Question

If A vector = i^ -2j^ + 3k^ and B vector =2i^+3j^-k^. Find area of parallelogram

Answer 1

Answer:

area = 12.124 units square

Explanation:

Area of parallelogram = magnitude of cross product of two vectors

A = x1 i^ -y1 j^ + z1 k^

B = x2 i^+y2 j^-z2 k^

here x1 = 1; y1 = -2; z1 = 3

x2 = 2; y2 = 3; z2 = -1

area =

$\sqrt{(y1 \times z2 - y2 \times z1) {}^{2} + ( {x1 \times z2 - x2 \times z1)}^{2} + ( {x1 \times y2 - x2 \times y1)}^{2} }$

now put the values in above expression we get

area = sqr root ((2-9)^2 + (-1-6)^2 + (3+4)^2)

or sqr root (49+49+49)

or sqr root 147

or 12.124

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