If $$ \vec{F}=\uparrow+k j $$ acts at point $$ A(2,3) $$ and the length of the perpendicular drawn from $$ B(2,1) $$ to the line of action of force $$ \vec{F} $$ equals $$ \frac{2 \sqrt{5}}{5} $$, then $$ k= $$ $$ \begin{array}{llll} ext { (a) } \pm 1 & ext { (b) } \pm 2 & ext { (c) } \pm 3 & ext { (d) } \pm 4\end{array} $$

Question

If $$ \vec{F}=\uparrow+k j $$ acts at point $$ A(2,3) $$ and the length of the perpendicular drawn from $$ B(2,1) $$ to the line of action of force $$ \vec{F} $$ equals $$ \frac{2 \sqrt{5}}{5} $$, then $$ k= $$ $$ \begin{array}{llll}	ext { (a) } \pm 1 & 	ext { (b) } \pm 2 & 	ext { (c) } \pm 3 & 	ext { (d) } \pm 4\end{array} $$

UpStudy AI · Accepted Answer

To determine the value of $$ k $$, let's analyze the given information step by step.

**Given:**
- The force vector is $$ \vec{F} = \mathbf{i} + k \mathbf{j} $$.
- The force acts at point $$ A(2, 3) $$.
- The perpendicular distance from point $$ B(2, 1) $$ to the line of action of the force $$ \vec{F} $$ is $$ \frac{2 \sqrt{5}}{5} $$.

**Steps to Solve:**

1. **Equation of the Line of Action:**
   
   The line of action of the force $$ \vec{F} $$ passing through point $$ A(2, 3) $$ can be expressed parametrically as:
   $$
   x = 2 + t, \quad y = 3 + kt
   $$
   Eliminating the parameter $$ t $$:
   $$
   t = x - 2 \implies y = 3 + k(x - 2)
   $$
   So, the equation of the line becomes:
   $$
   y = kx + (3 - 2k)
   $$
   Rearranging into standard form:
   $$
   kx - y + (3 - 2k) = 0
   $$

2. **Distance from Point $$ B(2, 1) $$ to the Line:**

The formula for the perpendicular distance $$ d $$ from a point $$ (x_0, y_0) $$ to the line $$ Ax + By + C = 0 $$ is:
   $$
   d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}
   $$
   Applying this to our line and point $$ B(2, 1) $$:
   $$
   d = \frac{|k \cdot 2 - 1 + (3 - 2k)|}{\sqrt{k^2 + (-1)^2}} = \frac{|2k - 1 + 3 - 2k|}{\sqrt{k^2 + 1}} = \frac{2}{\sqrt{k^2 + 1}}
   $$
   
   We know this distance is $$ \frac{2 \sqrt{5}}{5} $$:
   $$
   \frac{2}{\sqrt{k^2 + 1}} = \frac{2 \sqrt{5}}{5}
   $$
   
   Simplifying:
   $$
   \frac{1}{\sqrt{k^2 + 1}} = \frac{\sqrt{5}}{5} \implies \sqrt{k^2 + 1} = \frac{5}{\sqrt{5}} = \sqrt{5}
   $$
   
   Squaring both sides:
   $$
   k^2 + 1 = 5 \implies k^2 = 4 \implies k = \pm 2
   $$

**Conclusion:**
The value of $$ k $$ can be either $$ +2 $$ or $$ -2 $$.

**Answer:**  
Option (b) $$ \pm 2 $$
$$ k = \pm 2 $$

If acts at point and the length of the perpendicular drawn from
to the line of action of force equals , then

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