Home
Class 12
MATHS
The Cartesian equations of a line are...

The Cartesian equations of a line are `6x-2=3y+1=2z-2.` Find its direction ratios and also find a vector equation of the line.

Text Solution

Verified by Experts

The given line is `6x-2=3y+1=2z-2`
To put it in the symmetrical form, we must make the coeffcients `x, y and z` as 1. To do this, we divide each of the expressions in (i) by 6 and obtain `(x-(1//3))/(1)=(y+(1//3))/(2)=(z-1)/(3)`.
This shows that the given line passes through `(1//3, -1//3, 1)` and is parallel to the line whose direction ratios are 1, 2 and 3. Therefore, its vector equation is
`" "vecr=(1)/(3)hati-(1)/(3)hatj+hatk+lamda(hati+2hatj+3hatk)`
Promotional Banner

Similar Questions

Explore conceptually related problems

The cartesian equation of a line is (x+3)/2=(y-5)/4 =(z+6)/2 . Find its vector equation .

The Cartesian equation of a line is (x-3)/2=(y+1)/(-2)=(z-3)/5 . Find the vector equation of the line.

The cartesian equation of a line is (x-5)/3=(y+4)/7=(z-6)/2 . Write its vector form.

Consider the Cartesian equation of a line (x-3)/2=(y+1)/3=(z-5)/(-2) i.Find its vector equation. ii. Find its intersecting point with the plane 5x+2y-6z-7=0

The vector equation of the line 6x-3=3y+4=2z-2 is

The direction ratios of a line are 1,-2,5 , find the direction cosines of the line.

If the cartesian equation of the plane 2x-y+2z=3 find the vector equation of the plane in standard form.

A plane is at a distance 3 units from the origin. The direction ratios of normal to the plane are 1,-2,1. Find the vector and cartesian equation of the plane.

Find the direction consines of a line whose direction ratios are 1,3,-2

Find the equation of straight line joining the points of intersection of the lines 3x+2y+1=0 and x+y=3 to the intersection of the lines y-x=1 and 2x+y+2=0