equation of plane through the given intersecting planes is,

[ r → ⋅( 2 i ^ +2 j ^ −3 k ^ )−7 ]+λ[ r → ⋅( 2 i ^ +5 j ^ +3 k ^ )−9 ]=0 r → ⋅[ 2 i ^ +2 j ^ −3 k ^ +λ( 2 i ^ +5 j ^ +3 k ^ ) ]=7+9λ (1)

Since the point ( 2,1,3 ) lies on the plane, so the position vector of the point is,

r → =2 i ^ + j ^ +3 k ^

Substitute r → =2 i ^ + j ^ +3 k ^ in equation (1),

( 2 i ^ + j ^ +3 k ^ )⋅[ 2 i ^ +2 j ^ −3 k ^ +λ( 2 i ^ +5 j ^ +3 k ^ ) ]=7+9λ ( 2 i ^ + j ^ +3 k ^ )⋅( ( 2+2λ ) i ^ +( 2+5λ ) j ^ +( −3+3λ ) k ^ )=7+9λ 2( 2+2λ )+1( 2+5λ )+3( −3+3λ )=7+9λ 4+4λ+2+5λ−9+9λ=7+9λ

Solve further,

−3+18λ=7+9λ 9λ=10 λ= 10 9

Substitute λ= 10 9 in equation (1),

r → ⋅( 2 i ^ +2 j ^ −3 k ^ +( 10 9 )( 2 i ^ +5 j ^ +3 k ^ ) )=7+9( 10 9 ) r → ⋅( 2 i ^ +2 j ^ −3 k ^ +( 20 9 i ^ + 50 9 j ^ + 10 3 k ^ ) )=7+10 r → ⋅( 18 i ^ +18 j ^ −27 k ^ +20 i ^ +50 j ^ +30 k ^ 9 )=17 r → ⋅( 38 i ^ +68 j ^ +3 k ^ )=153

Therefore, the equation of plane passing through the given planes and the point is

r → ⋅( 38 i ^ +68 j ^ +3 k ^ )=153.

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\to r, (2^i + 2^j - 3^k) = 7, \to r (2^i + 5^j + 3^k) = 9

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\vec{r} \cdot (2 \hat{i} + \hat{j} + 3 \hat{k}) = 7, \vec{r} \cdot (2 \hat{i} + 5 \hat{j} + 3 \hat{k}) = 9

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