If the vectors 3i - 4j - k and 2i + 3j - 6k represent the diagonals of a rhombus, then the length of the side of the rhombus is
15
C.
Given,
AC = 3i - 4j - k
and BD = 2i + 3j - 6k
Let sides of a rhombus be a and b.
In a ABC,
AC = AB + BC
BD = BA + AD
On adding Eqs. (i) and (ii), we get
5i - j - 7k = 2b
A vector of magnitude 7 units, parallel to the resultant of the vectors a = 2 i - 3j - 2k and b = - i + 2j + k, is
7(i - j - k)
The point which divides the line joining the points (1, 3, 4) and (4, 3, 1) internally in the ratio 2 : 1, is
(2, - 3, 3)
(2, 3, 3)
(3, 3, 2)
The equation of the plane which is equidistant from the two parallel planes 2x - 2y + z + 3= 0 and 4x - 4y + 2z + 9 = 0 is
8x - 8y + 4z + 15 = 0
8x - 8y + 4z - 15 = 0
8x - 8y + 4z + 3 = 0
8x - 8y + 4z - 3 = 0