Question

Find the value of k so that the lines x = -y = kz and x - 2 = 2y + 1 = -z + 1 are perpendicular to each other.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given equation of lines are

$\rm :\longmapsto\:x = - y = kz$

and

$\rm :\longmapsto\:x - 2 = 2y + 1 = - z + 1$

Now, Consider,

$\rm :\longmapsto\:x = - y = kz$

Divide whole equation by k, we get

$\rm :\longmapsto\:\dfrac{x}{k} = \dfrac{y}{ - k} = \dfrac{z}{1}$

can be further rewritten as

$\rm :\longmapsto\:\dfrac{x - 0}{k} = \dfrac{y - 0}{ - k} = \dfrac{z - 0}{1}$

$\rm :\implies\:direction \: ratio \: of \: line \: is \: (k, - k,1)$

In vector form direction ratios of line (1) is represented as

$\bf\implies \:\vec{b_1} \: = \: k \hat{i} - k \hat{j} \: + \hat{k}$

Now, Consider second equation of line,

$\rm :\longmapsto\:x - 2 = 2y + 1 = - z + 1$

Divide whole equation by 2, we get

$\rm :\longmapsto\:\dfrac{x - 2}{2} = \dfrac{y + 0.5}{1} = \dfrac{ - z + 1}{2}$

can be rewritten as

$\rm :\longmapsto\:\dfrac{x - 2}{2} = \dfrac{y + 0.5}{1} = \dfrac{ z - 1}{ - 2}$

$\rm :\implies\:direction \: ratio \: of \: line \: is \: (2, 1, - 2)$

So,

In vector form, the direction ratios of the line is represented as

$\bf\implies \:\vec{b_2} \: = \: 2 \hat{i} + \hat{j} \: - 2\hat{k}$

Now, further it is given that,

Line 1 and Line 2 are perpendicular to each other.

$\rm :\implies\:\vec{b_1} \: . \: \vec{b_2} = 0$

$\rm :\longmapsto\:\: (k \hat{i} - k \hat{j} \: + \hat{k}) \: . \: (\: 2 \hat{i} + \hat{j} \: - 2\hat{k}) = 0$

$\rm :\longmapsto\:2k - k - 2 = 0$

$\rm :\longmapsto\:k - 2 = 0$

$\bf\implies \:k = 2$

Additional Information :-

Let us consider two lines in vector form as

$\rm :\longmapsto\:\vec{r} = \vec{a_1} + \alpha \vec{b_1}$

and

$\rm :\longmapsto\:\vec{r} = \vec{a_2} + \beta \vec{b_2}$

then

1. Two lines are parallel iff

$\boxed{ \rm \: \vec{b_1} = k\vec{b_2}}$

2. Two lines are perpendicular iff

$\boxed{ \rm \: \vec{b_1} \: . \: \vec{b_2} \: = \: 0}$

3. Angle between two lines is

$\boxed{ \rm \: cos \theta \: = \frac{\vec{b_1} \: . \: \vec{b_2}}{ |\vec{b_1}| \: |\vec{b_2}| } }$