1.

The equation of line passing through `(3,-1,2)` and perpendicular to the lines `vec(r)=(hat(i)+hat(j)-hat(k))+lamda(2hat(i)-2hat(j)+hat(k))` and`vec(r)=(2hat(i)+hat(j)-3hat(k))+mu(hat(i)-2hat(j)+2hat(k))` isA. `(x+3)/(2)=(y+1)/(3)=(z-2)/(2)`B. `(x-3)/(3)=(y+1)/(2)=(z-2)/(2)`C. `(x-3)/(2)=(y+1)/(3)=(z-2)/(2)`D. `(x-2)/(2)=(y+1)/(2)=(z-2)/(3)`

Answer» Correct Answer - C
Direction ration of the line perpendicular to the lines `vec(r)=vec(a_(1))+lamdavec(b_(1))andvec(r)=vec(a_(2))+muvec(b_(2))` is `alpha(vec(b_(1))xxvec(b_(2)))`
`:.` Direction ratio of line perpendicular to the lines
`vec(r)=(hat(i)+hat(j)-hat(k))+lamda(2hat(i)-2hat(j)+hat(k))`
`andvec(r)=(2hat(i)+hat(j)-3hat(k))+mu(hat(i)-2hat(j)+2hat(k))` is
`alpha|{:(hat(i),hat(j),hat(k)),(2,-2,1),(1,-2,2):}|`
`alpha[(-4+2)hat(i)-(4-1)hat(i)+(-4+2)hat(k)]`
`=alpha[-2hat(i)-3hat(j)-2hat(k)]`
Now, equation of line passing through (3,-1,2) and parallel to `-2hat(i)-3hat(j)-2hat(k)` is
`vec(r)=3hat(i)-hat(j)+2hat(k)+beta(2hat(i)+3hat(j)+2hat(k))`
Hence, cartesian from of the above equation is
`(x-3)/(2)=(y+1)/(3)=(z-2)/(2)`


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