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Question

Find the equation of the plane which is at a distance of 629 from the origin and its normal vector from the origin is 2^i3^j+4^k

A
2x+3y+4z=6
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B
3x4y+2z=6
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C
2x3y+4z=6
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D
2x3y4z=6
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Solution

The correct option is C 2x3y+4z=6
Let
n = 2ˆi3ˆj+4ˆkˆn = 2ˆi3ˆj+4ˆk4+9+16 = 2ˆi3ˆj+4ˆk29
And let r be a position vectorr=xˆi+yˆj+zˆk
Hence, the required equation of the plane is
r.ˆn=629(xˆi+yˆj+zˆk).(2ˆi3ˆj+4ˆk29)=6292x3y+4z29=6292x3y+4z=6
​​​​​​

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