Questions
Equation of the plane through the lines and is
a
2x+y−2z+5=0
b
2x+y−2z−5=0
c
x+2y−2z+7=0
d
x+2y+2z−5=0
detailed solution
Correct option is A
The Equation of the plane through the lines x−12=y+1−2=z−31 andx−11=y+12=z−32 is x−x1y−y1z−z1l1m1n1l2m2n2=0Hence the equation of the plane required is x−1y+1z−32−21122=0Simplify x−1−4−2−y+13+z−36=02x−1+y+1−2z−3=02x+y−2z+5=0 Therefore, the equation of the required plane is 2x+y−2z+5=0Talk to our academic expert!
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The line intersects the curve if c=
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