Q.
The vector equation of the plane passing through the intersection of the planes r→.i^+j^+k^=1 and r→.i^−2j^=−2 and the point 1,0,2 is
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a
r→.i^+7j^+3k^=73
b
r→.3i^+7j^+3k^=7
c
r→.i^−7j^+3k^=73
d
r→.i^+7j^+3k^=7
answer is D.
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Detailed Solution
The equation of the plane passing through the line of intersection of two planes x+y+z=1and x−2y+2=0 is x+y+z−1+λx−2y+2=0 It is passing through the point 1,0,2 Hence, 2+λ3=0=λ=−23 Therefore, the equation of the required plane is x+7y+3z−7=0 , this can be written as r→⋅i+7j+3k=7
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