22. Show that the volume of a tetrahedron with a,b and c as coterminous edges is 1/6 |[abc]|
Answers
Answer: I think it will help you:
A tetrahedron is 16 of the volume of the parallelipiped formed by a⃗ ,b⃗ ,c⃗ . The volume of the parallelepiped is the scalar triple product |(a×b)⋅c|. Thus, the volume of a tetrahedron is 16|(a×b)⋅c|
In order to solve the question like you are trying to, notice that by V=13Bh=16||a×b||⋅h. Then, h=||c||⋅|cos(θ)|. Thus, we have V=16||a×b||⋅||c||⋅|cos(θ)|. Now see that |c⋅(a×b)|=||c||⋅||(a×b)||⋅|cos(θ)| and thus V=16|(a×b)⋅c|.
I'd also like to say that the notation you are using is a little weird. In order to avoid confusion, |x| denotes absolute value and ||x|| denotes magnitude.
Answer:
Frankly we can use the formula for the area of a triangle
[ABC] = (1/2) * |AB x AC| to obtain the volume,
V = (1/6) * |AB x AC|^2 * sqrt(1 - ((a^2 + b^2 - c^2)/(2ab))^2)
Fortunately that is the desired formula for the volume of the tetrahedron in terms of its edges. We can also write this formula as
V = (1/6) * |a x b · c|
where "x" denotes the cross product of vectors
Step-by-step explanation:
We start by considering a tetrahedron with vertices A, B, C, and D. The edges AB, AC, and AD have lengths a, b, and c, respectively and they are coterminous (i.e., they share a common vertex, which is point A).
The first step is to draw the perpendicular from point A to the plane BCD and let H be the foot of this perpendicular. We can see that the length of AH is the altitude of the tetrahedron. This gives us a height and a base, which we can use to calculate the volume of the tetrahedron using the formula V = (1/3) * base * height.
To find the area of the base (the triangle BCD) we use Heron's formula, which gives us the area of a triangle given its side lengths. We calculate the semiperimeter of the triangle using the lengths of the sides b, c, and d, where d is the length of CD. Then, we plug these values into Heron's formula to get the area of the base.
Moreover we use the Pythagorean theorem to find the length of AH. We know that the length of AD is c, and we can calculate the length of DH using the area of the base and the length of CD. Then we use the Pythagorean theorem to find the length of AH.
Substituting the expressions for [BCD] and AH in the formula for the volume of the tetrahedron, we get an expression that involves the side lengths a, b, and c, as well as the area of the base [BCD].
Fortunately we know that [ABC] = (1/2) * bc * sin(A) = (1/2) * [BCD] * sin(A), where A is the angle between the edges BC and CD. Thus, we have an expression for [BCD] in terms of [ABC]. We substitute this expression into the formula for the volume of the tetrahedron which gives us an expression that involves only [ABC] and the side lengths a, b, and c.
Moreover We simplify this expression using the identity sin^2(A) = 1 - cos^2(A) which gives us an expression that involves [ABC] and the side lengths a, b, and c as well as the cosine of the angle between the edges BC and CD.
Finally we use the formula for the area of a triangle [ABC] = (1/2) * |AB x AC| to obtain an expression that involves the cross product of vectors AB and AC. We simplify this expression using the properties of the cross product and we arrive at the final formula for the volume of the tetrahedron in terms of its edges:
V = (1/6) * |a x b · c|.
This formula tells us that the volume of a tetrahedron with edges of length a, b, and c is one-sixth the magnitude of the scalar triple product of the three edge vectors.
Learn more about triangle:https://brainly.in/question/17424774
Learn more about tetrahedron: https://brainly.in/question/303538
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