Question

If foot of the perpendicular from a point \( P(a,b,0) \) on the line \( \dfrac{x-1}{2}=\dfrac{y-2}{1}=\dfrac{z-\alpha}{3} \) is \( A \) and mid-point of \( AP \) is \( \left(0,\dfrac{3}{4},-\dfrac{1}{4}\right) \), then the value of \( \left(a^2+b^2+\alpha^2\right) \) is -

Hint:

In 3D geometry, when a point is the foot of the perpendicular on a line, always use two conditions together: the point lies on the line, and the joining vector is perpendicular to the line's direction vector.

Answer 1

Correct Answer: 1

Step 1: Write the coordinates of point \( A \) using midpoint formula.
Let the foot of the perpendicular be \( A(x_1,y_1,z_1) \).
Since the midpoint of \( AP \) is \( \left(0,\dfrac{3}{4},-\dfrac{1}{4}\right) \), we use midpoint formula:
\[ \left(\frac{a+x_1}{2},\frac{b+y_1}{2},\frac{0+z_1}{2}\right)=\left(0,\frac{3}{4},-\frac{1}{4}\right) \] Comparing coordinates, we get
\[ a+x_1=0 \Rightarrow x_1=-a \] \[ b+y_1=\frac{3}{2} \Rightarrow y_1=\frac{3}{2}-b \] \[ z_1=-\frac{1}{2} \] So, coordinates of \( A \) are
\[ A\left(-a,\frac{3}{2}-b,-\frac{1}{2}\right) \]
Step 2: Parametric form of the given line.
The line is given by
\[ \frac{x-1}{2}=\frac{y-2}{1}=\frac{z-\alpha}{3}=t \] So any point on the line is
\[ x=1+2t,\quad y=2+t,\quad z=\alpha+3t \] Since \( A \) lies on the line, we have
\[ -a=1+2t \] \[ \frac{3}{2}-b=2+t \] \[ -\frac{1}{2}=\alpha+3t \] From the second equation,
\[ -b=2+t-\frac{3}{2}=\frac{1}{2}+t \] \[ b=-\frac{1}{2}-t \] From the first equation,
\[ a=-1-2t \]
Step 3: Use the perpendicular condition.
Since \( A \) is the foot of the perpendicular from \( P \) to the line, vector \( \overrightarrow{AP} \) is perpendicular to the direction vector of the line.
Direction vector of the line is
\[ \vec{d}=(2,1,3) \] Now,
\[ \overrightarrow{AP}=P-A=\left(a-(-a),\,b-\left(\frac{3}{2}-b\right),\,0-\left(-\frac{1}{2}\right)\right) \] \[ \overrightarrow{AP}=\left(2a,\,2b-\frac{3}{2},\,\frac{1}{2}\right) \] Since \( \overrightarrow{AP} \perp \vec{d} \), their dot product is zero:
\[ \left(2a,\,2b-\frac{3}{2},\,\frac{1}{2}\right)\cdot(2,1,3)=0 \] \[ 4a+\left(2b-\frac{3}{2}\right)+\frac{3}{2}=0 \] \[ 4a+2b=0 \] \[ 2a+b=0 \] Now substitute \( a=-1-2t \) and \( b=-\frac{1}{2}-t \):
\[ 2(-1-2t)+\left(-\frac{1}{2}-t\right)=0 \] \[ -2-4t-\frac{1}{2}-t=0 \] \[ -\frac{5}{2}-5t=0 \] \[ t=-\frac{1}{2} \]
Step 4: Find \( a \), \( b \), and \( \alpha \).
Substitute \( t=-\dfrac{1}{2} \) into the expressions:
\[ a=-1-2\left(-\frac{1}{2}\right)=-1+1=0 \] \[ b=-\frac{1}{2}-\left(-\frac{1}{2}\right)=0 \] Now from
\[ -\frac{1}{2}=\alpha+3t \] \[ -\frac{1}{2}=\alpha+3\left(-\frac{1}{2}\right) \] \[ -\frac{1}{2}=\alpha-\frac{3}{2} \] \[ \alpha=1 \]
Step 5: Calculate \( a^2+b^2+\alpha^2 \).
\[ a^2+b^2+\alpha^2=0^2+0^2+1^2=1 \]