Given vector u=3i + 5j and v=6i - 2j, find 3u + 2v and the dot product of u and v.

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Step-by-step explanation:

The desired sum is ...

  3u +2v = 3(3i +5j) +2(6i -2j) = (3·3 +2·6)i +(3·5+2(-2))j = 21i +11j

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The dot product is the sum of products of corresponding component values.

  u·v = (3)(6) +(5)(-2) = 18 -10 = 8

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.