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Let P_(1) denote the equation of a plane to which the vector (hati+hatj) is normal and which contains the line whose equation is vecr=hati+hatj+veck+lamda(hati-hatj-hatk)andP_(2) denote the equation of the plane containing the line L and a point with position vector j. Which of the following holds good? |
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Answer» The equation of `P_(1)" is"x+y=2.` `vecr=hati+hatj+hatk+lamda(hati-hatj-hatk)`, HENCE contains the point `hati+hatj+hatk` and is normal to vector `(hati+hatj)`. Hence, equation of plane is `""(vecr-(hati+hatj+hatk))*(hati+hatj)=0` or `""x+y=2` Plane `P_(2)` contains the line `""vecr=hati+hatj=hatk+lamda(hati-hatj-hatk)` and point `hatj` Hence, equation of plane is `|{:(x-0,,y-1,,z-0),(1-0,,1-1,,1-0),(1,,-1,,-1):}|=0` or`""x+2y-z=2` If `theta` is the acute angle between `P_(1) and P_(2)`, then `""costheta= (vec(n_1)*vec(n_2))/(|vec(n_1)||vec(n_2)|)=|((hati+hatj)*(hati+2hatj-hatk))/(SQRT2*sqrt6)|` `""= (3)/(sqrt2*sqrt6)= (sqrt(3))/(2)` `""theta=cos^(-1)""(sqrt(3))/(2)= (pi)/(6)` As `L` is containec in `P_(2) rArr theta=0` |
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