Show that the line whose vector equation is `vecr=(2 hati-2 hatj+3 hatk)+lambda (hati- hatj+4 hatk)` is parallel to the plane whose vector equation is `vecr.(hati+5 hatj+ hatk)=5` .Also, find the distance between them

State the condition for which the line vecr=veca+lambda vecb is parallel to the plane vecr. vecn=d . Prove that the line vecr=(hati+ hatj)+ lambda (2 hati+ hat j+4 hatk) is parallel to the plane vec r.(-2 hati+ hatk)=5 . Also , find the distance between the line and the plane.

If the line vecr=(2hati-hatj+3hatk)+lambda(2hati+hatj+2hatk) is parallel to the plane vecr*(3hati-2hatj+p hatk)=4 then the value of p is -

A vector perpendicular to both of (3hati+hatj+2hatk) and (2hati-2hatj+4hatk) is

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

Find the angle between the plane vecr.(2 hati-hatj+2 hat k)=6 and vecr.(3 hati+6 hatj-2 hatk) =9

Find the angle between the two vectors vecA = hati + 2hatj + 3hatk and vecB = 2hati+ hatj + 4hatk .

Find the angle between the two vectors vecA =hati-2hatj+3hatk and vecB=2hati+hatj+3hatk .

The shortest distance between the lines whose vector equations are vecr=hati-2hatj+3hatk+lambda(hati-hatj+hatk) and vecr=hati-hatj-hatk+mu(hati+2hatj-2hatk) is -