Verify that the points are the vertices of a parallelogram and find its area. (2,-1,1), (5, 1,4), (0,1,1), (3,3,4)
1 answer:
Answer:
Area = 13.15 square units
Step-by-step explanation:
Let the given vertices be represented as follows:
A(2, -1, 1) = 2i - j + k
B(5, 1, 4) = 5i + j + 4k
C(0, 1, 1) = 0i + j + k
D(3, 3, 4) = 3i + 3j + 4k
(i) Let's calculate the vectors of all the sides:
AB = B - A = (5i + j + 4k) - (2i - j + k)
AB = 5i + j + 4k - 2i + j - k [Collect like terms]
AB = 3i + 2j + 3k
BC = C - B = (0i + j + k) - (5i + j + 4k)
BC = 0i + j + k - 5i - j - 4k [Collect like terms]
BC = -5i + 0j - 3k
CD = D - C = (3i + 3j + 4k) - (0i + j + k)
CD = 3i + 3j + 4k - 0i - j - k [Collect like terms]
CD = 3i + 2j + 3k
DA = A - D = (2i - j + k) - (3i + 3j + 4k)
DA = 2i - j + k - 3i - 3j - 4k [Collect like terms]
DA = -i - 4j - 3k
AC = C - A = (0i + j + k) - (2i - j + k)
AC = 0i + j + k - 2i + j - k [Collect like terms]
AC = -2i + 2j
BD = D - B = (3i + 3j + 4k) - (5i + j + 4k)
BD = 3i + 3j + 4k - 5i - j - 4k [Collect like terms]
BD = -2i + 2j
(ii) From the results in (i) above, we can deduce that;
AB = CD This implies that AB || CD [AB is parallel to CD]
AC = BD This implies that AC || BD [AC is parallel to BD]
(iii) Therefore, ABDC is a parallelogram since opposite sides (AB and CD) are parallel. Hence, the points are vertices of a parallelogram
<u>Now let's calculate the area</u>
To find the area of the parallelogram, we find the magnitude of the cross product of any two adjacent sides.
In this case, we'll choose AB and AC
Area = |AB X AC|
Where;
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AB X AC = i(0 - 6) - j(0 + 6) + k(6 + 4)
AB X AC = - 6i - 6j + 10k
|AB X AC| =
|AB X AC| =
|AB X AC| = 13.15
Area = 13.15 square units.
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<u>PS: </u> ACBD is also a parallelogram. The diagram has also been attached to this response.
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Answer:
We conclude that the mean IQ of her students is different from the national average.
Step-by-step explanation:
We are given that the national mean (μ) IQ score from an IQ test is 100 with a standard deviation (s) of 15.
The dean of a college want to test whether the mean IQ of her students is different from the national average. For this, she administers IQ tests to her 144 students and calculates a mean score of 113
Let, Null Hypothesis, :
= 100 {means that the mean IQ of her students is same as of national average}
Alternate Hypothesis, :
100 {means that the mean IQ of her students is different from the national average}
(a) The test statistics that will be used here is One sample z-test statistics;
T.S. = ~ N(0,1)
where, Xbar = sample mean score = 113
s = population standard deviation = 15
n = sample of students = 144
So, test statistics =
= 10.4
Now, at 0.05 significance level, the z table gives critical value of 1.96. Since our test statistics is more than the critical value of z which means our test statistics will fall in the rejection region and we have sufficient evidence to reject our null hypothesis.
Therefore, we conclude that the mean IQ of her students is different from the national average.