Let L_1:(x-1)/(3)=(y-1)/(-1)=(z+1)/(0) and L_2:(x-2)/(2)=(y)/(0)=(z+4)/( ), R, be two line…

Question

Let ${L}_{1}:\frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}$ and ${L}_{2}:\frac{x-2}{2}=\frac{y}{0}=\frac{z+4}{\alpha },\alpha \in R$, be two lines, which intersect at the point $B$. If $P$ is the foot of perpendicular from the point $A(1,1,-1)$ on ${L}_{2}$, then the value of $26\alpha (PB{)}^{2}$ is \_\_\_\_\_\_\_\_\_ .

Answer: 216

Accepted Answer

Explanation

To find the value of $26\alpha (PB{)}^{2}$, we proceed as follows:

Intersection Point $B$ of ${L}_{1}$ and ${L}_{2}$

We are given the equations of the lines:

${L}_{1}:\frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}$

${L}_{2}:\frac{x-2}{2}=\frac{y}{0}=\frac{z+4}{\alpha }$

For ${L}_{1}$, $z+1=0$, which implies $z=-1$.

For ${L}_{2}$, since $y=0$, the direction ratios can be matched using the parameter $\mu $:

$\begin{gathered} x=3\lambda +1,\, y=-\lambda +1,\, z=-1, \ x=2\mu +2,\, y=0,\, z=\alpha \mu -4. \end{gathered}$

Setting the coordinates equal for intersection:

$\begin{gathered} 3\lambda +1=2\mu +2, \ -\lambda +1=0, \ -1=\alpha \mu -4. \end{gathered}$

From $-\lambda +1=0$, we find $\lambda =1$.

Substituting $\lambda =1$ into $3\lambda +1=2\mu +2$ gives:

$3(1)+1=2\mu +2\, \implies \, \mu =1.$

Also, substituting $\mu =1$ into $-1=\alpha \mu -4$ gives:

$-1=\alpha (1)-4\, \implies \, \alpha =3.$

So, the intersection point $B$ is:

$B(4,0,-1).$

Foot of Perpendicular from $A$ to ${L}_{2}$

The point $P$ on ${L}_{2}$ is given by:

$P(2\delta +2,0,3\delta -4).$

For vector $\overset{	o \leftarrow }{AP}=(2\delta +1,-1,3\delta -3)$, since $AP$ is perpendicular to ${L}_{2}$, the dot product should be zero:

$\begin{gathered} (2\delta +1)\cdot 2+(-1)\cdot 0+(3\delta -3)\cdot 3 =0. \end{gathered}$

Simplifying gives:

$4\delta +2+9\delta -9=0\, \implies \, 13\delta =7\, \implies \, \delta =\frac{7}{13}.$

So, the coordinates of point $P$ are:

$P(\frac{40}{13},0,\frac{-31}{13}).$

Distance $PB$ and Calculation

The vector $\overset{	o \leftarrow }{PB}$ is:

$\overset{	o \leftarrow }{PB}=(4-\frac{40}{13},0-0,-1-(\frac{-31}{13})).$

Calculating the components:

$\overset{	o \leftarrow }{PB}=(\frac{12}{13},0,\frac{18}{13}).$

The square of the distance $(PB{)}^{2}$ is:

${(\frac{12}{13})}^{2}+{(0)}^{2}+{(\frac{18}{13})}^{2}=\frac{144}{169}+\frac{324}{169}=\frac{468}{169}.$

Thus, $26\alpha (PB{)}^{2}$ is:

$26	imes 3	imes \frac{468}{169}=216.$

So, the final value is 216.

Solutions · 1