Distance between line `r=i+j+k+lambda(2i+j+4k)` and plane `rdot(i+2j-k)=3` is (1)`4/(sqrt(6))` (2) `5/(sqrt(6))` (3) `1/(sqrt(6))` (4)`0 `(5) `2/(sqrt(6))`

The distance of the line vec r=2 hat i-2 hat j+3 hat k+lambda( hat i- hat j+4 hat k) from the plane vec rdot(( hat i+5 hat j+ hat k))=5, is 5/(3sqrt(3)) b. (10)/(3sqrt(3)) c. (25)/(3sqrt(3)) d. none of these

The distance between the line vec r=2 hat i-2 hat j+3 hat k+lambda( hat i- hat j+4 hat k) and the plane vec rdot( hat i+5 hat j+ hat k)=5 is a. (10)/(3sqrt(3)) b. (10)/9 c. (10)/3 d. 3/(10)

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

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

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

Direction cosines of a line which passes through the points (1,2,3) and (-3,4,1) is A) (4/(2sqrt6), -2/(2sqrt6), 2/(2sqrt6)) B) (4/(2sqrt6), -2/(2sqrt6), -2/(2sqrt6)) C) (4/(2sqrt6), 4/(2sqrt6), 2/(2sqrt6)) D) (-4/(2sqrt6), 1/(2sqrt6), -2/(2sqrt6))

Let eveca=hati+hatj-hatk, vecb=hati-hatj+hatk and vecc be as unit vector perpendicular to veca and vecb the vecc= (A) 1/sqrt(j+k) (B) 1/sqrt(2)(j-k) (C) 1/sqrt(6) (i-2jk) (D) 1/sqrt(6) (2i-j+k)

A non-zero vector vec a is such that its projections along vectors ( hat i+ hat j)/(sqrt(2)),(- hat i+ hat j)/(sqrt(2)) and hat k are equal, then unit vector along vec a is a. (sqrt(2) hat j- hat k)/(sqrt(3)) b. ( hat j-sqrt(2) hat k)/(sqrt(3)) c. (sqrt(2))/(sqrt(3)) hat j+( hat k)/(sqrt(3)) d. ( hat j- hat k)/(sqrt(2))

Evaluate using binomial theorem: (i) (sqrt(2)+1)^(6) +(sqrt(2)-1)^(6) (ii) (sqrt(5)+sqrt(2))^(4)-(sqrt(5)-sqrt(2))^(4)

Find the angle between the line -> r=(2 hat i+3 hat j+9 hat k)+lambda(2 hat i+3 hat j+4 hat k) and the plane -> r(dot( hat i+ hat j+ hat k))=5.