Question

The distance of the point \[-\widehat{i}+2\widehat{j}+6\widehat{k}\]from the straight line through the point (2,3,−4)and parallel to \[6\widehat{i}+3\widehat{j}-4\widehat{k}\]is
A 7
B 10
C 9
D None of these

Answer 1

Hint: In this question, first we need to form an equation of the line that passes through the point (2, 3,−4) (let’s say A) and parallel to \[6\widehat{i}+3\widehat{j}-4\widehat{k}\] such as
     \[\left( x\text{ }\text{ }{{a}_{1}} \right)/{{c}_{1}}\text{ }=\text{ }\left( y\text{ }\text{ }{{a}_{2}} \right)/{{c}_{2}}\text{ }=\text{ }\left( z\text{ }\text{ }{{a}_{3}} \right)/{{c}_{3}}\text{ }=\text{ }\overrightarrow{r}\], where\[{{c}_{1}}\], \[{{c}_{2}}\], and \[{{c}_{3}}\] are the direction ratio of line and \[{{a}_{1}}\], \[{{a}_{2}}\], and \[{{a}_{_{3}}}\] are the coordinates of a point on the line. Then we need to assume a point on the line (let’s say Q). Now find the distance between the point Q and P (P is the point to which distance from a straight line is needed) such as\[Q\text{ }\text{ }P\].

Formula Used: The formula to find the distance between two points P and Q is given as
\[\overrightarrow{PQ}\text{ }=\text{ }\sqrt{{{\left( {{\text{q}}_{1}}\text{ }-\text{ }{{\text{p}}_{1}} \right)}^{2}}+\text{ }{{\left( {{\text{q}}_{2}}\text{ }-\text{ }{{\text{p}}_{2}} \right)}^{2}}+\text{ }{{\left( \text{ }{{\text{q}}_{3}}-\text{ }{{\text{p}}_{3}} \right)}^{2}}}\], where\[{{p}_{1}}\],\[{{p}_{2}}\], and \[{{p}_{3}}\] are the coordinates of point P and\[{{q}_{1}}\],\[{{q}_{2}}\] , and \[{{q}_{3}}\] are the coordinates of the point Q.

Complete step by step solution: In this question, we need to find the distance of a point from the straight line which is passing through a point and parallel to another line.
The given point (let’s say P) through which distance is to be measured is given as
\[-\widehat{i}+2\widehat{j}+6\widehat{k}\]
Forming the equation of the straight line
The general equation of the straight line which is passing through the point (2,3,−4)and is parallel to \[6\widehat{i}+3\widehat{j}-4\widehat{k}\] is given as
\[\left( x\text{ }\text{ }{{a}_{1}} \right)/{{c}_{1}}\text{ }=\text{ }\left( y\text{ }\text{ }{{a}_{2}} \right)/{{c}_{2}}\text{ }=\text{ }\left( z\text{ }\text{ }{{a}_{3}} \right)/{{c}_{3}}\text{ }=\text{ }\overrightarrow{r}\], where\[{{c}_{1}}\], \[{{c}_{2}}\], and \[{{c}_{3}}\] are the direction ratio of line and \[{{a}_{1}}\], \[{{a}_{2}}\], and \[{{a}_{_{3}}}\] are the coordinates of a point on the line. As the point (2,3,−4) (let’s say S) is at the line so it represents \[{{a}_{1}}\], \[{{a}_{2}}\], \[{{a}_{_{3}}}\]and as the straight line is parallel to line \[6\widehat{i}+3\widehat{j}-4\widehat{k}\] so, its direction ratio (6, 3, -4) is the direction ratio (\[{{c}_{1}}\], \[{{c}_{2}}\], and \[{{c}_{3}}\]) of the straight line to which the distance of the point needs to determine.
\[\left( x\text{ }\text{ }2 \right)/6\text{ }=\text{ }\left( y\text{ }\text{ }3 \right)/3\text{ }=\text{ }\left( z\text{ }+\text{ }4 \right)/-4\text{ }=\text{ }\overrightarrow{r}\]

Take any arbitrary point on the straight line
Now assume any arbitrary point on the line, let’s say Q which is equal to (\[6\overrightarrow{r}\text{ }+2\], \[3\overrightarrow{r}\text{ }+\text{ }3\], \[-4\overrightarrow{r}\text{ }\text{ }4\]).

Finding the direction ratio of line joining points Q and P
Formula to find direction ratio of any line is given as
     \[\] \[Q\text{ }\text{ }P\text{ }=\text{ }6\overrightarrow{r}\text{ }+\text{ }3,\text{ }3\overrightarrow{r}\text{ }+\text{ }1,\text{ }-4\overrightarrow{r}\text{ }\text{ }10\](Direction ratio of line joining two points Q and P such as\[{{d}_{1}}\], \[{{d}_{2}}\], \[{{d}_{3}}\]).

8. Finding the value of \[\overrightarrow{r}\]
As the relationship between the direction ratios of two lines that are perpendicular to each other is given as
\[{{\text{c}}_{1}}{{\text{d}}_{1}}\text{ }+\text{ }{{\text{c}}_{2}}{{\text{d}}_{2}}\text{ }+\text{ }{{\text{c}}_{3}}{{\text{d}}_{3}}\text{ }=\text{ }0\]
\[6\left( 6\overrightarrow{r}\text{ }+\text{ }3 \right)\text{ }+\text{ }3\left( 3\overrightarrow{r}\text{ }+\text{ }1 \right)\text{ }\text{ }4\left( -\text{ }4\overrightarrow{r}\text{ }-\text{ }10 \right)\text{ }=\text{ }0\]
\[36\overrightarrow{r}\text{ }+\text{ }18\text{ }+\text{ }9\overrightarrow{r}\text{ }+\text{ }3\text{ }+\text{ }16\overrightarrow{r}\text{ }+\text{ }40\text{ }=\text{ }0\]
\[61\overrightarrow{r}\text{ }+\text{ }61\text{ }=\text{ }0\]
\[\overrightarrow{r}\text{ }=\text{ }-1\]
Putting the value of \[\overrightarrow{r}\]to find the point Q coordinates such as
\[6\overrightarrow{r}\text{ }+2\], \[3\overrightarrow{r}\text{ }+\text{ }3\], \[-4\overrightarrow{r}\text{ }\text{ }4\] (In point 2nd)
-4, 0, 0 are the coordinates of point Q.

     5. Finding the distance between the points P and Q
As we got the coordinate of points P and Q thus the distance between two points or separating between point and the straight line is given as
\[\overrightarrow{PQ}\text{ }=\text{ }\sqrt{{{\left( {{\text{q}}_{1}}\text{ }-\text{ }{{\text{p}}_{1}} \right)}^{2}}+\text{ }{{\left( {{\text{q}}_{2}}\text{ }-\text{ }{{\text{p}}_{2}} \right)}^{2}}+\text{ }{{\left( \text{ }{{\text{q}}_{3}}-\text{ }{{\text{p}}_{3}} \right)}^{2}}}\]
\[\overrightarrow{PQ}\text{ }=\text{ }\sqrt{{{\left( \text{-4 + 1} \right)}^{2}}+\text{ }{{\left( -2 \right)}^{2}}+\text{ }{{\left( \text{ -6} \right)}^{2}}}\]
\[\overrightarrow{PQ}\text{ }=\text{ }7\]

Option ‘A’ is correct

Note: It is important to note that the sum of the product of the direction ratio of lines that are perpendicular to each other is equal to zero whereas the sum of the product of the direction ratio of lines that are parallel to each other is equal to one because in case of the perpendicular cross product of the unit vector take place and cross product between same unit vector is always zero unlike in case of parallel lines.