Show that the vectors 2i^−j^+k^,i^−3j^−5k^and 3i^−4j^−4k^form the vertices of a right angled triangle.
Vectors a=i^+2j^+3k^,b=2i^−j^+k^ and c=3i^+j^+4k^, are so placed that the end point of one vector is the starting point of the next vector. Then the vector are (A) not coplanar (B) coplanar but cannot form a triangle (C) coplanar and form a triangle (D) coplanar and can form a right angled triangle
Vectors a=−4i^+3k^;b=14i^+2j^−5k^ are laid off from one point. Vector d^ , which is being laid of from the same point dividing the angle between vectors aandb in equal halves and having the magnitude 6, is a. i^+j^+2k^ b. i^−j^+2k^ c. i^+j^−2k^ d. 2i^−j^−2k^
Statement 1: Let a,b,candd be the position vectors of four points A,B,CandD and 3a−2b+5c−6d=0. Then points A,B,C,andD are coplanar. Statement 2: Three non-zero, linearly dependent coinitial vector (PQ,PRandPS) are coplanar. Then PQ=λPR+μPS,whereλandμ are scalars.
The vector a has the components 2p and 1 w.r.t. a rectangular Cartesian system. This system is rotated through a certain angel about the origin in the counterclockwise sense. If, with respect to a new system, a has components (p+1)and1 , then p is equal to a. −4 b. −1/3 c. 1 d. 2
Find the vector of magnitude 3, bisecting the angle between the vectors a=2i^+j^−k^ and b=i^−2j^+k^˙