Find the dot product of the given vectors.
u=2i−6j
v=3i+9j

u = 2i - 6j

v = 3i + 9j

Take the dot product:

u • v = (2i - 6j) • (3i + 9j)

u • v = (2i • 3i) + (2i • 9j) + (-6j • 3i) + (-6j • 9j)

u • v = 6 (i • i) + 18 (i • j) - 18 (j • i) - 54 (j • j)

The dot product is defined so that for the orthogonal unit vectors i and j, we have i • i = j • j = 1, and i • j = 0. So the above reduces to

u • v = 6 + 0 - 0 - 54

u • v = -48

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐

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