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

Question

Let a straight line $L$ pass through the point $P(2, -1, 3)$ and be perpendicular to the lines $ \frac{x - 1}{2} = \frac{y + 1}{1} = \frac{z - 3}{-2} $ and $ \frac{x - 3}{1} = \frac{y - 2}{3} = \frac{z + 2}{4} $. If the line $L$ intersects ...

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:

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

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 $Q$:

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

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

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

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

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

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