The distance between the line `vecr = 2i - 2j + 3k + lamda(i - j + 4k)` and the plane `vecr. (i + 5j + k) = 5` is

The distance between the line vec r = 2i+2j+3k + lambda (i-j+4k) and the plane vec r . (i+5j+k) = 5 is

The distance between the line vec r=2i-2j+3k+lambda(i-j+4k) and the plane vec r*(i+5j+k)=5 is

If theta is the between the line r = ( i + 2j- k) + lambda (i - j + 2k) , lambda in R and the plane r. (2i - j + k ) = 4. then a value of cos theta is :

Find the distance between the line barr=i+j+lambda(2i+j+4k) and the plane barr.(-2i+k)=5 .

The distance between the planes r. (i+ 2j - 2k ) + 5 =0 and r. (2i + 4j - 4k ) - 16 =0 is

Find the shortest distance between the skew lines ⃗ r = ( ⃗ i + 2 ⃗ j + ⃗ k ) + λ ( ⃗ i − ⃗ j + ⃗ k ) and ⃗ r = ( 2 ⃗ i − ⃗ j − ⃗ k ) + η ( 2 ⃗ i + ⃗ j + 2 ⃗ k )