Question

The distance of the point having position vector $-\hat{i}+2\hat{j}+6\hat{k}$ from the straight line passing through the point ( 2, 3, -4 ) and parallel to the vector $6\hat{i}+3\hat{j}-4\hat{k}$ is:
(a) 7
(b) $4\sqrt{3}$
(c) $2\sqrt{13}$
(d) 6

Answer 1

to solve this question, we will use the concept of the condition of parallel lines that their direction cosines are equal and the concept of perpendicular vectors that their dot product is zero. And, using formulas such as $\left( a\hat{i}+b\hat{j}+c\hat{k} \right)\cdot \left( d\hat{i}+e\hat{j}+f\hat{k} \right)=ad+be+cf$ and distance formula $XY=\sqrt{{{(d-a)}^{2}}+{{(e-b)}^{2}}+{{(f-c)}^{2}}}$ we will find out the distance of point A from line.

Complete step by step answer:
As, point is given with position vector $-\hat{i}+2\hat{j}+6\hat{k}$, writing point in coordinate form, we get point A( -1, 2, 6 ).
Now, we are given a line on which point ( 2, 3, -4 ) lies and has a parallel vector $6\hat{i}+3\hat{j}-4\hat{k}$.
Now, we know two parallel lines have the same direction cosines. So, equation of line through on which point ( 2, 3, -4 ) lies and has parallel vector $6\hat{i}+3\hat{j}-4\hat{k}$ will be $\dfrac{x-2}{6}=\dfrac{y-3}{3}=\dfrac{z+4}{-4}$ as we know that, the standard form of equation $\dfrac{x-a}{d}=\dfrac{y-b}{e}=\dfrac{z-c}{f}$, where d, e, f are direction ratios and ( a, b, c ) lying on line.
Now, let assume there is point B on line $\dfrac{x-2}{6}=\dfrac{y-3}{3}=\dfrac{z+4}{-4}=k$ with parameter k, and AB be the perpendicular.
Then, we are asked to find the distance of point A from line $\dfrac{x-2}{6}=\dfrac{y-3}{3}=\dfrac{z+4}{-4}=k$ and we know distance of any point from a line is the perpendicular distance.
So, the distance of point A( -1, 2, 6 ) will be the length of line AB.
So, coordinates of point B in terms of k will be B ( 6k + 2, 3k + 3, -4k – 4 )
Then, $AB=\vec{B}-\vec{A}$
Or, $AB=(6k+2+1)\hat{i}+(3k+3-2)\hat{j}+(-4k-4-6)\hat{k}$
On simplifying, we get
$AB=(6k+3)\hat{i}+(3k+1)\hat{j}+(-4k-10)\hat{k}$
Now, as line AB and $6\hat{i}+3\hat{j}-4\hat{k}$are perpendicular to each other, then we know that the dot product of two perpendicular lines is 0.
So, $\left( (6k+3)\hat{i}+(3k+1)\hat{j}+(-4k-10)\hat{k} \right)\cdot \left( 6\hat{i}+3\hat{j}-4\hat{k} \right)=0$
We know that, $\left( a\hat{i}+b\hat{j}+c\hat{k} \right)\cdot \left( d\hat{i}+e\hat{j}+f\hat{k} \right)=ad+be+cf$
So, $\left( (6k+3)6+(3k+1)3+(-4k-10)(-4) \right)=0$
On simplifying, we get
36k + 18 + 9k +3 + 16k + 40 = 0
On solving we get
61k + 61 = 0
k = - 1
Now, we have B ( 6k + 2, 3k + 3, -4k – 4 )
Putting, k = - 1 in B ( 6k + 2, 3k + 3, -4k – 4 ), we get
B ( - 6 + 2, - 3 + 3, 4 – 4 )
Or, B( -4, 0, 0 ) and we have A( -1, 2, 6),
If we have two vectors, $X=a\hat{i}+b\hat{j}+c\hat{k}$ and $Y=d\hat{i}+e\hat{j}+f\hat{k}$ then distance $XY=\sqrt{{{(d-a)}^{2}}+{{(e-b)}^{2}}+{{(f-c)}^{2}}}$
So, $AB=\sqrt{{{(-4+1)}^{2}}+{{(0-2)}^{2}}+{{(0-6)}^{2}}}$
$AB=\sqrt{49}$
AB = 7 units

So, the correct answer is “Option A”.

Note: Always convert the equation of line firstly to standard line equation which is $\dfrac{x-a}{d}=\dfrac{y-b}{e}=\dfrac{z-c}{f}$, where d, e, f are direction ratios and ( a, b, c ) lying on line. Always remember that if we have two vectors, $X=a\hat{i}+b\hat{j}+c\hat{k}$ and $Y=d\hat{i}+e\hat{j}+f\hat{k}$ then distance $XY=\sqrt{{{(d-a)}^{2}}+{{(e-b)}^{2}}+{{(f-c)}^{2}}}$. Avoid calculation error while solving the question.