
If the co-ordinates of the points P \[\left( {2,3,4} \right)\] and Q $\left( {1, - 2,1} \right)$ and O be the origin, then
A. OP $ = $ OQ
B. OP $ \bot $ OQ
C. OP $\parallel $ OQ
D. None of these
Answer
164.1k+ views
Hint: In order to solve this type of question, first we will write the formula for finding the angle between the lines. Then, we will find the direction ratios of both the lines separately. Next, we will substitute the values obtained in the formula for finding the angle between the lines to get the correct answer.
Formula used:
Angle between a pair of lines having direction ratios ${a_1},{b_1},{c_1}$ and ${a_2},{b_2},{c_2}$ is given by,
$\cos \theta = \left| {\dfrac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {{a_1}^2 + {b_1}^2 + {c_1}^2} \sqrt {{a_2}^2 + {b_2}^2 + {c_2}^2} }}} \right|$
Direction ratios of the line passing through a line (say, $\overrightarrow {AB} $) $A\left( {{x_1},{y_1},{z_1}} \right)$ and $B\left( {{x_2},{y_2},{z_2}} \right)$ is given by,
$\left( {{x_2} - {x_1}} \right),\left( {{y_2} - {y_1}} \right),\left( {{z_2} - {z_1}} \right)$
Complete step by step solution:
Angle between a pair of lines having direction ratios ${a_1},{b_1},{c_1}$ and ${a_2},{b_2},{c_2}$ is given by,
$\cos \theta = \left| {\dfrac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {{a_1}^2 + {b_1}^2 + {c_1}^2} \sqrt {{a_2}^2 + {b_2}^2 + {c_2}^2} }}} \right|$ ………………..equation $\left( 1 \right)$
For $\overrightarrow {OP} $,
$O\left( {0,0,0} \right)$ and $P\left( {2,3,4} \right)$
Direction ratios of $\overrightarrow {OP} $,
$\left( {{x_2} - {x_1}} \right),\left( {{y_2} - {y_1}} \right),\left( {{z_2} - {z_1}} \right)$
Substituting the values,
$\left( {2 - 0} \right),\left( {3 - 0} \right),\left( {4 - 0} \right) = 2,3,4$
$\therefore {a_1} = 2,\;{b_1} = 3,\;{c_1} = 4$ ………………..equation $\left( 2 \right)$
For $\overrightarrow {OQ} $,
$O\left( {0,0,0} \right)$ and $Q\left( {1, - 2,1} \right)$
Direction ratios of $\overrightarrow {OQ} $,
$\left( {{x_2} - {x_1}} \right),\left( {{y_2} - {y_1}} \right),\left( {{z_2} - {z_1}} \right)$
Substituting the values,
$\left( {1 - 0} \right),\left( { - 2 - 0} \right),\left( {1 - 0} \right) = 1, - 2,1$
$\therefore {a_2} = 1,\;{b_2} = - 2,\;{c_2} = 1$ ………………..equation $\left( 3 \right)$
Substituting equations $\left( 2 \right)$ and $\left( 3 \right)$ in equation $\left( 1 \right)$,
$\cos \theta = \left| {\dfrac{{\left( {2 \times 1} \right) + \left( {3 \times \left( { - 2} \right)} \right) + \left( {4 \times 1} \right)}}{{\sqrt {{2^2} + {3^2} + {4^2}} \sqrt {{1^2} + {{\left( { - 2} \right)}^2} + {1^2}} }}} \right|$
Solving it,
$\cos \theta = \left| {\dfrac{{2 - 6 + 4}}{{\sqrt {29} \sqrt 6 }}} \right|$
$\cos \theta = 0$
Finding the value of $\theta ,$
$\cos \theta = \cos {90^ \circ }$
On comparing both the sides,
$\theta = {90^ \circ }$
Thus, we conclude that $\overrightarrow {OP} $ $ \bot $ $\overrightarrow {OQ} $.
$\therefore $ The correct option is (B).
Note: The direction ratios are quite helpful in finding the relationship between two lines or vectors as they can be used to find the direction cosines of a line or the angle between the two lines. Also, they are of great help in finding the dot product between the two vectors.
Formula used:
Angle between a pair of lines having direction ratios ${a_1},{b_1},{c_1}$ and ${a_2},{b_2},{c_2}$ is given by,
$\cos \theta = \left| {\dfrac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {{a_1}^2 + {b_1}^2 + {c_1}^2} \sqrt {{a_2}^2 + {b_2}^2 + {c_2}^2} }}} \right|$
Direction ratios of the line passing through a line (say, $\overrightarrow {AB} $) $A\left( {{x_1},{y_1},{z_1}} \right)$ and $B\left( {{x_2},{y_2},{z_2}} \right)$ is given by,
$\left( {{x_2} - {x_1}} \right),\left( {{y_2} - {y_1}} \right),\left( {{z_2} - {z_1}} \right)$
Complete step by step solution:
Angle between a pair of lines having direction ratios ${a_1},{b_1},{c_1}$ and ${a_2},{b_2},{c_2}$ is given by,
$\cos \theta = \left| {\dfrac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {{a_1}^2 + {b_1}^2 + {c_1}^2} \sqrt {{a_2}^2 + {b_2}^2 + {c_2}^2} }}} \right|$ ………………..equation $\left( 1 \right)$
For $\overrightarrow {OP} $,
$O\left( {0,0,0} \right)$ and $P\left( {2,3,4} \right)$
Direction ratios of $\overrightarrow {OP} $,
$\left( {{x_2} - {x_1}} \right),\left( {{y_2} - {y_1}} \right),\left( {{z_2} - {z_1}} \right)$
Substituting the values,
$\left( {2 - 0} \right),\left( {3 - 0} \right),\left( {4 - 0} \right) = 2,3,4$
$\therefore {a_1} = 2,\;{b_1} = 3,\;{c_1} = 4$ ………………..equation $\left( 2 \right)$
For $\overrightarrow {OQ} $,
$O\left( {0,0,0} \right)$ and $Q\left( {1, - 2,1} \right)$
Direction ratios of $\overrightarrow {OQ} $,
$\left( {{x_2} - {x_1}} \right),\left( {{y_2} - {y_1}} \right),\left( {{z_2} - {z_1}} \right)$
Substituting the values,
$\left( {1 - 0} \right),\left( { - 2 - 0} \right),\left( {1 - 0} \right) = 1, - 2,1$
$\therefore {a_2} = 1,\;{b_2} = - 2,\;{c_2} = 1$ ………………..equation $\left( 3 \right)$
Substituting equations $\left( 2 \right)$ and $\left( 3 \right)$ in equation $\left( 1 \right)$,
$\cos \theta = \left| {\dfrac{{\left( {2 \times 1} \right) + \left( {3 \times \left( { - 2} \right)} \right) + \left( {4 \times 1} \right)}}{{\sqrt {{2^2} + {3^2} + {4^2}} \sqrt {{1^2} + {{\left( { - 2} \right)}^2} + {1^2}} }}} \right|$
Solving it,
$\cos \theta = \left| {\dfrac{{2 - 6 + 4}}{{\sqrt {29} \sqrt 6 }}} \right|$
$\cos \theta = 0$
Finding the value of $\theta ,$
$\cos \theta = \cos {90^ \circ }$
On comparing both the sides,
$\theta = {90^ \circ }$
Thus, we conclude that $\overrightarrow {OP} $ $ \bot $ $\overrightarrow {OQ} $.
$\therefore $ The correct option is (B).
Note: The direction ratios are quite helpful in finding the relationship between two lines or vectors as they can be used to find the direction cosines of a line or the angle between the two lines. Also, they are of great help in finding the dot product between the two vectors.
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