find the area of triangle constructed on the vectors P=6I-4J+4K and Q=-4I+2J-3K
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Given, P = 6i - 4j + 4k
and Q = -4i + 2j -3k
taking dot product :
P.Q = P Q cos θ
-24-8-12= 62 + (-4)2 + 42 (-4)2 + 22+ (-3)2
-44 = 68 29 cos θ
θ = cos (-0.99)
θ = 171.89
Since, area of triangle is given by the formula :
= 12P Q sin θ
=12 68 29 sin (171.890) = 3.12 units
and Q = -4i + 2j -3k
taking dot product :
P.Q = P Q cos θ
-24-8-12= 62 + (-4)2 + 42 (-4)2 + 22+ (-3)2
-44 = 68 29 cos θ
θ = cos (-0.99)
θ = 171.89
Since, area of triangle is given by the formula :
= 12P Q sin θ
=12 68 29 sin (171.890) = 3.12 units
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