Answer
Verified
387.6k+ views
Hint: Here we use the concept of median, which will give us the midpoint of the base of the triangle where the median meets the base. From point A and the midpoint, we find the direction ratios of the line and since the median is equally inclined to coordinate axes so we take the directional cosines as equal and equating them will give the values of \[\lambda ,\mu \].
* Midpoint of a line joining two points \[({x_1},{y_1},{z_1}), ({x_2},{y_2},{z_2})\] is given by \[\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2},\dfrac{{{z_1} + {z_2}}}{2}} \right)\]
*Directional ratios of a line joining points \[P({x_1},{y_1},{z_1}),Q({x_2},{y_2},{z_2})\] is given by
\[{x_2} - {x_1},{y_2} - {y_1},{z_2} - {z_1}\]
* Directional cosines of a line joining points \[P({x_1},{y_1},{z_1}), Q({x_2},{y_2},{z_2})\] is given by
\[\dfrac{{{x_2} - {x_1}}}{{PQ}},\dfrac{{{y_2} - {y_1}}}{{PQ}},\dfrac{{{z_2} - {z_1}}}{{PQ}}\] where \[PQ = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2} + {{({z_2} - {z_1})}^2}} \]
Complete step-by-step answer:
First we draw a triangle \[\vartriangle ABC\], where we draw a median from vertex A to the base BC, the midpoint of BC is D.
We find the coordinates of point D using the formula for midpoint of a line.
Midpoint of a line joining two points \[({x_1},{y_1},{z_1}), ({x_2},{y_2},{z_2})\] is given by \[\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2},\dfrac{{{z_1} + {z_2}}}{2}} \right)\]
Here the points are \[B( - 1,3,2), C(\lambda ,5,\mu )\]
Mid-point becomes \[\left( {\dfrac{{ - 1 + \lambda }}{2},\dfrac{{3 + 5}}{2},\dfrac{{2 + \mu }}{2}} \right)\]
So the midpoint of BC is \[D\left( {\dfrac{{ - 1 + \lambda }}{2},4,\dfrac{{2 + \mu }}{2}} \right)\].
Now we look at the line joining the points A and D. Since, we have coordinates of the two points, we can write their directional ratios.
Directional ratios of a line joining points \[P({x_1},{y_1},{z_1}), Q({x_2},{y_2},{z_2})\] is given by \[{x_2} - {x_1},{y_2} - {y_1},{z_2} - {z_1}\]
Here points are \[A(2,3,5),D(\dfrac{{ - 1 + \lambda }}{2},4,\dfrac{{2 + \mu }}{2})\]
So the directional ratios of AD are \[(\dfrac{{ - 1 + \lambda }}{2} - 2),(4 - 3),(\dfrac{{2 + \mu }}{2} - 5)\]
Taking LCM in the each term
\[
(\dfrac{{ - 1 + \lambda - 4}}{2}),(1),(\dfrac{{2 + \mu - 10}}{2}) \\
(\dfrac{{\lambda - 5}}{2}),(1),(\dfrac{{\mu - 8}}{2}) \\
\]
So the directional ratios are \[\dfrac{{\lambda - 5}}{2},1,\dfrac{{\mu - 8}}{2}\] … (1)
Since, the median through A is equally inclined to the coordinate axes then it means the directional cosines of the line are equal.
We know directional cosines of a line joining points \[P({x_1},{y_1},{z_1}), Q({x_2},{y_2},{z_2})\] is given by
\[\dfrac{{{x_2} - {x_1}}}{{PQ}},\dfrac{{{y_2} - {y_1}}}{{PQ}},\dfrac{{{z_2} - {z_1}}}{{PQ}}\] where \[PQ = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2} + {{({z_2} - {z_1})}^2}} \]
Since, the directional cosines are equal we can write
\[\dfrac{{{x_2} - {x_1}}}{{PQ}} = \dfrac{{{y_2} - {y_1}}}{{PQ}} = \dfrac{{{z_2} - {z_1}}}{{PQ}}\]
Since the length PQ is same in all denominators, we can remove it
\[{x_2} - {x_1} = {y_2} - {y_1} = {z_2} - {z_1}\] … (2)
From equation (2) we can say the directional ratios of lines are equal, So we equate the directional ratios from equation (1).
\[ \Rightarrow \dfrac{{\lambda - 5}}{2} = 1 = \dfrac{{\mu - 8}}{2}\]
Taking first two terms we get
\[ \Rightarrow \dfrac{{\lambda - 5}}{2} = 1\]
Cross multiplying the values
\[
\Rightarrow \lambda - 5 = 1 \times 2 \\
\Rightarrow \lambda - 5 = 2 \\
\]
Shift all constants to one side of the equation
\[
\Rightarrow \lambda = 2 + 5 \\
\Rightarrow \lambda = 7 \\
\]
Taking last two terms we get
\[ \Rightarrow \dfrac{{\mu - 8}}{2} = 1\]
Cross multiplying the values
\[
\Rightarrow \mu - 8 = 1 \times 2 \\
\Rightarrow \mu - 8 = 2 \\
\]
Shift all constants to one side of the equation
\[
\Rightarrow \mu = 2 + 8 \\
\Rightarrow \mu = 10 \\
\]
Now when we take ratio of the values \[\lambda ,\mu \]
\[ \Rightarrow \dfrac{\lambda }{\mu } = \dfrac{7}{{10}}\]
Cross multiplying the values
\[ \Rightarrow 10\lambda = 7\mu \]
Shift all values to one side of the equation
\[ \Rightarrow 10\lambda - 7\mu = 0\]
So, the correct answer is “Option C”.
Note: Students many times make the mistake of finding the length in the denominator of directional cosines which is not required as the directional cosines are equal. The equal denominators will cancel out when we equate two values. It will make the solution more complex as the length will also be in the form of \[\lambda ,\mu \] which are unknown to us.
* Midpoint of a line joining two points \[({x_1},{y_1},{z_1}), ({x_2},{y_2},{z_2})\] is given by \[\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2},\dfrac{{{z_1} + {z_2}}}{2}} \right)\]
*Directional ratios of a line joining points \[P({x_1},{y_1},{z_1}),Q({x_2},{y_2},{z_2})\] is given by
\[{x_2} - {x_1},{y_2} - {y_1},{z_2} - {z_1}\]
* Directional cosines of a line joining points \[P({x_1},{y_1},{z_1}), Q({x_2},{y_2},{z_2})\] is given by
\[\dfrac{{{x_2} - {x_1}}}{{PQ}},\dfrac{{{y_2} - {y_1}}}{{PQ}},\dfrac{{{z_2} - {z_1}}}{{PQ}}\] where \[PQ = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2} + {{({z_2} - {z_1})}^2}} \]
Complete step-by-step answer:
First we draw a triangle \[\vartriangle ABC\], where we draw a median from vertex A to the base BC, the midpoint of BC is D.
We find the coordinates of point D using the formula for midpoint of a line.
Midpoint of a line joining two points \[({x_1},{y_1},{z_1}), ({x_2},{y_2},{z_2})\] is given by \[\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2},\dfrac{{{z_1} + {z_2}}}{2}} \right)\]
Here the points are \[B( - 1,3,2), C(\lambda ,5,\mu )\]
Mid-point becomes \[\left( {\dfrac{{ - 1 + \lambda }}{2},\dfrac{{3 + 5}}{2},\dfrac{{2 + \mu }}{2}} \right)\]
So the midpoint of BC is \[D\left( {\dfrac{{ - 1 + \lambda }}{2},4,\dfrac{{2 + \mu }}{2}} \right)\].
Now we look at the line joining the points A and D. Since, we have coordinates of the two points, we can write their directional ratios.
Directional ratios of a line joining points \[P({x_1},{y_1},{z_1}), Q({x_2},{y_2},{z_2})\] is given by \[{x_2} - {x_1},{y_2} - {y_1},{z_2} - {z_1}\]
Here points are \[A(2,3,5),D(\dfrac{{ - 1 + \lambda }}{2},4,\dfrac{{2 + \mu }}{2})\]
So the directional ratios of AD are \[(\dfrac{{ - 1 + \lambda }}{2} - 2),(4 - 3),(\dfrac{{2 + \mu }}{2} - 5)\]
Taking LCM in the each term
\[
(\dfrac{{ - 1 + \lambda - 4}}{2}),(1),(\dfrac{{2 + \mu - 10}}{2}) \\
(\dfrac{{\lambda - 5}}{2}),(1),(\dfrac{{\mu - 8}}{2}) \\
\]
So the directional ratios are \[\dfrac{{\lambda - 5}}{2},1,\dfrac{{\mu - 8}}{2}\] … (1)
Since, the median through A is equally inclined to the coordinate axes then it means the directional cosines of the line are equal.
We know directional cosines of a line joining points \[P({x_1},{y_1},{z_1}), Q({x_2},{y_2},{z_2})\] is given by
\[\dfrac{{{x_2} - {x_1}}}{{PQ}},\dfrac{{{y_2} - {y_1}}}{{PQ}},\dfrac{{{z_2} - {z_1}}}{{PQ}}\] where \[PQ = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2} + {{({z_2} - {z_1})}^2}} \]
Since, the directional cosines are equal we can write
\[\dfrac{{{x_2} - {x_1}}}{{PQ}} = \dfrac{{{y_2} - {y_1}}}{{PQ}} = \dfrac{{{z_2} - {z_1}}}{{PQ}}\]
Since the length PQ is same in all denominators, we can remove it
\[{x_2} - {x_1} = {y_2} - {y_1} = {z_2} - {z_1}\] … (2)
From equation (2) we can say the directional ratios of lines are equal, So we equate the directional ratios from equation (1).
\[ \Rightarrow \dfrac{{\lambda - 5}}{2} = 1 = \dfrac{{\mu - 8}}{2}\]
Taking first two terms we get
\[ \Rightarrow \dfrac{{\lambda - 5}}{2} = 1\]
Cross multiplying the values
\[
\Rightarrow \lambda - 5 = 1 \times 2 \\
\Rightarrow \lambda - 5 = 2 \\
\]
Shift all constants to one side of the equation
\[
\Rightarrow \lambda = 2 + 5 \\
\Rightarrow \lambda = 7 \\
\]
Taking last two terms we get
\[ \Rightarrow \dfrac{{\mu - 8}}{2} = 1\]
Cross multiplying the values
\[
\Rightarrow \mu - 8 = 1 \times 2 \\
\Rightarrow \mu - 8 = 2 \\
\]
Shift all constants to one side of the equation
\[
\Rightarrow \mu = 2 + 8 \\
\Rightarrow \mu = 10 \\
\]
Now when we take ratio of the values \[\lambda ,\mu \]
\[ \Rightarrow \dfrac{\lambda }{\mu } = \dfrac{7}{{10}}\]
Cross multiplying the values
\[ \Rightarrow 10\lambda = 7\mu \]
Shift all values to one side of the equation
\[ \Rightarrow 10\lambda - 7\mu = 0\]
So, the correct answer is “Option C”.
Note: Students many times make the mistake of finding the length in the denominator of directional cosines which is not required as the directional cosines are equal. The equal denominators will cancel out when we equate two values. It will make the solution more complex as the length will also be in the form of \[\lambda ,\mu \] which are unknown to us.
Recently Updated Pages
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Which type of bond is stronger ionic or covalent class 12 chemistry CBSE
What organs are located on the left side of your body class 11 biology CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
How fast is 60 miles per hour in kilometres per ho class 10 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
When people say No pun intended what does that mea class 8 english CBSE