UGC NET CSE | December 2012 | Part 3 | Question: 12

Question

Let $V_1 = 2I-J+k$ and $V_2 = I+J-K$, then the angle between $V_1$ and $V_2$ and a vector perpendicular to both $V_1$ and $V_2$ shall be

• A. $90^o \text{ and } (-2I+J-3K)$
• B. $60^o \text{ and } (2I+J+3K)$
• C. $90^o \text{ and } (2I+J-3K)$
• D. $90^o \text{ and } (-2I-J+3K)$

Comments

I cant understand how this comes (−2I−J+3K) can you please explain

Answer 1

Here all options are incorrect hence marks to ALL

V1=2I−J+k  

V2=I+J-K    

Their dot product is V1.V2=|V1||V2|cos ( theta) =0 if they are perpendiclar to each other

now (2i-j+k).(i+j-k)=2-1-1=0 hence theta =90 degree  so choice B is out

To find vector perpendicular to V1and V2 we need a cross product of these two as below

 i      j      k  

2    -1      1    solving this determinant  we get  0i+3j+3k which does not match any choice 

1     1     -1

Aleternatively u can perform dot product of vectors given by choice A,C,D to V1 ,V2 result

of no choice will give u 0 hence  they are not having 90degree angle with V1 and V2I+J−KI+J−KI+J−KI+J−KI+J−K