The direction ratios of line 3x+1 =6y-2=1-z are ……

Find the direction cosines of the sides of the triangle whose vertices are (3, 5, –4), (–1, 1, 2) and (–5, –5, –2).

Let A(3, 5, – 4), B(–1, 1, 2), C(–5, –5, –2) be the vertices of ΔABC.

Direction ratios of AB are – 1 – 3, 1 – 5, 2 + 4 i.e. – 4, – 4, 6
          Dividing each by 

The direction ratios of line 3x+1 =6y-2=1-z are ……

The direction ratios of line 3x+1 =6y-2=1-z are ……

cosines of the line AB as 
The direction ratios of line 3x+1 =6y-2=1-z are ……

    i.e.       
The direction ratios of line 3x+1 =6y-2=1-z are ……

The direction ratios of line 3x+1 =6y-2=1-z are ……

Direction ratios of BC are – 5 + 1, –5 –1, –2 –2 i.e. – 4, –6, –4.


               Dividing each by
The direction ratios of line 3x+1 =6y-2=1-z are ……
     
           direction ratios of the line BC as 
The direction ratios of line 3x+1 =6y-2=1-z are ……
  Direction ratios of CA are 3+5, 5+5,  -4+2 i.e., 8, 10 -2.

  Dividing each by 

The direction ratios of line 3x+1 =6y-2=1-z are ……


direction ratios of the line CA as 
The direction ratios of line 3x+1 =6y-2=1-z are ……

Choose correct alternatives :

The direction ratios of the line 3x + 1 = 6y – 2 = 1 – z are ______ 

  • 2, 1, 6

  • 2, 1, – 6

  • 2, – 1, 6

  • – 2, 1, 6

The direction ratios of the line 3x + 1 = 6y – 2 = 1 – z are 2, 1, – 6

Concept: Distance Between Skew Lines and Parallel Lines

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Given equations of the line are:3x -1 = 6y +2 = 1 - z

Rewriting the above equation, we have,

`3(x-1/3)=6(y+2/6)=-(z-1)`

`((x-1/3))/(1/3)=((y+1/3))/(1/6)=((z-1))/-1 ....(1)`

Now consider the general equation of the line:

`(x-a)/l=(y-b)/m=(z-c)/n...(2)`

where, l,m and n are the direction ratios of the line and the point (a,b,c) lies on the line.
Compare the equation (1), with the general equation (2),

we have l=1/3, m=1/6 and n=-1

Also, a=1/3, b=-1/3 and c=1

This shows that the given line passes through (1/3, -1/3, 1)

Therefore, the given line passes through the point having

position vector `bara=1/3hati-i/3hatj+hatk` and is parallel to the
vector `barb=1/3hati+1/6hatj-hatk`

So its vector equation is

`barr=(1/3hati-1/3hatj+hatk)+lambda(1/3hati+1/6hatj-hatk)`