Answer
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Hint: Use the equation of plane and the line given to find the position vector of the point of intersection of the both and hence find the point of intersection of the plane and the line. Then use the distance formula in three-dimensional space to find the distance from the given point.
Complete step-by-step answer:
A plane is a two-dimensional doubly ruled surface spanned by two linearly independent vectors.
A vector equation of plane in normal form is represented as \[\vec r.\vec n = d\], where \[\vec r\] is the position vector of a point lying on the plane, \[\vec n\] is the unit vector normal to the plane joining it to the origin and $d$ is the perpendicular distance of the plane from the origin in three dimensional cartesian space.
Given the problem, a line intersects a plane in three-dimensional space.
Equation of line is given as:
\[\vec r = 2\hat i - \hat j + 2\hat k + \lambda \left( {3\hat i + 4\hat j + 2\hat k} \right){\text{ (1)}}\]
And the equation of plane is given as:
$\vec r.\left( {\hat i - \hat j + \hat k} \right) = 5{\text{ (2)}}$
We need to find the distance of the point of intersection of the above line and plane from the point
$A$ whose coordinates in three dimensions are given as $( - 1, - 5, - 10)$.
To find the intersection point of line and the plane, we need to put the value of \[\vec r\] from equation (1) of the line into equation (2) of plane.
$ \Rightarrow \left( {2\hat i - \hat j + 2\hat k + \lambda \left( {3\hat i + 4\hat j + 2\hat k} \right)} \right).\left( {\hat i - \hat j + \hat k} \right) = 5{\text{ (3)}}$
If $\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k$ and $\vec b = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k$, then dot product of the vectors is given by
$\vec a.\vec b = \left( {{a_1}\hat i + {a_2}\hat j + {a_3}\hat k} \right).\left( {{b_1}\hat i + {b_2}\hat j + {b_3}\hat k} \right) = {a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3}$
Using above in equation (3), we get
$
\Rightarrow \left( {2\hat i - \hat j + 2\hat k + \lambda \left( {3\hat i + 4\hat j + 2\hat k} \right)} \right).\left( {\hat i - \hat j + \hat k} \right) = 5 \\
\Rightarrow \left( {\left( {2 + 3\lambda } \right)\hat i + \left( { - 1 + 4\lambda } \right)\hat j + \left( {2 + 2\lambda } \right)\hat k} \right).\left( {\hat i - \hat j + \hat k} \right) = 5 \\
\Rightarrow \left( {\left( {2 + 3\lambda } \right).1 + \left( { - 1 + 4\lambda } \right).\left( { - 1} \right) + \left( {2 + 2\lambda } \right).1} \right) = 5 \\
\Rightarrow \lambda + 5 = 5 \\
\Rightarrow \lambda = 0 \\
$
Hence the value of $\lambda $ obtained when line and the plane intersect is equal to 0.
Using the above result in equation (1), we get
\[\vec r = 2\hat i - \hat j + 2\hat k{\text{ (4)}}\]
This is the equation of line intersecting the plane and also the position vector of the point of intersection of line and plane lying on the plane.
This position vector joins origin and the point it points to. If $\left( {x,y,z} \right)$ is the point, the position vector is pointing to, then the same is represented as:
$\vec r = \left( {x - 0} \right)\hat i + \left( {y - 0} \right)\hat j + \left( {z - 0} \right)\hat k = x\hat i + y\hat j + z\hat k$
Comparing the above with the position vector of the point of intersection of line and the plane as given in equation (4), we get the point of intersection as:
$
\vec r = x\hat i + y\hat j + z\hat k = \vec r = 2\hat i - \hat j + 2\hat k \\
\Rightarrow \left(
x = 2 \\
y = - 1 \\
z = 2 \\
$
Hence the point of intersection of line and the plane is $B\left( {2, - 1,2} \right)$.
We know that the distance between two points $\left( {{x_1},{y_1},{z_1}} \right)$ and $\left( {{x_2},{y_2},{z_3}} \right)$in three-dimensional space is given by $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} $ units.
Hence using above formula to find the distance between point $( - 1, - 5, - 10)$ and the point of intersection of line and the plane is $B\left( {2, - 1,2} \right)$, we get
$
{\text{distance}} = \sqrt {{{\left( { - 1 - 2} \right)}^2} + {{\left( { - 5 + 1} \right)}^2} + {{\left( { - 10 - 2} \right)}^2}} = \sqrt {{{\left( { - 3} \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( { - 12} \right)}^2}} \\
\Rightarrow {\text{distance}} = \sqrt {9 + 16 + 144} = \sqrt {169} = 13{\text{ units}} \\
$
Hence the distance of the point of intersection of the given line and plane from the point A whose coordinates in three dimensions are given as $( - 1, - 5, - 19)$ is 13 units.
Note: The equation of plane and line need to be kept in mind in all forms in order to solve problems like above. The intersection of a line and a plane in three-dimensional space can be the empty set, a point, or a line. Above problem is the one in which line and a plane have only one pint of intersection. If a line intersects a plane at a unique point, then the position vector of a point on the plane must be equal to the equation of line. This condition has been used to solve the above problem. The method can vary depending upon the form in which the equations of line and plane are given.
Complete step-by-step answer:
A plane is a two-dimensional doubly ruled surface spanned by two linearly independent vectors.
A vector equation of plane in normal form is represented as \[\vec r.\vec n = d\], where \[\vec r\] is the position vector of a point lying on the plane, \[\vec n\] is the unit vector normal to the plane joining it to the origin and $d$ is the perpendicular distance of the plane from the origin in three dimensional cartesian space.
Given the problem, a line intersects a plane in three-dimensional space.
Equation of line is given as:
\[\vec r = 2\hat i - \hat j + 2\hat k + \lambda \left( {3\hat i + 4\hat j + 2\hat k} \right){\text{ (1)}}\]
And the equation of plane is given as:
$\vec r.\left( {\hat i - \hat j + \hat k} \right) = 5{\text{ (2)}}$
We need to find the distance of the point of intersection of the above line and plane from the point
$A$ whose coordinates in three dimensions are given as $( - 1, - 5, - 10)$.
To find the intersection point of line and the plane, we need to put the value of \[\vec r\] from equation (1) of the line into equation (2) of plane.
$ \Rightarrow \left( {2\hat i - \hat j + 2\hat k + \lambda \left( {3\hat i + 4\hat j + 2\hat k} \right)} \right).\left( {\hat i - \hat j + \hat k} \right) = 5{\text{ (3)}}$
If $\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k$ and $\vec b = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k$, then dot product of the vectors is given by
$\vec a.\vec b = \left( {{a_1}\hat i + {a_2}\hat j + {a_3}\hat k} \right).\left( {{b_1}\hat i + {b_2}\hat j + {b_3}\hat k} \right) = {a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3}$
Using above in equation (3), we get
$
\Rightarrow \left( {2\hat i - \hat j + 2\hat k + \lambda \left( {3\hat i + 4\hat j + 2\hat k} \right)} \right).\left( {\hat i - \hat j + \hat k} \right) = 5 \\
\Rightarrow \left( {\left( {2 + 3\lambda } \right)\hat i + \left( { - 1 + 4\lambda } \right)\hat j + \left( {2 + 2\lambda } \right)\hat k} \right).\left( {\hat i - \hat j + \hat k} \right) = 5 \\
\Rightarrow \left( {\left( {2 + 3\lambda } \right).1 + \left( { - 1 + 4\lambda } \right).\left( { - 1} \right) + \left( {2 + 2\lambda } \right).1} \right) = 5 \\
\Rightarrow \lambda + 5 = 5 \\
\Rightarrow \lambda = 0 \\
$
Hence the value of $\lambda $ obtained when line and the plane intersect is equal to 0.
Using the above result in equation (1), we get
\[\vec r = 2\hat i - \hat j + 2\hat k{\text{ (4)}}\]
This is the equation of line intersecting the plane and also the position vector of the point of intersection of line and plane lying on the plane.
This position vector joins origin and the point it points to. If $\left( {x,y,z} \right)$ is the point, the position vector is pointing to, then the same is represented as:
$\vec r = \left( {x - 0} \right)\hat i + \left( {y - 0} \right)\hat j + \left( {z - 0} \right)\hat k = x\hat i + y\hat j + z\hat k$
Comparing the above with the position vector of the point of intersection of line and the plane as given in equation (4), we get the point of intersection as:
$
\vec r = x\hat i + y\hat j + z\hat k = \vec r = 2\hat i - \hat j + 2\hat k \\
\Rightarrow \left(
x = 2 \\
y = - 1 \\
z = 2 \\
$
Hence the point of intersection of line and the plane is $B\left( {2, - 1,2} \right)$.
We know that the distance between two points $\left( {{x_1},{y_1},{z_1}} \right)$ and $\left( {{x_2},{y_2},{z_3}} \right)$in three-dimensional space is given by $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} $ units.
Hence using above formula to find the distance between point $( - 1, - 5, - 10)$ and the point of intersection of line and the plane is $B\left( {2, - 1,2} \right)$, we get
$
{\text{distance}} = \sqrt {{{\left( { - 1 - 2} \right)}^2} + {{\left( { - 5 + 1} \right)}^2} + {{\left( { - 10 - 2} \right)}^2}} = \sqrt {{{\left( { - 3} \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( { - 12} \right)}^2}} \\
\Rightarrow {\text{distance}} = \sqrt {9 + 16 + 144} = \sqrt {169} = 13{\text{ units}} \\
$
Hence the distance of the point of intersection of the given line and plane from the point A whose coordinates in three dimensions are given as $( - 1, - 5, - 19)$ is 13 units.
Note: The equation of plane and line need to be kept in mind in all forms in order to solve problems like above. The intersection of a line and a plane in three-dimensional space can be the empty set, a point, or a line. Above problem is the one in which line and a plane have only one pint of intersection. If a line intersects a plane at a unique point, then the position vector of a point on the plane must be equal to the equation of line. This condition has been used to solve the above problem. The method can vary depending upon the form in which the equations of line and plane are given.
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