Question

Find the coordinates of the foot of perpendicular and the length of the perpendicular drawn from a point $P\left( 5,4,2 \right)$ to the line $\vec{r}=-\hat{i}+3\hat{j}+\hat{k}+\mu \left( 2\hat{i}+3\hat{j}-\hat{k} \right)$. Also find the image of $P$ in this line.

Answer 1

Hint: Write the equation of line in the form $\dfrac{x-{{x}_{1}}}{l}=\dfrac{y-{{y}_{1}}}{m}=\dfrac{z-{{z}_{1}}}{n}=\mu$ and take any general point of the line which is the foot of perpendicular drawn from the point $P\left( 5,4,2 \right)$. Write the direction ratios of the line joining the point $P\left( 5,4,2 \right)$ to its foot of perpendicular on the line. Use the condition relating the direction ratios of two perpendicular lines to find the value of parameter $\mu$. Solve the equation to find the foot of perpendicular and then use distance formula to find the length of perpendicular. To find the image of point $P\left( 5,4,2 \right)$, use the fact that the foot of perpendicular is the midpoint of the point and its image in the line.

We have a line whose equation is $\vec{r}=-\hat{i}+3\hat{j}+\hat{k}+\mu \left( 2\hat{i}+3\hat{j}-\hat{k} \right)$ and point $P\left( 5,4,2 \right)$. We have to find the foot of perpendicular from this point to the line and the length of this perpendicular. We also have to find the image of this point in the line.
We can rewrite the equation of line as $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}=\left( 2\mu -1 \right)\hat{i}+\left( 3+3\mu \right)\hat{j}+\left( 1-\mu \right)\hat{k}$. By eliminating the parameter $\mu$, we have $\dfrac{x+1}{2}=\dfrac{y-3}{3}=\dfrac{z-1}{-1}=\mu$. Any point on this line can be of the form $\left( 2\mu -1,3+3\mu ,1-\mu \right)$.
Let’s assume the foot of perpendicular drawn from point $P\left( 5,4,2 \right)$ to this line is of form $Q\left( 2\mu -1,3+3\mu ,1-\mu \right)$. We will write the direction ratios of line $PQ$.
We know that direction ratios of any line joining points $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ are ${{x}_{2}}-{{x}_{1}},{{y}_{2}}-{{y}_{1}},{{z}_{2}}-{{z}_{1}}$.
Substituting ${{x}_{1}}=2\mu -1,{{y}_{1}}=3+3\mu ,{{z}_{1}}=1-\mu ,{{x}_{2}}=5,{{y}_{2}}=4,{{z}_{2}}=2$ in the above equation, we have $6-2\mu ,1-3\mu ,1+\mu$ as the direction ratios of line $PQ$.
We know that the direction ratios of line of the form $\dfrac{x-{{x}_{1}}}{l}=\dfrac{y-{{y}_{1}}}{m}=\dfrac{z-{{z}_{1}}}{n}=\mu$ is $l,m,n$. Thus, the direction ratios of line $\dfrac{x+1}{2}=\dfrac{y-3}{3}=\dfrac{z-1}{-1}=\mu$ is $2,3,-1$. This line is perpendicular to line $PQ$.
We know that if two lines with direction ratios ${{a}_{1}},{{b}_{1}},{{c}_{1}}$ and ${{a}_{2}},{{b}_{2}},{{c}_{2}}$ are perpendicular, then we have ${{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}=0$.
As line $\dfrac{x+1}{2}=\dfrac{y-3}{3}=\dfrac{z-1}{-1}=\mu$ is perpendicular to line $PQ$, we have $2\left( 6-2\mu \right)+3\left( 1-3\mu \right)-1\left( 1+\mu \right)=0$.
Simplifying the above equation, we have $14-14\mu =0$.
$\Rightarrow \mu =1$
Substituting the value $\mu =1$ in the coordinates of point $Q\left( 2\mu -1,3+3\mu ,1-\mu \right)$, we have $Q\left( 2\mu -1,3+3\mu ,1-\mu \right)=Q\left( 1,6,0 \right)$.
Thus, the foot of perpendicular drawn from point $P\left( 5,4,2 \right)$ to the line $\dfrac{x+1}{2}=\dfrac{y-3}{3}=\dfrac{z-1}{-1}=\mu$ is $Q\left( 1,6,0 \right)$.
We will now find the length of $PQ$.
We know that distance between two points of the form $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ is $\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}+{{\left( {{z}_{1}}-{{z}_{2}} \right)}^{2}}}$ units.
Substituting ${{x}_{1}}=1,{{y}_{1}}=6,{{z}_{1}}=0,{{x}_{2}}=5,{{y}_{2}}=4,{{z}_{2}}=2$ in the above formula, we have $\sqrt{{{\left( 1-5 \right)}^{2}}+{{\left( 6-4 \right)}^{2}}+{{\left( 0-2 \right)}^{2}}}=\sqrt{{{\left( -4 \right)}^{2}}+{{\left( 2 \right)}^{2}}+{{\left( -2 \right)}^{2}}}=\sqrt{24}=2\sqrt{6}$ units as the distance between points $P\left( 5,4,2 \right)$ and $Q\left( 1,6,0 \right)$.
We have to find the image of point $P\left( 5,4,2 \right)$ in the line $\dfrac{x+1}{2}=\dfrac{y-3}{3}=\dfrac{z-1}{-1}=\mu$. Let’s assume that the image of point $P\left( 5,4,2 \right)$ is $R\left( x,y,z \right)$. We observe that the point $Q\left( 1,6,0 \right)$ is the midpoint of line $PR$.
We know that midpoint of two points of the form $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)$ is $\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2},\dfrac{{{z}_{1}}+{{z}_{2}}}{2} \right)$.
Substituting ${{x}_{1}}=x,{{y}_{1}}=y,{{z}_{1}}=z,{{x}_{2}}=5,{{y}_{2}}=4,{{z}_{2}}=2$ in the above equation, we have $\left( \dfrac{x+5}{2},\dfrac{y+4}{2},\dfrac{z+2}{2} \right)$ as the midpoint of line $PR$ which is point $Q\left( 1,6,0 \right)$.
Thus, we have $\left( \dfrac{x+5}{2},\dfrac{y+4}{2},\dfrac{z+2}{2} \right)=\left( 1,6,0 \right)$.
Comparing the terms on both sides, we have $\dfrac{x+5}{2}=1,\dfrac{y+4}{2}=6,\dfrac{z+2}{2}=0$.
Simplifying the above equations by rearranging the terms, we have $x=-3,y=8,z=-2$.
Thus, the coordinates of $R\left( x,y,z \right)$ is $R\left( x,y,z \right)=R\left( -3,8,-2 \right)$.
Hence, the foot of perpendicular drawn from point $P\left( 5,4,2 \right)$ to the line is $Q\left( 1,6,0 \right)$ and the length of this perpendicular is $2\sqrt{6}$ units. Also, the image of point $P\left( 5,4,2 \right)$ in the line is $R\left( -3,8,-2 \right)$.

Note: One needs to be careful while taking points on the line and using the condition relating direction ratios of two perpendicular lines. We can’t solve this question without using the relation between direction ratios of two perpendicular lines. Also, one needs to keep in mind that the foot of perpendicular from the point on the line is the midpoint of the point and its image on the line.