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This is one of our many quizzes on Intersection points. Click on this to access more such competitive quizzes on Pair Of Straight Lines.

Intersection points Quiz 2

mathematics

 Mark the correct intersection point for the equation of lines given $3y + 2x + 2 = 0$and $12x + 3y + 7 = 0$. 

 Apply the formula $x = \dfrac{{{b_1}{c_2} - {b_2}{c_1}}}{{{a_1}{b_2} - {a_2}{b_1}}}$ and $y = \dfrac{{{a_2}{c_1} - {a_1}{c_2}}}{{{a_1}{b_2} - {a_2}{b_1}}}$.

 The two lines given are: $ \\ 2x + 3y + 2 = 0 -----(1) \\ 12x + 3y + 7 = 0 -----(2) \\ $ Applying the formula $x = \dfrac{{{b_1}{c_2} - {b_2}{c_1}}}{{{a_1}{b_2} - {a_2}{b_1}}}$and $y = \dfrac{{{a_2}{c_1} - {a_1}{c_2}}}{{{a_1}{b_2} - {a_2}{b_1}}} \\ \\ $ On substituting the values in the formula to find $x$ we get: $ \\ x = \dfrac{{3(7) - 3(2)}}{{2(3) - 12(3)}}  \\ \Rightarrow x = \dfrac{{15}}{{ - 30}}  \\   \Rightarrow x =  - \dfrac{1}{2}  \\ $ Now to get the values of $y$, $ \\ y = \dfrac{{12(2) - 2(7)}}{{2(3) - 12(3)}}  \\ \Rightarrow x = \dfrac{{10}}{{ - 30}}  \\   \Rightarrow x =  - \dfrac{1}{3}  \\  $

 Choose the correct equation of the plane which is passing through the intersection of two planes whose equation are $2x + y + z = 1$ and $2x + 3y + 2z = 5$ which is perpendicular to the other plane $x - y + 2z = 0$.

 Use the condition ${a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = 0$ as the plane is perpendicular to other plane. 

 Let the equation of the plane passing through the intersection is: $ \\ (2x + y + z - 1) + \lambda (2x + 3y + 2z - 5) = 0 \\ (2\lambda  + 2)x + (3\lambda  + 1)y + (2\lambda  + 1)z - (5\lambda  + 1) = 0 -----(1) \\ $ Let ${a_1},{b_1},{c_1}$ be the direction ratios of this plane $(2\lambda  + 2),(3\lambda  + 1),(2\lambda  + 1) \\ \\ $ The plane which is present in equation is perpendicular to the $x - y + 2z = 0$. So its direction ratios ${a_2},{b_2},{c_2}$ are $(1, - 1, 2)$. $ \\ $ Since the planes are been perpendicular so we will apply the condition: $ \\ {a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = 0 \\ $ Substituting the values we get, $  (2\lambda  + 2)1 + (3\lambda  + 1)( - 1) + (2\lambda  + 1)(2) = 0  \\   \Rightarrow (2\lambda  + 2) - (3\lambda  + 1) + (4\lambda  + 2) = 0  \\   \Rightarrow 3\lambda  + 3 = 0  \\   \Rightarrow \lambda  =  - 1  \\ $ Substituting the value in equation $(1)$ we get, $  ( - 2 + 2)x + ( - 3 + 1)y + ( - 2 + 1)z - ( - 5 + 1) = 0  \\   \Rightarrow  - 2y - z + 4 = 0  \\  $

 Choose the correct equation of plane which contains the line of intersection of the planes given $\mathop r\limits^ \to  .(2\widehat i + 2\widehat j + 2\widehat k) = 2$,$\mathop r\limits^ \to  .(2\widehat i + 3\widehat j - 2\widehat k) + 4 = 0$ and is perpendicular to the plane $\mathop r\limits^ \to  .(4\widehat i + 3\widehat j - 2\widehat k) - 8 = 0$.

 Calculate the value of $\lambda $ using the given equations of the plane. 

 $\mathop r\limits^ \to  .(2\widehat i + 2\widehat j + 2\widehat k) - 2 = 0 ----(1) \\ \mathop r\limits^ \to  .(2\widehat i + 3\widehat j - 2\widehat k) + 4 = 0 ----(2) \\ $ The equation of the plane which is passing through the line intersection of the plane can be given by equation $1$ and equation $2$ as: $ \\ [\mathop r\limits^ \to  .(2\widehat i + 2\widehat j + 2\widehat k) - 2] + \lambda [\mathop r\limits^ \to  .(2\widehat i + 3\widehat j - 2\widehat k) + 4] = 0$$ \Rightarrow \mathop r\limits^ \to  .[(2\lambda  + 2)\widehat i + (3\lambda  + 2)\widehat j + (2 - 2\lambda )\widehat k] + (4\lambda  - 2) = 0 ----(3) \\ $ The plane in the above equation $3$is perpendicular to the plane $\mathop r\limits^ \to  .(4\widehat i + 3\widehat j - 2\widehat k) - 8 = 0$ Therefore, $ \\ 4(2\lambda  + 2) + 3(3\lambda  + 2) - 2(2 - 2\lambda ) = 0  \\  \Rightarrow 21\lambda  + 10 = 0  \\  \Rightarrow \lambda  =  - \dfrac{{10}}{{21}}  \\  $<br>Substituting the value of $\lambda $ in equation $3$ we get, $ \\ \mathop r\limits^ \to  .\left[\left(2\left( { - \dfrac{{10}}{{21}}}\right) + 2\right)\widehat i + \left(3\left( - \dfrac{{10}}{{21}}\right) + 2\right)\widehat j + \left(2 - 2\left( - \dfrac{{10}}{{21}}\right)\right)\widehat k\right] + \left(4\left( - \dfrac{{12}}{{21}}\right) - 2\right) = 0  \\   \Rightarrow \mathop r\limits^ \to \cdot (22\widehat i + 12\widehat j + 62\widehat k) - 30 = 0  \\  $

 Calculate the point of intersection for the lines given $\dfrac{{x - 6}}{4} = \dfrac{{3y - 6}}{{ - 9}} = \dfrac{{z + 1}}{3}$and $\dfrac{{x + 6}}{2} = \dfrac{{2y - 6}}{{10}} = \dfrac{{z + 1}}{4}$.

 Put one equation of line equal to $\lambda$ and another equation of line equal to $\alpha $.

 The two equation given are: $ \\ \dfrac{{x - 6}}{4} = \dfrac{{3y - 6}}{{ - 9}} = \dfrac{{z + 1}}{3} ----(1) \\ \dfrac{{x + 6}}{2} = \dfrac{{2y - 6}}{{10}} = \dfrac{{z + 1}}{4} ----(2) \\ $ Putting equation $1$ equal to $\lambda $. We get, $ \\ \dfrac{{x - 6}}{4} = \dfrac{{3y - 6}}{{ - 9}} = \dfrac{{z + 1}}{3} = \lambda   \\   \Rightarrow x = 4\lambda  + 6  \\   \Rightarrow y =  - 3\lambda  + 2  \\   \Rightarrow z = 3\lambda  - 1  \\  $ Let this line be $P$ and its coordinates are $(4\lambda  + 6, - 3\lambda  + 2,3\lambda  - 1) \\ \\ $ Putting the equation $2$ equal to $\alpha $. We get, $ \\  \dfrac{{x + 6}}{2} = \dfrac{{2y - 6}}{{10}} = \dfrac{{z + 1}}{4} = \alpha   \\  \Rightarrow x = 2\alpha  - 6  \\   \Rightarrow y = 4\alpha  + 3  \\   \Rightarrow z = 4\alpha  - 1  \\  $ Let this line be $Q$ and its coordinates are $(2\alpha  - 6,4\alpha  + 3,4\alpha  - 1)$ As we know that these two lines are intersecting with each other this means that their coordinates will be same so we can write: $ \\  x = 4\lambda  + 6 = 2\alpha  - 6  \\ y =  - 3\lambda  + 2 = 4\alpha  + 3  \\  z = 3\lambda  - 1 = 4\alpha  - 1  \\  $ We can get the value of $\lambda $ from above equation of $z$. We get, $ \\  3\lambda  - 1 = 4\alpha  - 1  \\   \Rightarrow 3\lambda  = 4\alpha   \\   \Rightarrow \lambda  = \dfrac{{4\alpha }}{3}  \\  $ Substituting the value of $\lambda $ in equation of $y$ to get the value of $\alpha $. We get, $ \\ - 3\left(\dfrac{{4\alpha }}{3}\right) + 2 = 4\alpha  + 3  \\   \Rightarrow 8\alpha  =  - 1  \\   \Rightarrow \alpha  =  - \dfrac{1}{8}  \\  $ Substituting the value of $\alpha $in the equations to get the coordinates. We get, $ \\  x = 2\alpha  - 6 = 2\left( - \dfrac{1}{8}\right) - 6 =  - \dfrac{{25}}{4}  \\ y = 4\alpha  + 3 = 4\left( - \dfrac{1}{8}\right) + 3 = \dfrac{5}{2}  \\ z = 4\alpha  - 1 = 4\left( - \dfrac{1}{8}\right) - 1 =  - \dfrac{3}{2}  \\  $ The coordinates are: $\left( - \dfrac{{25}}{4},\dfrac{5}{2}, - \dfrac{3}{2}\right)$

 Mark the distance of the point $( - 2, - 6, - 10)$ from intersection point of the line $\mathop r\limits^ \to   = 2\widehat i - \widehat j + 4\widehat k + \lambda (3\widehat i + 4\widehat j + 3\widehat k)$ and the plane $\mathop r\limits^ \to  .(\widehat i - 2\widehat j + \widehat k) = 4$.

 The formula used for calculating distance between two points is $d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2} + {{({z_2} - {z_2})}^2}} $

 The equation of the line given is: $ \\ \mathop r\limits^ \to   = 2\widehat i - \widehat j + 4\widehat k + \lambda (3\widehat i + 4\widehat j + 3\widehat k) -----(1) \\ $ The equation of the plane given is: $ \\ \mathop r\limits^ \to  .(\widehat i - 2\widehat j + \widehat k) = 4 ----(2) \\ $ Substituting the value of $\mathop r\limits^ \to  $ from equation $1$ in equation $2$ we get, $ \\ [2\widehat i - \widehat j + 4\widehat k + \lambda (3\widehat i + 4\widehat j + 3\widehat k)].(\widehat i - 2\widehat j + \widehat k) = 4  \\   \Rightarrow [(3\lambda  + 2)\widehat i + (4\lambda  - 1)\widehat j + (3\lambda  + 4)].(\widehat i - 2\widehat j + \widehat k) = 4  \\   \Rightarrow (3\lambda  + 2) - (8\lambda  - 2) + (3\lambda  + 4) = 4  \\   \Rightarrow  - 2\lambda  + 4 = 0  \\   \Rightarrow \lambda  = 2  \\  $ Substituting the value in equation $1$ we get another equation of line as: $ \\ \mathop r\limits^ \to   = 8\widehat i + 7\widehat j + 10\widehat k \\ $ This means that the intersection point of the given line and given plane coordinates are $(8,7,10)$. $ \\ $ For calculating the distance between two points  $(8,7,10)$and $( - 2, - 6, - 10)$ is $ \\ d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2} + {{({z_2} - {z_2})}^2}}  \\  d = \sqrt {{{( - 2 - 8)}^2} + {{( - 6 - 7)}^2} + {{( - 10 - 10)}^2}}   \\   \Rightarrow d = \sqrt {100 + 169 + 400}   \\   \Rightarrow d = \sqrt {669}  = 25.8  \\  $