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Question

Find the equation of the plane containing the point (2,4.1) and perpendicular to each of the plane 3x4y+2z7=0 and x+2y3z+2=0.

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Solution

Given planes are :
3x4y+2z7=0 and
x+2y3z+2=0
normal of the planes are
¯n1=3^i4^j+2^k and
n2=^i+2^j3^k
The plane is perpendicular to both these planes.
its normal will also be perpendicular to both the normal.
i.e. n=n1×n2=∣ ∣ ∣^i^j^k342123∣ ∣ ∣
=(124)^i+(11)(^j)+10^k
=8^i+11^j+10^k
Equation of the plane is given by 8x+11y+10z=k
it pass through the point (2,4,1)
8(2)+11(4)+10(1)=k
k=16+4410=6010=50
plane is 8x+11y+10z=50

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