Question

If the foot of perpendicular drawn from the point $(1,0,3)$ on a line passing through $(\alpha ,7,1)$ is $\left(\frac{5}{3},\frac{7}{3},\frac{17}{3}\right)$ then $\alpha$ is equal to:

Answer 1

Step 1: Solve for the direction ratios of the lines

Let the points be $P(1,0,3)$, $Q\left(\frac{5}{3},\frac{7}{3},\frac{17}{3}\right)$ and $A(\alpha ,7,1)$

We know that the direction ratios of a vector line joining two points $\left(x_1,y_1,z_1\right)$and $\left(x_2,y_2,z_2\right)$are $\left(x_2-x_1,y_2-y_1,z_2-z_1\right)$

$\therefore$Direction ratios of line $PQ=\left(\frac{5}{3}-1,\frac{7}{3}-0,\frac{17}{3}-3\right)=\left(\frac{2}{3},\frac{7}{3},\frac{8}{3}\right)$

Direction ratios of line $AQ=\left(\frac{5}{3}-\alpha ,\frac{7}{3}-7,\frac{17}{3}-1\right)=\left(\frac{5-3\alpha }{3},\frac{-14}{3},\frac{14}{3}\right)$

Step 2: Solve for the value of $\alpha$

Given that $PQ\perp AQ\Rightarrow \overrightarrow{PQ}.\overrightarrow{AQ}=0$

$\begin{array}{crcl}\Rightarrow & \left(\frac{2}{3}\right)\left(\frac{5-3\alpha }{3}\right)+\left(\frac{7}{3}\right)\left(\frac{-14}{3}\right)+\left(\frac{8}{3}\right)\left(\frac{14}{3}\right)& =& 0\\ \Rightarrow & \frac{10-6\alpha -98+112}{9}& =& 0\\ \Rightarrow & 6\alpha & =& 112-98+10\\ \Rightarrow & 6\alpha & =& 24\\ \Rightarrow & \alpha & =& \frac{24}{6}\\ \Rightarrow & \alpha & =& 4\end{array}$

Hence, the value of $\alpha$ is $4$