Browse Questions

# Find the perpendicular distance of the points (2,3,4) from the line $\large\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$, Also, find the coordinates of the foot of the perpendicular.

Toolbox:
• Distance between two points is $\sqrt {(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$
Step 1:
The given line $l$ is
$\large\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$
$\Rightarrow\large\frac{x-4}{-2}=\frac{y}{6}=\frac{z-1}{-3}$
Any point on $Q$ on the line $l$ is given by
$\Rightarrow\large\frac{x-4}{-2}=\frac{y}{6}=\frac{z-1}{-3}$$=\lambda x=-2\lambda+4,y=6\lambda,z=-3\lambda+1 Step 2: Direction ratios of PQ are -2\lambda+4-2,6\lambda-3,-3\lambda+1-4 Or -2\lambda+2,6\lambda-3,-3\lambda-3 Since PQ\perp l a_1a_2+b_1b_2+c_1c_2=0 -2(-2\lambda+2)+6(6\lambda-3)-3(-3\lambda-3)=0 \Rightarrow 49\lambda-13=0 \Rightarrow \lambda=\large\frac{13}{49} Step 3: Hence the coordinates of Q are x=(-2)(\large\frac{13}{49}$$)+4$
$\;\;=\large\frac{170}{49}$
$y=(6)(\large\frac{13}{49})$
$\;\;=\large\frac{78}{49}$
$x=(-3)(\large\frac{13}{49}$$)+4$
$\;\;=\large\frac{10}{49}$
Step 4:
The distance PQ is given by
$\sqrt{(2-\large\frac{170}{49})^2+(3-\large\frac{78}{49})^2+(4-\large\frac{10}{49})^2}$
$\sqrt{(\large\frac{72}{49})^2+(\large\frac{115}{49})^2+(\large\frac{186}{49})^2}$
On simplifying we get,
$\Rightarrow 4$units