Question

Find the vector equation of a line passing through the point (-1, -1, 2) and parallel to the line 2x - 2 = 3y + 1 = 6z - 2.

Answer 1

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Answer 2

Answer:

Vector equation of line is given by

$\vec r =-\hat i-\hat j+2\hat k+\lambda(\frac{1}{2}\hat i+\frac{1}{3}\hat j+\frac{1}{6}\hat k)$

Step-by-step explanation:

To find the vector equation of a line passing through the point  (-1, -1, 2) and parallel to the line 2x - 2 = 3y + 1 = 6z - 2.

As we know that Eq of a line passing through a point(x1,y1,z1) is given by

position vector $\vec a= -\hat i-\hat j+2\hat k$

vector equation is given by $\vec r=\vec\:a+\lambda\:\vec b$

to find the vector b we must find direction ratio,as direction ratio's of parallel lines are in proportion

thus

$2x-2=3y+1=6z-2\\\\\frac{x-1}{1/2}=\frac{y+1/3}{1/3}=\frac{z-1/3}{1/6}$

so direction ratio's of line are (1/2,1/3,1/6)

$\vec b=\frac{1}{2}\hat i+\frac{1}{3}\hat j+\frac{1}{6}\hat k$

So vector equation is given by

$\vec r =\vec\:a+\lambda\:\vec b\\\\\vec r =-\hat i-\hat j+2\hat k+\lambda(\frac{1}{2}\hat i+\frac{1}{3}\hat j+\frac{1}{6}\hat k)$