If a=i+j+k,b=i-j+2k and c=xi+(x-2)j-k and if the vector c lies in the plane of vectors a and ab, then x is equal to?
0
1
-2
2
Finding the value for :
Given,
a=i+j+k,b=i-j+2kandc=xi+(x-2)j-k[a,b,c]=0
Now,
1111-12xx-2-1=0
⇒1(1-2x+4)-1(-1-2x)+(x-2+x)=0
⇒ 5-2x+1+2x+2x-2=0
⇒ 4+2x=0
⇒ 2x=-4
⇒ x=-42
∴x=-2
Hence, option(C) is correct.
Let c→be a vector perpendicular to the vectors a=i+j-kand b=i+2j+k. If c·(i+j+3k)=8, then the value of c·(axb) is equal to