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The dr. of normal to the plane through (...

The dr. of normal to the plane through `(1,0,0), (0,1,0)` which makes an angle `pi/4` with plane , `x+y=3` are

A

`lt1,sqrt2,1gt`

B

`lt1,1,sqrt2gt`

C

`lt1,1,2gt`

D

`ltsqrt2,1,1gt`

Text Solution

Verified by Experts

The correct Answer is:
b
Any plane through (1,0,0) is
`a(x-1)+by+cz=0" "(i)`
It passes through (0,1,0). Therefore,
`a(0-1)+b+(1)+c(0)=0or-a+b=0" "(ii)`
(i) make an angle of `(pi)/(4)` with `x+y=3`, therefore
`cos""(pi)/(4)=(a(1)+b(1)c+(0))/(sqrt(a^(2)+b^(2)+c^(2))sqrt(01+1+0))`
or `(1)/(sqrt2)=(a+b)/(sqrt2sqrt(a^(2)+b^(2)+c^(2)))`
or `a+b=sqrt(a^(2)+b^(2)+c^(2))`
Squaring, we get
`a^(2)b^(2)+2ab=a^(2)+b^(2)+c^(2)`
or `2ab=c^(2)or2a^(2)=c^(2)" "["using"(ii)]`
or `c=sqrt2a`
Hence, `a:b:c=a:a:sqrt2a`
`=1:1:sqrt2`
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Knowledge Check

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