Find the shortest distance between the two skew lines L and M given by their symmetric equations.

Question

ScanSolve AI Tutor · Accepted Answer

To find the shortest distance between the two skew lines L and M, we use the formula:

D = \frac{|(\vec{P_L P_M}) \cdot (\vec{d_L} \times \vec{d_M})|}{\|\vec{d_L} \times \vec{d_M}\|}

where $\vec{d_L}$ and $\vec{d_M}$ are the direction vectors of lines L and M, respectively, and $\vec{P_L P_M}$ is a vector connecting a point $P_L$ on line L to a point $P_M$ on line M.

From the given equations:
Line L: $\frac{x+8}{2} = \frac{y-10}{3} = \frac{z-6}{1}$
Line M: $\frac{x-1}{-1} = \frac{y-1}{2} = \frac{z-1}{4}$

Identify direction vectors and points.
For line L: $\vec{d_L} = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix}$ and $P_L = (-8, 10, 6)$ .
For line M: $\vec{d_M} = \begin{pmatrix} -1 \\ 2 \\ 4 \end{pmatrix}$ and $P_M = (1, 1, 1)$ .
Calculate the vector $\vec{P_L P_M}$ .
$\vec{P_L P_M} = P_M - P_L = (1 - (-8), 1 - 10, 1 - 6) = (9, -9, -5)$
Calculate the cross product of the direction vectors, $\vec{d_L} \times \vec{d_M}$ .
$\vec{d_L} \times \vec{d_M} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & 1 \\ -1 & 2 & 4 \end{vmatrix}$ $= \mathbf{i}((3)(4) - (1)(2)) - \mathbf{j}((2)(4) - (1)(-1)) + \mathbf{k}((2)(2) - (3)(-1))$ $= \mathbf{i}(12 - 2) - \mathbf{j}(8 - (-1)) + \mathbf{k}(4 - (-3))$ $= 10\mathbf{i} - 9\mathbf{j} + 7\mathbf{k}$
So, $\vec{d_L} \times \vec{d_M} = \begin{pmatrix} 10 \\ -9 \\ 7 \end{pmatrix}$ .
Calculate the magnitude of the cross product, $\|\vec{d_L} \times \vec{d_M}\|$ .
$\|\vec{d_L} \times \vec{d_M}\| = \sqrt{10^2 + (-9)^2 + 7^2}$ $= \sqrt{100 + 81 + 49}$ $= \sqrt{230}$
Calculate the scalar triple product, $(\vec{P_L P_M}) \cdot (\vec{d_L} \times \vec{d_M})$ .
$(\vec{P_L P_M}) \cdot (\vec{d_L} \times \vec{d_M}) = (9)(10) + (-9)(-9) + (-5)(7)$ $= 90 + 81 - 35$ $= 171 - 35$ $= 136$
Calculate the shortest distance.
$D = \frac{|136|}{\sqrt{230}}$ $D = \frac{136}{\sqrt{230}}$
To rationalize the denominator, multiply the numerator and denominator by $\sqrt{230}$ :
$D = \frac{136\sqrt{230}}{230}$
Simplify the fraction:
$D = \frac{68\sqrt{230}}{115}$
The shortest distance between the lines L and M is $\frac{68\sqrt{230}}{115}$ .

Find the shortest distance between the two skew lines L and M given by their symmetric equations.

Answer

Step-by-step solution

Identify direction vectors and points.

Calculate the vector $\vec{P_L P_M}$ .

Calculate the cross product of the direction vectors, $\vec{d_L} \times \vec{d_M}$ .

Calculate the magnitude of the cross product, $\|\vec{d_L} \times \vec{d_M}\|$ .

Calculate the scalar triple product, $(\vec{P_L P_M}) \cdot (\vec{d_L} \times \vec{d_M})$ .

Calculate the shortest distance.

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Find the shortest distance between the two skew lines L and M given by their symmetric equations.

Answer

Step-by-step solution

Identify direction vectors and points.

Calculate the vector PLPM⃗\vec{P_L P_M}PL​PM​​.

Calculate the cross product of the direction vectors, dL⃗×dM⃗\vec{d_L} \times \vec{d_M}dL​​×dM​​.

Calculate the magnitude of the cross product, ∥dL⃗×dM⃗∥\|\vec{d_L} \times \vec{d_M}\|∥dL​​×dM​​∥.

Calculate the scalar triple product, (PLPM⃗)⋅(dL⃗×dM⃗)(\vec{P_L P_M}) \cdot (\vec{d_L} \times \vec{d_M})(PL​PM​​)⋅(dL​​×dM​​).

Calculate the shortest distance.

Try one yourself

Still stuck on this one?

More Mathematics Questions

Calculate the vector $\vec{P_L P_M}$ .

Calculate the cross product of the direction vectors, $\vec{d_L} \times \vec{d_M}$ .

Calculate the magnitude of the cross product, $\|\vec{d_L} \times \vec{d_M}\|$ .

Calculate the scalar triple product, $(\vec{P_L P_M}) \cdot (\vec{d_L} \times \vec{d_M})$ .