Question

If the distance of the point \(P(4\alpha,\alpha,\beta)\), \(\beta<0\), from the line \[ \vec r = 4\hat i-\hat k+\mu(2\hat i+3\hat k),\ \mu\in\mathbb{R}, \] along a line with direction ratios \(3,-1,0\) is \(\dfrac{13}{\sqrt{10}}\), then \(\alpha^2+\beta^2\) is equal to _______.

Hint:

When distance is measured along a given direction, always project the joining vector on that direction.

Answer 1

Correct Answer: 2.4

Step 1: Data from the problem

A point on the given line:
\[ A(4,0,-1) \]
Direction vector of the line:
\[ \vec d_1=(2,0,3) \]
Direction along which distance is measured:
\[ \vec d_2=(3,-1,0),\quad |\vec d_2|=\sqrt{10} \]
Point:
\[ P(4\alpha,\alpha,\beta) \]

Step 2: Vector joining line to point

\[ \vec{AP}=(4\alpha-4,\alpha,\beta+1) \]
Distance along \( \vec d_2 \):
\[ \frac{|\vec{AP}\cdot \vec d_2|}{|\vec d_2|} =\frac{13}{\sqrt{10}} \Rightarrow |\vec{AP}\cdot \vec d_2|=13 \]
\[ |(4\alpha-4)3+\alpha(-1)|=13 \Rightarrow |11\alpha-12|=13 \]
Since \( \beta<0 \),
\[ 11\alpha-12=-13 \Rightarrow \alpha=-\frac{1}{11} \]

Step 3: Perpendicularity condition

\[ \vec{AP}\cdot\vec d_1=0 \]
\[ (4\alpha-4)2+(\beta+1)3=0 \]
Substitute \( \alpha=-\frac{1}{11} \):
\[ \beta=-\frac{17}{11} \]

Step 4: Required value

\[ \alpha^2+\beta^2=\frac{1}{121}+\frac{289}{121} =\frac{290}{121} \]

Final Answer:
\[ \boxed{\dfrac{290}{121}} \]