
The direction cosine of a line which is perpendicular to both the lines whose direction ratios are \[(1, - 2, - 2)\] and \[(0,2,1)\] are
A. \[(\dfrac{2}{3}, - \dfrac{1}{3},\dfrac{2}{3})\]
B. \[(\dfrac{2}{3},\dfrac{1}{3},\dfrac{2}{3})\]
C. \[(\dfrac{2}{3},\dfrac{1}{3}, - \dfrac{2}{3})\]
D. \[( - \dfrac{2}{3},\dfrac{1}{3},\dfrac{2}{3})\]
Answer
577.5k+ views
Hint: To find out the direction cosines of a line perpendicular to the lines whose direction ratios are \[(1, - 2, - 2)\] and \[(0,2,1)\], we find the cross product of \[(1, - 2, - 2)\] and \[(0,2,1)\]. This will give us the direction ratio of the line required, now we use \[\dfrac{l}{{\sqrt {{l^2} + {m^2} + {n^2}} }},\dfrac{m}{{\sqrt {{l^2} + {m^2} + {n^2}} }},\dfrac{n}{{\sqrt {{l^2} + {m^2} + {n^2}} }}\] to find the direction cosines of the required line.
Complete step by step answer:
Given direction ratios are \[(1, - 2, - 2)\] and \[(0,2,1)\]
The vectors that can be formed by given direction ratios are ,
\[\widehat i - 2\widehat j - 2\widehat k\]and \[2\widehat j + \widehat k\]
Now, we need to calculate the vector perpendicular to both the vectors so it can be given as ,
\[
= \left| {\begin{array}{*{20}{c}}
{\widehat i}&{\widehat j}&{\widehat k} \\
1&{ - 2}&{ - 2} \\
0&2&1
\end{array}} \right| \\
= ( - 2 + 4)\widehat i - (1 - 0)\widehat j + (2 - 0)\widehat k \\
= 2\widehat i - \widehat j + 2\widehat k \\
\]
Hence, the direction ratio of vector perpendicular to both the given direction ratios is \[(2, - 1,2)\]
Hence, the direction cosines are
\[\dfrac{l}{{\sqrt {{l^2} + {m^2} + {n^2}} }},\dfrac{m}{{\sqrt {{l^2} + {m^2} + {n^2}} }},\dfrac{n}{{\sqrt {{l^2} + {m^2} + {n^2}} }}\]
\[
= \dfrac{2}{{\sqrt {{2^2} + {{( - 1)}^2} + {2^2}} }},\dfrac{{ - 1}}{{\sqrt {{2^2} + {{( - 1)}^2} + {2^2}} }},\dfrac{2}{{\sqrt {{2^2} + {{( - 1)}^2} + {2^2}} }} \\
= \dfrac{2}{3},\dfrac{{ - 1}}{3},\dfrac{2}{3} \\
\]
So, option (a) is our required correct answer.
Note: Any number proportional to the direction cosine is known as the direction ratio of a line. These direction numbers are represented by a, b and c. We can conclude that some of the squares of the direction cosines of a line is \[1\].
Once, direction ratios are known you can use the formula as per \[\dfrac{l}{{\sqrt {{l^2} + {m^2} + {n^2}} }},\dfrac{m}{{\sqrt {{l^2} + {m^2} + {n^2}} }},\dfrac{n}{{\sqrt {{l^2} + {m^2} + {n^2}} }}\]to calculate given direction cosines. Calculate without any mistake.
Complete step by step answer:
Given direction ratios are \[(1, - 2, - 2)\] and \[(0,2,1)\]
The vectors that can be formed by given direction ratios are ,
\[\widehat i - 2\widehat j - 2\widehat k\]and \[2\widehat j + \widehat k\]
Now, we need to calculate the vector perpendicular to both the vectors so it can be given as ,
\[
= \left| {\begin{array}{*{20}{c}}
{\widehat i}&{\widehat j}&{\widehat k} \\
1&{ - 2}&{ - 2} \\
0&2&1
\end{array}} \right| \\
= ( - 2 + 4)\widehat i - (1 - 0)\widehat j + (2 - 0)\widehat k \\
= 2\widehat i - \widehat j + 2\widehat k \\
\]
Hence, the direction ratio of vector perpendicular to both the given direction ratios is \[(2, - 1,2)\]
Hence, the direction cosines are
\[\dfrac{l}{{\sqrt {{l^2} + {m^2} + {n^2}} }},\dfrac{m}{{\sqrt {{l^2} + {m^2} + {n^2}} }},\dfrac{n}{{\sqrt {{l^2} + {m^2} + {n^2}} }}\]
\[
= \dfrac{2}{{\sqrt {{2^2} + {{( - 1)}^2} + {2^2}} }},\dfrac{{ - 1}}{{\sqrt {{2^2} + {{( - 1)}^2} + {2^2}} }},\dfrac{2}{{\sqrt {{2^2} + {{( - 1)}^2} + {2^2}} }} \\
= \dfrac{2}{3},\dfrac{{ - 1}}{3},\dfrac{2}{3} \\
\]
So, option (a) is our required correct answer.
Note: Any number proportional to the direction cosine is known as the direction ratio of a line. These direction numbers are represented by a, b and c. We can conclude that some of the squares of the direction cosines of a line is \[1\].
Once, direction ratios are known you can use the formula as per \[\dfrac{l}{{\sqrt {{l^2} + {m^2} + {n^2}} }},\dfrac{m}{{\sqrt {{l^2} + {m^2} + {n^2}} }},\dfrac{n}{{\sqrt {{l^2} + {m^2} + {n^2}} }}\]to calculate given direction cosines. Calculate without any mistake.
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