Let a straight line L pass through the point P(2,-1,3) and be perpendicular to the lines x…

Question

Let a straight line $L$ pass through the point $P(2,-1,3)$ and be perpendicular to the lines $\dfrac{x-1}{2}=\dfrac{y+1}{1}=\dfrac{z-3}{-2}$ and $\dfrac{x-3}{1}=\dfrac{y-2}{3}=\dfrac{z+2}{-4}$. If the line $L$ intersects the $yz$-plane at the point $Q$, then the distance between the points $P$ and $Q$ is:

Accepted Answer

To find the distance between the points $ P $ and $ Q $, let's consider the following steps:

Determine the direction vector of line $ L $:

For the line $ L $ to be perpendicular to both given lines, we find the direction vector of $ L $ using the cross product of the direction vectors of the given lines.

The first line has direction vector $ \langle 2, 1, -2 angle $ and the second line has direction vector $ \langle 1, 3, 4 angle $. The cross product is:

$    \begin{vmatrix}    \hat{i} & \hat{j} & \hat{k} \    2 & 1 & -2 \    1 & 3 & 4    \end{vmatrix}    = 10\hat{i} - 10\hat{j} + 5\hat{k}    = 5(2\hat{i} - 2\hat{j} + \hat{k})    $

Thus, the direction vector of line $ L $ is $ \langle 2, -2, 1 angle $.

Equation of line $ L $:

Since $ L $ passes through point $ P(2, -1, 3) $ with direction vector $ \langle 2, -2, 1 angle $, its parametric form is:

$    \frac{x-2}{2} = \frac{y+1}{-2} = \frac{z-3}{1} = \lambda    $

Find the intersection with the $ yz $-plane:

On the $ yz $-plane, $ x = 0 $. Substitute into the line equation to find $ \lambda $:

$    2\lambda + 2 = 0 \implies \lambda = -1    $

Substitute $ \lambda = -1 $ back into the parametric equations to find $ Q $:

$    x = 2(-1) + 2 = 0, \quad y = -2(-1) - 1 = 1, \quad z = (-1) + 3 = 2    $

Therefore, the intersection point $ Q $ is $ (0, 1, 2) $.

Calculate the distance between $ P $ and $ Q $:

$    	ext{d}(P, Q) = \sqrt{(2-0)^2 + (-1-1)^2 + (3-2)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3    $

Thus, the distance between points $ P $ and $ Q $ is 3.

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