The shortest distance between the lines (x-4)/(1)=(y-3)/(2)=(z-2)/(-3) and (x+2)/(2)=(y-6)…

Question

The shortest distance between the lines $\frac{x-4}{1}=\frac{y-3}{2}=\frac{z-2}{-3}$ and $\frac{x+2}{2}=\frac{y-6}{4}=\frac{z-5}{-5}$ is:

Accepted Answer

For the two lines

$\frac{x-4}{1}=\frac{y-3}{2}=\frac{z-2}{-3}$

and

$\frac{x+2}{2}=\frac{y-6}{4}=\frac{z-5}{-5},$

we use the NCERT formula for the shortest distance between two skew lines:

$=\frac{|(\vec{{a}_{2}}-\vec{{a}_{1}})\cdot (\vec{{b}_{1}}	imes \vec{{b}_{2}})|}{|\vec{{b}_{1}}	imes \vec{{b}_{2}}|}$

where:

$\vec{{a}_{1}}$ is a point on the first line

$\vec{{a}_{2}}$ is a point on the second line

$\vec{{b}_{1}},\vec{{b}_{2}}$ are direction vectors of the two lines

Identify points and direction vector: 1: Identify points and direction vectors

From the first line,

$\frac{x-4}{1}=\frac{y-3}{2}=\frac{z-2}{-3}$

a point on it is

$A(4,3,2)$

and its direction vector is

$\vec{{b}_{1}}=\hat{i}+2\hat{j}-3\hat{k}$

From the second line,

$\frac{x+2}{2}=\frac{y-6}{4}=\frac{z-5}{-5}$

a point on it is

$B(-2,6,5)$

and its direction vector is

$\vec{{b}_{2}}=2\hat{i}+4\hat{j}-5\hat{k}$

Find $\vec{AB}: 2: Find $\vec{AB}$

$\vec{AB}=B-A=(-2-4,6-3,5-2)=(-6,3,3)$

So,

$\vec{AB}=-6\overset{\^}{i}+3\overset{\^}{j}+3\overset{\^}{k}$

Find $\vec{{b}\_{1}}	imes \vec{{b}\_{2}}: 3: Find $\vec{{b}_{1}}	imes \vec{{b}_{2}}$

$\vec{{b}_{1}}	imes \vec{{b}_{2}}=\begin{vmatrix} \hat{i} \quad  \hat{j} \quad  \hat{k} \ 1 \quad  2 \quad  -3 \ 2 \quad  4 \quad  -5 \end{vmatrix}$

Expanding,

$=\hat{i}(2(-5)-(-3)(4))-\hat{j}(1(-5)-(-3)(2))+\hat{k}(1\cdot 4-2\cdot 2)$

$=\hat{i}(-10+12)-\hat{j}(-5+6)+\hat{k}(4-4)$

$=2\hat{i}-\hat{j}+0\hat{k}$

So,

$\vec{{b}_{1}}	imes \vec{{b}_{2}}=(2,-1,0)$

Its magnitude is

$|\vec{{b}_{1}}	imes \vec{{b}_{2}}|=\sqrt{{2}^{2}+(-1{)}^{2}+{0}^{2}}=\sqrt{5}$

Find scalar triple produc: 4: Find scalar triple product

$\vec{AB}\cdot (\vec{{b}_{1}}	imes \vec{{b}_{2}})=(-6,3,3)\cdot (2,-1,0)$

$=(-6)(2)+(3)(-1)+(3)(0)$

$=-12-3=-15$

So,

$|\vec{AB}\cdot (\vec{{b}_{1}}	imes \vec{{b}_{2}})|=15$

Find shortest distanc: 5: Find shortest distance

$d=\frac{15}{\sqrt{5}}=3\sqrt{5}$

Hence, the shortest distance is

$3\sqrt{5}$

So, the correct option is:

Solutions · 1