Let `L_(1):bar(r)=(i-j)+t_(1)(2i+3j+k) ,L_(2):bar(r)=(-i+2j+2k)+t_(2)(5i+j)` ,then The shortest distance between `L_(1)` and `L_(2)` is

Let L_(1):bar(r)=(i-j)+t_(1)(2i+3j+k),L_(2):bar(r)=(-i+2j+2k)+t_(2)(5i+j) , then The distance of origin from the plane passing through the point (1,-1,1) and whose normal is perpendicular to both L_(1) and L_(2) is

Let L_(1)=(x+3)/(-4)=(y-6)/(3)=(z)/(2) and L_(2)=(x-2)/(-4)=(y+1)/(1)=(z-6)/(1) Which of these are incorrect? (A) L_(1), L_(2) are coplanar. (B) L_(1),L_(2) are skew lines (C) Shortest distance between L_(1),L_(2) is 9 (D) (hat i-4hat j+8hat k) is a vector perpendicular to both L_(1),L_(2)

Let L_(1):r=(i+5j+5k)+t(4i-4j+5k)" and "L_(2): r=(2i+4j+5k)+t(8i-3j+k) be two lines then

If bar(a).i+bar(a).(i+j)+bar(a).(i+j+k)=1 then bar(a)=

Let vec a be vector parallel to line of intersection of planes P_(1) and P_(2) through origin.If P_(1) is parallel to the vectors 2bar(j)+3bar(k) and 4bar(j)-3bar(k) and P_(2) is parallel to bar(j)-bar(k) and 3bar(I)+3bar(j), then the angle between vec a and 2bar(i)+bar(j)-2bar(k) is :

The shortest distance between the lines r=(4i-j)+lambda(i+2j-3k) and r=(i-j+2k)+mu(2i+4j-5k) is

The vector equations of two lines L_1 and L_2 are respectively, L_1:r=2i+9j+13k+lambda(i+2j+3k) and L_2: r=-3i+7j+pk +mu(-i+2j-3k) Then, the lines L_1 and L_2 are

Let L_(1):vec r=(hat i-hat j+2 hat k)+lambda(hat i-hat j+2 hat k),lambda in R L_(2):vec r=(hat j-hat k)+mu(3 hat i+hat j+p hat k),mu in R ,and L_(3):vec r=delta(l hat i+m hat j+n hat k),delta in R be three lines such that L_(1) is perpendicular to L_(2) and L_(3) is perpendicular to both L_(1) and L_(2) .Then,the point which lies on L_(3) is

The shortest distance between the line L_1=(hat i - hat j hat k) lambda (2 hat i - 14 hat j 5 hat k) and L_2=(hat j hat k) mu (-2 hat i - 4 hat 7 hat) then L_1 and L_2 is