The line L_1 is parallel to the vector a=-3i+2j+4k and passes through the point (7,6,2) an…

Question

The line ${L}_{1}$ is parallel to the vector $\overset{	o }{a}=-3\hat{i}+2\hat{j}+4\hat{k}$ and passes through the point $(7,6,2)$ and the line ${L}_{2}$ is parallel to the vector $\overset{	o }{b}=2\hat{i}+\hat{j}+3\hat{k}$ and passes through the point $(5,3,4)$. The shortest distance between the lines ${L}_{1}$ and ${L}_{2}$ is :

Accepted Answer

The line ${L}_{1}$ is parallel to the vector $\overset{	o }{a}=-3\hat{i}+2\hat{j}+4\hat{k}$ and passes through the point $(7,6,2)$. The line ${L}_{2}$ is parallel to the vector $\overset{	o }{b}=2\hat{i}+\hat{j}+3\hat{k}$ and passes through the point $(5,3,4)$.

Equations of the Lines:

Equation of ${L}_{1}$:

$r=7\hat{i}+6\hat{j}+2\hat{k}+\lambda (-3\hat{i}+2\hat{j}+4\hat{k})$

Equation of ${L}_{2}$:

$r=5\hat{i}+3\hat{j}+4\hat{k}+\mu (2\hat{i}+\hat{j}+3\hat{k})$

Shortest Distance Between ${L}_{1}$ and ${L}_{2}$:

The formula for the shortest distance between two skew lines is given by:

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

Calculations:

Vector between the points:

${\vec{a}}_{2}-{\vec{a}}_{1}=(5\hat{i}+3\hat{j}+4\hat{k})-(7\hat{i}+6\hat{j}+2\hat{k})=-2\hat{i}-3\hat{j}+2\hat{k}$

Cross product of direction vectors:

${\vec{b}}_{1}=-3\hat{i}+2\hat{j}+4\hat{k},\, {\vec{b}}_{2}=2\hat{i}+\hat{j}+3\hat{k}$

${\vec{b}}_{1}	imes {\vec{b}}_{2}=\begin{vmatrix} \hat{i} \quad  \hat{j} \quad  \hat{k} \ -3 \quad  2 \quad  4 \ 2 \quad  1 \quad  3 \end{vmatrix}=2\hat{i}+17\hat{j}-7\hat{k}$

Magnitude of the cross product:

$|{\vec{b}}_{1}	imes {\vec{b}}_{2}|=\sqrt{{2}^{2}+{17}^{2}+(-7{)}^{2}}=\sqrt{342}$

Dot product:

$({\vec{a}}_{2}-{\vec{a}}_{1})\cdot ({\vec{b}}_{1}	imes {\vec{b}}_{2})=(-2\hat{i}-3\hat{j}+2\hat{k})\cdot (2\hat{i}+17\hat{j}-7\hat{k})$

$=(-2)(2)+(-3)(17)+(2)(-7)=-4-51-14=-69$

Shortest Distance:

$=|\frac{-69}{\sqrt{342}}|=\frac{23}{\sqrt{38}}$

Solutions · 1