Question

Find the distances between the following pairs of parallel lines
(a) r · (i + j) + 7 = 0, r · (i + j) − 11 = 0;
(b) r · (2i − 3j) + 6 = 0, r · (4i − 6j) + 5 = 0;
(c) r = i + j + s(4i − j), r = 4i + 5j + t(8i − 2j).

Answer 1

(a)

converting equation into general form

then given lines are-

$$x+y+7=0\\and\\x+y-11=0\\distance=\frac{|d_1-d_2|}{\sqrt{a^2+b^2}}=\frac{|7+11|}{\sqrt{1^2+1^2}}=\frac{18}{\sqrt{2}}=9\sqrt{2}=12.728$$

(b)

Similarly given lines are -

$$4x-6y+12=0\\and\\4x-6y+5=0\\distance=\frac{|d_1-d_2|}{\sqrt{a^2+b^2}}=\frac{|12-5|}{\sqrt{4^2+(-6)^2}}=\frac{7}{\sqrt{52}}=0.97$$

(c)

passing points are $(1,1)\ and\ (4,5)$

distance betwwen passing point of these two lines is-

$$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}=\sqrt{(4-1)^2+(5-1)^2}\\d=5$$

vector between these two points= A = $3\hat i+4\hat j$

vector parallel to any of the two line =B = $4\hat i-\hat j$

Distance between two lines = $d\sin\theta$

where $\theta$ is the angle between vector A and B

$now,\\\cos\theta=\frac{\overset{\to}A.\overset{\to}B}{|\overset{\to}A||\overset{\to}B|}=\frac{12-4}{5\times\sqrt{17}}=\frac{8}{5\sqrt{17}}\\then,\sin\theta=0.922$

So,

distance between two lines = $d\sin\theta=5\times0.922=4.61$