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Miscellaneous Exercise 6A · Q39

Q.Find the vector and Cartesian equations of the line passing through the point (−1,−1,2)(-1, -1, 2) and parallel to the line 2x−2=3y+1=6z−22x - 2 = 3y + 1 = 6z - 2.

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Rewrite 2x−2=3y+1=6z−22x-2=3y+1=6z-2 in standard symmetric form by factoring each expression:

2(x−1)=3(y+13)=6(z−13)2(x-1)=3\left(y+\dfrac13\right)=6\left(z-\dfrac13\right)

x−11/2=y+1/31/3=z−1/31/6\dfrac{x-1}{1/2}=\dfrac{y+1/3}{1/3}=\dfrac{z-1/3}{1/6}

Multiplying all three denominators by 66 (which does not change the line, since scaling all direction ratios by the same nonzero constant gives an equivalent direction), the direction ratios are

(6⋅12, 6⋅13, 6⋅16)=(3,2,1).\left(6\cdot\dfrac12,\ 6\cdot\dfrac13,\ 6\cdot\dfrac16\right)=(3,2,1).

The required line is parallel to this, i.e. has direction b⃗=3i^+2j^+k^\vec b=3\hat i+2\hat j+\hat k, and passes through (−1,−1,2)(-1,-1,2) (position vector a⃗=−i^−j^+2k^\vec a=-\hat i-\hat j+2\hat k). …

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