InterviewSolution
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InterviewSolution
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The vector equation of the line passing through the point (-1, -1 ,2) and parallel to the line 2x - 2= 3y +1 = 6z -2 is |
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Answer» Equation of the line is `2x-2=3y+1=6z-2` Dividing throughout by 6, we get `(2(x-))/(6)=(3(y+(1)/(3)))/(6)=(6(z-(2)/(6)))/(6)` `:.(x-1)/(3)=(y+(1)/(3))/(2)=(z-(1)/(3))/(1)` `:.` Direction ratio of the line are 3, 2, 1. Let `vec(b)` be the vector parallel to this line `:.vec(b)=-hati+2hatj+hatk" "......(i)` Let `vec(a)` be the position vector of the point (-1,-1,2) `:.vec(a)=-hati-hatj+2hatk" "......(ii)` The vector equation of a line passing through the piont (-1, -1, 2) with position vector `vec(a)` and parallel to `vec(b)" is "vec(r)=vec(a)+lamdavec(b)` `:.` Vector equation of the required line is `vec(r)=(-hati-hatj+2hatk)+lamda(3hatj+2hatj+hatk)` .......(using (i) and (ii) |
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