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20.) Find the area of the parallelogram with vertices P1, P2,

P3, and P4. P1 = (7, -5, 5), P2 = (2, -8, 4), P3 = (9, 8, 4), P4 =

(4, 5, 3)

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02:39

Find the area of the parallelogram whose vertices are $P_{1}, P_{2}, P_{3},$ and $P_{4}$.$P_{1}=(-1,1,1) ; \quad P_{2}=(-1,2,2) ; \quad P_{3}=(-3,5,-4) ;$$P_{4}=(-3,4,-5)$

02:25

Find the area of the parallelogram whose vertices are $P_{1}, P_{2}, P_{3},$ and $P_{4}$.$P_{1}=(0,0,0) ; \quad P_{2}=(-1,2,0) ; \quad P_{3}=(1,5,-4) ;$$P_{4}=(2,3,-4)$

03:56

Find the area of the parallelogram with vertices $P_{1}, P_{2}, P_{3},$ and $P_{4}$.$$\begin{array}{l}P_{1}=(-1,1,1), \quad P_{2}=(-1,2,2), \quad P_{3}=(-3,4,-5), \\P_{4}=(-3,5,-4)\end{array}$$

03:14

Find the area of the parallelogram with vertices $P_{1}, P_{2}, P_{3},$ and $P_{4}$.$P_{1}=(-2,0,2), \quad P_{2}=(2,1,-1), \quad P_{3}=(2,-1,2)$

03:30

Find the area of the parallelogram with vertices $P_{1}, P_{2}, P_{3},$ and $P_{4}$.$P_{1}=(1,2,0), \quad P_{2}=(-2,3,4), \quad P_{3}=(0,-2,3)$

Transcript

So this question is asked that the four corners of a parallelogram are given Ok? For is given. Now we have to find the area of the so we know that the area of the parallelogram is nothing but the cross the magnitude of the cross product of any two consecutive victims. Okay, that is. Let us take this to the authorities originally originating from the same point. So we take this letter I suppose you vector and this vector. Okay, so we have to take the cross product of them. Now let us find the U vector. That is this is the U vector. Okay, this is the U. Vector and this is the director. So you’ve explained nothing. But the difference between the current 34 minus two, so 25 minus of minus one. That is 13, 3 minus four is minus one. Okay, so you trace this similarly the view will be seven minus negative five uh minus five minus of minus is that? There is minus…

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