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kolezko [41]
3 years ago
9

Find the area of the parallelogram with vertices: P(0,0,0), Q(-4,3,-1), R(-4,2,-2), S(-8,5,-3).

Mathematics
1 answer:
-BARSIC- [3]3 years ago
4 0

Answer:

Step-by-step explanation:

Co-ordinate of point P(0,0,0)

Q (-4,3,-1)

R (-4,2,-2)

S (-8,5,-3)

\vec{PQ}=

\vec{PS}=

Area of Parallelogram =|\vec{PQ}\times \vec{PS}|

Area=\begin{vmatrix}i &j &k \\ -4 &3 &-1 \\ -8 &5 &-3 \end{vmatrix}

A=|\hat{i}(-9+5)-\hat{j}(-12-8)+\hat{k}(-20+24)|

A=|-4\hat{i}+20\hat{j}+4\hat{k}|

A=\sqrt{(-4)^2+(20)^2+(4)^2}

A=\sqrt{432}

A=20.78\ unit^2

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Required

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c = 0.01

At 0.01 level (check row 20 and column 24), the critical value is:

<em></em>f_{0.01,24,20} = 2.86<em> --- the upper bound</em>

At 0.01 level (check row 24 and column 20), the critical value is:

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f_{0.99,20,24} = \frac{1}{f_{0.01,20,24}} = \frac{1}{2.74} =0.365 ---- the lower bound

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0.365 < 0.923 < 2.86

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<em>This implies that, there is no sufficient evidence.</em>

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