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Lines whose equation are `(x-3)/2=(y-2)/3=(z-1)/(lamda)` and `(x-2)/3=(y-3)/2=(z-2)/3` lie in same plane, then. Angle between the plane containing both lines and the plane `4x+y+2z=0` isA. `(pI)/(3)`B. `(pi)/(2)`C. `(pi)/(6)`D. `cos^(-1)((1)/(sqrt(186)))` |
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Answer» Correct Answer - B normal vector to plane containing lines is `vecn=|{:(hati,hatj,hatk),(2,3,-1),(3,2,3):}|=hati+6hatj+5hatk` So angle between planes `costheta=(4+6-10)/(sqrt(21)sqrt(62))=0impliestheta=(pi)/(2)` |
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