Find the equation of the plane through the intersection of the planes- r-3-k-9 and r- 2-k-3 and passing through the origin-

Plane P_1: r⃗.(î+3ĵ k̂)=9 Cartesian equation of plane P1: x+3y z 9=0 Plane P_2: r⃗.(2î ĵ+k̂)=3 Cartesian equation of plane P_2: 2x y+z 3=0 E ...

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Byju's Answer

Standard XII

Mathematics

Family of Planes Passing through the Intersection of Two Planes

Find the equa...

Question

Find the equation of the plane through the intersection of the planes.
$\to r \cdot (^i + 3^j -^k) = 9$ and $\to r \cdot (2^i -^j +^k) = 3$ and passing through the origin.

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Solution

Plane $P_{1} : \to r . (^i + 3^j -^k) = 9$
Cartesian equation of plane $P 1 : x + 3 y - z - 9 = 0$
Plane $P_{2} : \to r . (2^i -^j +^k) = 3$
Cartesian equation of plane $P_{2} : 2 x - y + z - 3 = 0$
Equation of plane passing through the intersection of two planes is:
$P : P_{1} + k (P_{2}) = 0 \Rightarrow x + 3 y - z - 9 + k (2 x - y + z - 3) = 0....... (1)$
$P$ passes through origin
$∴$ put $(0, 0, 0)$ in the eq(1),
$- 9 + k (- 3) = 0 \Rightarrow k = - 3$
$∴ P : x + 3 y - z - 9 + (- 3) (2 x - y + z - 3) = 0 \Rightarrow - 5 x + 6 y - 4 z = 0 \Rightarrow 5 x - 6 y + 4 z = 0$

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Find the equation of the plane through the intersection of the planes.→r⋅(^i+3^j−^k)=9 and →r⋅(2^i−^j+^k)=3 and passing through the origin.

Find the equation of the plane through the intersection of the planes.
$\to r \cdot (^i + 3^j -^k) = 9$ and $\to r \cdot (2^i -^j +^k) = 3$ and passing through the origin.