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Question

Show that the equation of the plane through the line x13=y+64=z+12 and parallel to x22=y13=z+45

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Solution

Consider the problem

Since, required plane is passes through

x13=y+64=z+12...(1)

So, it passes through point (1,6,1)

Since, it is passes through above line (1) and parallel to

x22=y13=z+45

So, normal to the plane is perpendicular to vectors

a=3^i+4^j+2^kandb=2^i+3^j+5^k

Equation of plane passing through (1,6,1) is

A(x1)+B(y+6)+C(z+1)=0...(1)

since, it is to vectors aandb

So,

3A+4B+2C=0...(2)2A+3B+5C=0.....(3)

Now, Solving (2) and (3) we get

A206=B415=C98A14=B11=C1=k

A=14k,B=11k,C=k

Now putting values of A, B and C in (1) we get,

14k(x1)+(11k)(y+6)+k(z+1)=014(x1)11(y+6)+z+1=014x1411y66+z+1=014x11y+z79=0


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