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The distance between the line `x=2+t,y=1+t,z=-1/2-t/2` and the plane `vecr.(hati+2hatj+6hatk)=10`, isA. `1/6`B. `1/(sqrt(41))`C. `1/7`D. `9/(sqrt(41))` |
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Answer» Correct Answer - D The equation of the lines is `(x-2)/1=(y-1)/1=(z+1/2)/(-1//2)=t` Clearly, it is parallel to the plane `vecr.(hati+2hatj+6hatk)=10`. So required distance is equal to the length of perpendicular form `(2,1,-1/2)` to the plane `vecr.(hati+2hatj-6hatk)-10=0` and is given by `d=|((2hati+hatj-1/2hatk).(hati+2hatj+6hatk)-10)/(sqrt(1+4+36))|=9/(sqrt(41))` |
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