Question

A point on a line has coordinates \[\left( {p + 1,p - 3,\sqrt 2 p} \right)\] where \[p\] is any real number. What are the direction cosines of the line?
(A) \[\dfrac{1}{2},\dfrac{1}{2},\dfrac{1}{{\sqrt 2 }}\]
(B) \[\dfrac{1}{2},\dfrac{1}{2},\dfrac{1}{2}\]
(C) \[\dfrac{1}{2},\dfrac{1}{2}, - \dfrac{1}{2}\]
(D) cannot be determined due to insufficient data

Answer 1

Hint: To solve this question, first we let the given coordinates as \[x,y\] and \[z\] and find the value of \[p\] and put it as equation \[\left( 1 \right),\left( 2 \right)\] and \[\left( 3 \right)\]. Then we compare all the equations. After that, we will find the direction ratios using the concept if a cartesian equation of line is given i.e., \[\dfrac{{x - {x_1}}}{a} = \dfrac{{y - {y_1}}}{b} = \dfrac{{z - {z_1}}}{c}\] then \[a,b\] and \[c\] are direction ratios. After that, we will use the formula of direction cosines to get the result.
Formula used:
\[l = \dfrac{a}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\] , \[m = \dfrac{b}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\] , \[n = \dfrac{c}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]
where \[l,m\] and \[n\] are the direction cosines.

Complete step-by-step solution:
Given coordinates of point, \[\left( {p + 1,p - 3,\sqrt 2 p} \right)\]
Now, let \[p + 1 = x\]
\[ \Rightarrow p = x - 1{\text{ }} - - - \left( 1 \right)\]
Similarly, \[p - 3 = y\]
\[ \Rightarrow p = y + 3{\text{ }} - - - \left( 2 \right)\]
Also, \[\sqrt 2 p = z\]
\[ \Rightarrow p = \dfrac{z}{{\sqrt 2 }}{\text{ }} - - - \left( 3 \right)\]
From equation \[\left( 1 \right),\left( 2 \right)\] and \[\left( 3 \right)\] if we compare, we see that all the left sides are equal, so the right sides will also get equal.
\[ \Rightarrow x - 1 = y + 3 = \dfrac{z}{{\sqrt 2 }} = p\]
\[ \Rightarrow \dfrac{{x - 1}}{1} = \dfrac{{y + 3}}{1} = \dfrac{z}{{\sqrt 2 }} = p\]
which can also be written as,
\[ \Rightarrow \dfrac{{x - 1}}{1} = \dfrac{{y + 3}}{1} = \dfrac{{z - 0}}{{\sqrt 2 }} = p{\text{ }} - - - \left( 4 \right)\]
Now, we know that
If \[\dfrac{{x - {x_1}}}{a} = \dfrac{{y - {y_1}}}{b} = \dfrac{{z - {z_1}}}{c}\] then \[a,b\] and \[c\] are direction ratios
\[\therefore \] from equation \[\left( 4 \right)\]
\[a = 1,{\text{ }}b = 1,{\text{ }}c = \sqrt 2 \]
So, direction ratios are \[1,{\text{ }}1,{\text{ }}\sqrt 2 \]
Now we will find the direction cosines using,
\[l = \dfrac{a}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\] , \[m = \dfrac{b}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\] , \[n = \dfrac{c}{{\sqrt {{a^2} + {b^2} + {c^2}} }}\]
where \[l,m\] and \[n\] are the direction cosines.
\[\therefore l = \dfrac{1}{{\sqrt {{1^2} + {1^2} + {{\left( {\sqrt 2 } \right)}^2}} }},{\text{ }}m = \dfrac{1}{{\sqrt {{1^2} + {1^2} + {{\left( {\sqrt 2 } \right)}^2}} }},{\text{ }}n = \dfrac{{\sqrt 2 }}{{\sqrt {{1^2} + {1^2} + {{\left( {\sqrt 2 } \right)}^2}} }}\]
\[ \Rightarrow l = \dfrac{1}{{\sqrt {1 + 1 + 2} }},{\text{ }}m = \dfrac{1}{{\sqrt {1 + 1 + 2} }},{\text{ }}n = \dfrac{{\sqrt 2 }}{{\sqrt {1 + 1 + 2} }}\]
On simplifying, we get
\[ \Rightarrow l = \dfrac{1}{{\sqrt 4 }},{\text{ }}m = \dfrac{1}{{\sqrt 4 }},{\text{ }}n = \dfrac{{\sqrt 2 }}{{\sqrt 4 }}\]
\[ \Rightarrow l = \dfrac{1}{2},{\text{ }}m = \dfrac{1}{2},{\text{ }}n = \dfrac{{\sqrt 2 }}{2}\]
which can also be written as,
\[ \Rightarrow l = \dfrac{1}{2},{\text{ }}m = \dfrac{1}{2},{\text{ }}n = \dfrac{1}{{\sqrt 2 }}\]
Hence, direction cosines of the line are \[l = \dfrac{1}{2},{\text{ }}m = \dfrac{1}{2},{\text{ }}n = \dfrac{1}{{\sqrt 2 }}\]
Hence, option (1) is correct.

Note: To solve this question, we can also use another formula for direction cosines i.e., \[l = \dfrac{x}{r},{\text{ }}m = \dfrac{y}{r},{\text{ }}n = \dfrac{z}{r}\] where \[x,{\text{ }}y,{\text{ }}z\] are the coordinates of point and \[r\] is the distance of point from origin.
While solving by this method, first we find the distance of point from origin, then we put the values in the formula. After that, with the help of options we find the value of \[l,{\text{ }}m,{\text{ }}n\] But this will be a very lengthy process and chances of mistakes will be high. That’s why we prefer not to use this formula.
And also remember, the direction cosines of a line are unique, but it may have infinitely many sets of direction ratios.