The value of `a` so that the volume of parallelepiped formed by ` hat i+a hat j+ hat k , hat j+a hat ka n da hat i+ hat k` is minimum is `-3` b. `3` c. `1//sqrt(3)` d. `sqrt(3)`

Find the value of a so that the volume of the parallelepiped formed by vectors hat i+ahat j+k,hat j+ahat k and ahat i+hat k becomes minimum.

The unit vector perpendicular to vec A = 2 hat i + 3 hat j + hat k and vec B = hat i - hat j + hat k is

The value of p so that vectors 2hat i-hat j+hat k , hat i+2hat j-3hat k and 3hat i+phat j+5hat k are coplanar is

underset with a = 2hat i + hat j-3hat k, vec b = -hat i + hat j + 2hat k, vec c = 4hat i + 3hat k

Show that the points A(-2 hat i+3 hat j+5 hat k), B( hat i+2 hat j+3 hat k) and C(7 hat i-3 hat k) are collinear.

If the vectors 3 hat i +3 hat j +9 hat k and hat i +a hat j +3 hat k are parallel, then a =

If theta is the angle between the vectors 2 hat i-2 hat j+4 hat k\ a n d\ 3 hat i+ hat j+2 hat k , then sintheta= 2/3 b. 2/(sqrt(7)) c. (sqrt(2))/7 d. sqrt(2/7)

hat i.(2hat j times3hat k)+hat j.(2hat k times3hat i)+hat k.(2hat i times3hat j) is equal to

Find the area of the parallelogram determined by the vectors: 2hat i and 3hat j2hat i+hat j+3hat k and hat i-hat j3hat i+hat j-2hat k and hat i-3hat j+4hat khat i-3hat j+hat k and hat i+hat j+hat k

The value of hat i*(hat j xxhat k)+hat j*(hat i+hat k)+hat k*(hat i xxhat j) is (A) 0 (B) (C) 1 (D) 3