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lyudmila [28]
3 years ago
5

Determine whether the lines l1 and l2 are parallel, skew, or intersecting. If they intersect, find the point of intersection.l1:

x=3-2t, y=7+4t, z=-3+8tl2: x=-1-u, y=18+3u, z=7+2u
Mathematics
1 answer:
xeze [42]3 years ago
8 0

Answer:

The lines ared skew

Step-by-step explanation:

First we check if they are parallel. To do this we compare the ratios of the slopes of each component, and this should be the same:

\frac{L1_{x} }{L2_{x} } =\frac{L1_{y} }{L2_{y} }=  \frac{L1_{z} }{L2_{z} }

\frac{-2}{-1} =\frac{4}{3} =\frac{8}{2}

As we can see the ratios are different so they are not parallel lines.

Now we test for intersesction:

we have to equate each component and solve it as a system:

  • 3-2t=-1-u
  • 7+4t=18+3u
  • -3+8t=7+2u

From equation 1: u=2t-4 we replace this in eq. 2

From equation 2: t=1/2 replacing both in equation 3 should give us an identity

From equation 3: we reach 1=3 as this is not true, out hypothesis that the lines intersect is wrong.

Since this lines are not paralel and don't intersect we must conclude that the are skew

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Answer:

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Step-by-step explanation:

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3 years ago
The vector (a) is a multiple of the vector (2i +3j) and (b) is a multiple of (2i+5j) The sum (a+b) is a multiple of the vector (
kow [346]

Answer:

\|a\| = 5\sqrt{13}.

\|b\| = 3\sqrt{29}.

Step-by-step explanation:

Let m,n, and k be scalars such that:

\displaystyle a = m\, (2\, \vec{i} + 3\, \vec{j}) = m\, \begin{bmatrix}2 \\ 3\end{bmatrix}.

\displaystyle b = n\, (2\, \vec{i} + 5\, \vec{j}) = n\, \begin{bmatrix}2 \\ 5\end{bmatrix}.

\displaystyle (a + b) = k\, (8\, \vec{i} + 15\, \vec{j}) = k\, \begin{bmatrix}8 \\ 15\end{bmatrix}.

The question states that \| a + b \| = 34. In other words:

k\, \sqrt{8^{2} + 15^{2}} = 34.

k^{2} \, (8^{2} + 15^{2}) = 34^{2}.

289\, k^{2} = 34^{2}.

Make use of the fact that 289 = 17^{2} whereas 34 = 2 \times 17.

\begin{aligned}17^{2}\, k^{2} &= 34^{2}\\ &= (2 \times 17)^{2} \\ &= 2^{2} \times 17^{2} \end{aligned}.

k^{2} = 2^{2}.

The question also states that the scalar multiple here is positive. Hence, k = 2.

Therefore:

\begin{aligned} (a + b) &= k\, (8\, \vec{i} + 15\, \vec{j}) \\ &= 2\, (8\, \vec{i} + 15\, \vec{j}) \\ &= 16\, \vec{i} + 30\, \vec{j}\\ &= \begin{bmatrix}16 \\ 30 \end{bmatrix}\end{aligned}.

(a + b) could also be expressed in terms of m and n:

\begin{aligned} a + b &= m\, (2\, \vec{i} + 3\, \vec{j}) + n\, (2\, \vec{i} + 5\, \vec{j}) \\ &= (2\, m + 2\, n) \, \vec{i} + (3\, m + 5\, n) \, \vec{j} \end{aligned}.

\begin{aligned} a + b &= m\, \begin{bmatrix}2\\ 3 \end{bmatrix} + n\, \begin{bmatrix} 2\\ 5 \end{bmatrix} \\ &= \begin{bmatrix}2\, m + 2\, n \\ 3\, m + 5\, n\end{bmatrix}\end{aligned}.

Equate the two expressions and solve for m and n:

\begin{cases}2\, m + 2\, n = 16 \\ 3\, m + 5\, n = 30\end{cases}.

\begin{cases}m = 5 \\ n = 3\end{cases}.

Hence:

\begin{aligned} \| a \| &= \| m\, (2\, \vec{i} + 3\, \vec{j})\| \\ &= m\, \| (2\, \vec{i} + 3\, \vec{j}) \| \\ &= 5\, \sqrt{2^{2} + 3^{2}} = 5 \sqrt{13}\end{aligned}.

\begin{aligned} \| b \| &= \| n\, (2\, \vec{i} + 5\, \vec{j})\| \\ &= n\, \| (2\, \vec{i} + 5\, \vec{j}) \| \\ &= 3\, \sqrt{2^{2} + 5^{2}} = 3 \sqrt{29}\end{aligned}.

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Consider the following sets:
bagirrra123 [75]
U = {points on the coordinate plane}
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4 0
3 years ago
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