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Find the equation of the plane through the line of intersection of the planes `x+y+z=1 and 2x+3y+4z=5` which is perpendicular to the plane `x-y+z=0`A. `r.(hati-hatk)-2=0`B. `rxx(hati+hatk)+2=0`C. `rxx(hati-hatk)+2=0`D. `r.(hati-hatk)+2=0` |
Answer» Correct Answer - D Since, equation of planes passes through the line of intersection of the planes `x+y+z=1` and `2x+3y+4z=5,` is `(x+y+z-1)+lambda(2x+3y+4z-5)=0` `implies" "(1+2lambda)x+(1+3lambda)y+(1+4lambda)z-(1+5lambda)=0" "...(i)` `because` The plane (i) is perpendicular to the plane `x-y+z=0`. `:.(1+2lambda)-(1+3lambda)+(1+4lambda)=0` [`because` if plane `a_(1)x+b_(1)y+c_(1)z+d_(1)=0` is perpendicular to plane `a_(2)x+b_(2)y+c_(2)z+d_(2)=0," then"a_(1)a_(2)+b_(1)b_(2)+c_(1)c_(2)=0`] `implies" "3lambda+1=0` `implies" "lambda=-(1)/(3)" "...(ii)` So, the equation of required plane, is `(1-(2)/(3))x+(1-(3)/(3))y+(1-(4)/(3))z-(1-(5)/(3))=0` `implies(1)/(3)x-(1)/(3)z+(2)/(3)=0impliesx-z+2=0` Now, vector form is `r.(hati-hatk)+2=0` |
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