Family of Planes Passing through the Intersection of Two Planes
Let p1 : x - ...
Question
Let p1:x−2y+3z=5 and p2:2x+3y+z+4=0 be two planes. If P is the foot of the perpendicular dropped from the origin O to the line of intersection of the planes, then
A
−−→OP⋅((^i−2^j+3^k)×(2^i+3^j+^k))=0
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B
if Q is another point on the line of intersection of planes, then equation of the plane containing a triangle OPQ is 14x+7y+17z=0.
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C
equation of the plane perpendicular to the line of intersection of the planes and passing through (1,1,1) is 11x−5y−7z+1=0.
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D
If N1 and N2 are foot of the perpendicular from the origin O to the planes p1 and p2 respectively, then ON1+ON2 is equal to 9√14.
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Solution
The correct option is D If N1 and N2 are foot of the perpendicular from the origin O to the planes p1 and p2 respectively, then ON1+ON2 is equal to 9√14. Let →r=x^i+y^j+z^k p1≡→r⋅(^i−2^j+3^k)=5p2≡→r⋅(2^i+3^j+^k)=−4
DR's of intersection line (L) of p1 and p2 is ∣∣
∣
∣∣^i^j^k1−23231∣∣
∣
∣∣=−11^i+5^j+7^k
Now, OP is ⊥ to L
and ^i−2^j+3^k,2^i+3^j+^k are also perpendicular to L.
This means −−→OP⋅((^i−2^j+3^k)×(2^i+3^j+^k))=0
Equation of planes passing through intersection of p1 and p2 is p1+λp2=0(x−2y+3z−5)+λ(2x+3y+z+4)=0x(1+2λ)+y(3λ−2)+z(3+λ)−5+4λ=0⋯(1)
Put (0,0,0) 0+0+0+4λ−5=0⇒λ=54
Putting this value in equation (1), x(1+2(54))+y(3(54)−2)+z(3+54)+0=0⇒14x+7y+17z=0
Vector along the line of intersection of planes is ∣∣
∣
∣∣^i^j^k1−23231∣∣
∣
∣∣=−11^i+5^j+7^k