Let a line l pass through the origin and be perpendicular to the lines l_1:r=(-11-7k)+ (i+…

Question

Let a line $l$ pass through the origin and be perpendicular to the lines

${l}_{1}:\vec{r}=(\hat{\imath}-11\hat{\jmath}-7\hat{k})+\lambda (\hat{i}+2\hat{\jmath}+3\hat{k}),\lambda \in R$ and

${l}_{2}:\vec{r}=(-\hat{\imath}+\hat{k})+\mu (2\hat{\imath}+2\hat{\jmath}+\hat{k}),\mu \in R$.

If $P$ is the point of intersection of $l$ and ${l}_{1}$, and $Q(\propto ,\beta ,\gamma )$ is the foot of perpendicular from P on ${l}_{2}$, then $9(\alpha +\beta +\gamma )$ is equal to \_\_\_\_\_\_\_\_\_\_\_\_\_.

Answer: 5

Accepted Answer

We have,

${l}_{1}:\vec{r}=(\hat{\imath}-11\hat{\jmath}-7\hat{k})+\lambda (\hat{i}+2\hat{\jmath}+3\hat{k}),\lambda \in R$ and

${l}_{2}:\vec{r}=(-\hat{\imath}+\hat{k})+\mu (2\hat{\imath}+2\hat{\jmath}+\hat{k}),\mu \in R$.

Let direction ratio of line $l$ be $a,b$ and $c$ Equation of line $l$

$\begin{gathered} \vec{r} =(0\hat{i}+0\hat{j}+0\hat{k})+\delta (a\hat{i}+b\hat{j}+c\hat{k}) \ \ =\delta (a\hat{i}+b\hat{j}+c\hat{k}) \end{gathered}$

As, line $l$ is perpendicular to ${l}_{1}$ and ${l}_{2}$,

$a\hat{i}+b\hat{j}+c\hat{k}=\begin{vmatrix} \hat{i} \quad  \hat{j} \quad  \hat{k} \ 1 \quad  2 \quad  3 \ 2 \quad  2 \quad  1 \end{vmatrix}=-4\hat{i}+5\hat{j}-2\hat{k}$

$	herefore $ Equation of line $l:\vec{r}=\delta (-4\hat{i}+5\hat{j}-2\hat{k})$

As, $P$ is the intersecting point of $l$ and ${l}_{1}$

$-4\delta =1+\lambda ,5\delta =-11+2\lambda ,-2\delta =-7+3\lambda $

After solving the above three equation, we get

$\delta =-1\lambda =3$

$	herefore $ Co-ordinate of point $P$ is $(4,-5,2)$.

$Q$ is a point on line ${l}_{2}$

Let co-ordinate of $Q$ be $(-1+2\mu ,2\mu ,1+\mu )$

$\begin{gathered} \overset{	o \leftarrow }{PQ}=(-5+2\mu )\hat{i}+(2\mu +5)\hat{j}+(\mu -1)\hat{k} \ \overset{	o \leftarrow }{PQ}\cdot (2\hat{i}+2\hat{j}+\hat{k})=0[\because \overset{	o \leftarrow }{PQ}\perp {l}_{2}] \ 2(-5+2\mu )+2(2\mu +5)+\mu -1=0 \ 9\mu -1=0\Rightarrow \mu =\frac{1}{9} \end{gathered}$

$\begin{gathered} 	herefore \alpha =-1+\frac{2}{9}=\frac{-7}{9},\beta =2	imes \frac{1}{9}=\frac{2}{9},r=1+\frac{1}{9}=\frac{10}{9} \ 9(\alpha +\beta +\gamma )=9(-\frac{7}{9}+\frac{2}{9}+\frac{10}{9})=5 \end{gathered}$

Solutions · 1