The cartesian equation of the line barr ...

The cartesian equation of the line `barr = (hati + hatj + hatk)+ lambda(hatj + hatk)` is

A

(a) `x=1, y=z`

B

(b) `(x-1)/1 = (y-1)/2 = (z-1)/2`

C

(c) `x=y=z`

D

(c) `x-1=y=z `

Verified by Experts

The correct Answer is:

A

Similar Questions

Explore conceptually related problems

Cartesian form of the eqution of line barr=3hati-5hatj+7hatk+lambda(2hati+hatj-3hatk) is

The acute angle between the line barr=(3hati-hatj-hatk)+lambda(hati-hatj+hatk) and the plane barr.(3hati-4hatk)=4 is

The equation of the plane containing lines barr=hati+2hatj-hatk+lambda(hati+2hatj-hatk) and barr=hati+2hatj-hatk+mu(hati+hatj+3hatk) is

The vector equation of the plane bar r=(3hati+hatj)+lambda(-hatj+hatk)+mu(hati+2hatj+3hatk) in scalar product form is

The vector equation of the plane passing through the intersection of the planes barr.(hati-hatj+2hatk)=3" and "barr.(3hati-hatj-hatk)=4 is

The angle between the line barr=(2hati+3hatj+hatk)+lambda(hati+2hatj-hatk) and the plane barr.(2hati-hatj+hatk)=4 is

The line of intersection of the planes barr.(3hati-hatj+hatk)=1" and " barr.(hati+4hatj-2hatk)=2 is parallel to the vector

The shortest distance between the lines barr=(4hati-hatj)+lambda(hati+2hatj-3hatk)andbarr=(hati-hatj+2hatk)+mu(2hati+4hatj-5hatk) is

The equation of the plane through the intersection of the planes barr*(hati+2hatj+3hatk)= -3,barr*(hati+hatj+hatk)=4 and the point (1,1,1) is

Cosine of the angle between the lines barr=5hati-hatj+4hatk+lambda(hati+2hatj+2hatk) and barr=7hati+2hatj+2hatk+mu(3hati+2hatj+6hatk) is