Question

What is the distance of the point $B$ with P.V. $\hat i + 2\hat j + 3\hat k$ from the line through A with P.V. $4\hat i + 2\hat j + 2\hat k$ and parallel to the vector $2\hat i + 3\hat j + 6\hat k$?
A) $\sqrt {10}$
B) $\sqrt 5$
C) $\sqrt 6$
D) $2$

Answer 1

In this question, we have to find the distance between a point and a line. First, find the equation of the line using the given passing point and parallel vector. Then, assume a foot of perpendicular and find an equation joining the point B and foot of perpendicular. This line and the given line are now perpendicular. Use dot product to find the value of $\lambda$. Put this in the equation and find the required distance.

Complete step-by-step solution:
We have to find the distance between a point and a line. Position vector of the point B is given and we have to form an equation of the line.
For line, we are given a passing through point and a vector to which the line is parallel. To form an equation, we need a passing point and direction. We will use the direction of the parallel line. Using this information, we can form the equation of the line using the formula –
$\Rightarrow \vec r = \vec a + \lambda \vec d$, where $\vec a$ is the passing point of the line and $\vec d$ is the line parallel to the required line.
Here, $\vec a = 4\hat i + 2\hat j + 2\hat k$ and $\vec d = 2\hat i + 3\hat j + 6\hat k$. Putting in the formula –
$\Rightarrow \vec r = 4\hat i + 2\hat j + 2\hat k + \lambda \left( {2\hat i + 3\hat j + 6\hat k} \right)$
Now, we have an equation of a line and a point. We have to find the distance between the two of them.
Let D be the foot of perpendicular and B be the point from which the distance is to be found.
$\Rightarrow$ $\vec B$ $= \hat i + 2\hat j + 3\hat k$
$\Rightarrow \vec B\vec D =$ equation of line – position vector of B
Substitute we get,
$\Rightarrow \vec B\vec D = 4\hat i + 2\hat j + 2\hat k + \lambda \left( {2\hat i + 3\hat j + 6\hat k} \right) - \left( {\hat i + 2\hat j + 3\hat k} \right)$
-$\Rightarrow \vec B\vec D = 3\hat i - \hat k + \lambda \left( {2\hat i + 3\hat j + 6\hat k} \right)$ ………………... (1)
Now, our line $\vec r$ is perpendicular to $\vec B\vec D$. Hence, their dot product will be 0.
$\Rightarrow \vec B\vec D.\vec r = 0$
$\Rightarrow 3\hat i - \hat k + \lambda \left( {2\hat i + 3\hat j + 6\hat k} \right).\left( {2\hat i + 3\hat j + 6\hat k} \right) = 0$
Simplifying we get,
$\Rightarrow \left( {3 + 2\lambda } \right)2 + \left( {3\lambda } \right)3 + \left( { - 1 + 6\lambda } \right)6 = 0$
On simplifying we will get the value of $\lambda$,
$\Rightarrow 6 + 4\lambda + 9\lambda - 6 + 36\lambda = 0$
Add and subtract the terms,
$\Rightarrow 49\lambda = 0$
Hence,
$\Rightarrow \lambda = 0$
Putting this in equation (1),
$\Rightarrow \vec B\vec D = 3\hat i - \hat k$
Now, BD is the required distance. Finding its magnitude,
$\Rightarrow \left| {\left. {\vec B\vec D} \right|} \right. = \left| {\left. {\sqrt {{3^2} + {{\left( { - 1} \right)}^2}} } \right|} \right.$
Hence,
$\Rightarrow \left| {\left. {\vec B\vec D} \right|} \right. = \sqrt {10}$

$\therefore$ The required distance is option A) $\sqrt {10}$.

Note: We have to know that, in order to find the foot of perpendicular, you can use the following formula –
$a - \left( {\dfrac{{\left( {a - p} \right).b}}{{|b{|^2}}}} \right)b$, where distance is to be found between the point P with position vector p and a line $a = r + \lambda b$. After this, you just have to find the distance between the point and foot of the perpendicular.