Question

Find the area of the parallelogram with vertices A(-3, 3), B(-1,

Answer 1

Question 1 is incomplete. Only one vertex is given. So I have solved the remaining questions.
sol A = (2,-1,4) and b =( 7, 2, 1) from the definition of cross product we have a no axb = (as b3 - 83 b) î+ (azb, - aiba 9 +(a, b, a, b ) ê from the vector a=(a,,a2, 93)=(2, -1,4) & bei bz (bs, bz, bz) = (7, 2, 1), or do we have 9.2, A2=-1, Q3=4 & b = 7, b2=2, b31 ax b= (-1X1 – 442)+ (4X7 - 203) +(272-(1997) & -gî + 26 ŷt 11 å axb= (-9, 26, 11) Anu Yow from the properties of cross product we know axba - bxa .. bxaca axb - L9,-26,-11) An

ash+29-4% , b= 2 +5% Writing ng the above given rector in co-oredinate as a = 21,2,-4) 2 ball; 0,5%. Now, s 3 from the definition of cross product axb= g . 2 - 4 5 • Â (245 - 0xcu))+ (185 – (41)x(u)) ?f + (1x0 –2864)] & - 10 î + (5-4)3 + 2a secretario d zlo +3 +2K.. axba L10, 1, 2) =) - Cat .) Proton e d