
Let the plane \[\begin{array}{*{20}{c}}
{ax + by + cz + d}& = &0
\end{array}\]bisects the line joining the points (4, -3, 1) and (2, 3, -5) at the right angles. If a, b, c, and d are the integers, then the minimum value of \[({a^2} + {b^2} + {c^2} + {d^2})\] is
Answer
162k+ views
Hint: First of all, find the equation of the plane. When we find the equation of the plane, then we compare the equation of the plane with the given equation of the plane. After comparing, we will get the value of the a, b, c, and c respectively.
Complete step by step solution:
Let us assume that p, q, and r are the direction ratios respectively.
Here, according to the question, we have given the equation of the plane that is,
\[\begin{array}{*{20}{c}}
{ \Rightarrow ax + by + cz + d}& = &0
\end{array}\]…………. (a)
Here, a line bisects the plane normally whose coordinates are (4, -3, 1) and (2, 3, -5). And the midpoint of this line will lie in the plan.
Now, we will draw a figure accordingly,

Image: Plane perpendicular to a line
Now, we will have to find the direction ratio, for that purpose, we will find the difference between the coordinates between the M(4, -3, 1) and O(2, 3, -5). Therefore, the direction ratio is,
\[ \Rightarrow \](1, -3, 3).
Now we got the direction ratios that is.
\[\begin{array}{*{20}{c}}
{ \Rightarrow p}& = &1
\end{array}\], \[\begin{array}{*{20}{c}}
q& = &{ - 3}
\end{array}\] and \[\begin{array}{*{20}{c}}
r& = &3
\end{array}\]
Now, we know that the equation of the plane when the plane coordinates and the direction ratios are given,
Therefore, we Can write.
\[\begin{array}{*{20}{c}}
{p(x - {x_1}) + q(y - {y_1}) + r(z - {z_1})}& = &0
\end{array}\]
Now,
\[ \Rightarrow \begin{array}{*{20}{c}}
{1(x - 3) - 3(y - 0) + 3(z + 2)}& = &0
\end{array}\]
\[ \Rightarrow \begin{array}{*{20}{c}}
{x - 3y - 3z + 3}& = &0
\end{array}\] …………………. (b)
Now compare the equation (a) and (b). we will get,
\[ \Rightarrow \begin{array}{*{20}{c}}
a& = &1
\end{array}\], \[\begin{array}{*{20}{c}}
b& = &{ - 3}
\end{array}\] \[\begin{array}{*{20}{c}}
c& = &{ - 3}
\end{array}\] and \[\begin{array}{*{20}{c}}
d& = &3
\end{array}\]
Therefore, the minimum of \[{a^2} + {b^2} + {c^2} + {d^2}\]will be,
\[ \Rightarrow {(1)^2} + {( - 3)^2} + {( - 3)^2} + {(3)^2}\]
\[ \Rightarrow 28\]
Hence correct answer is 28.
Note: The first point to keep in mind is that the midpoint of the line lies in the plan. And with the help of the midpoint of the line and point M, we can determine the direction ratio. After that, we can find the equation of the plane with the help of the midpoint of the line which lies on the plane and the direction ratio.
Complete step by step solution:
Let us assume that p, q, and r are the direction ratios respectively.
Here, according to the question, we have given the equation of the plane that is,
\[\begin{array}{*{20}{c}}
{ \Rightarrow ax + by + cz + d}& = &0
\end{array}\]…………. (a)
Here, a line bisects the plane normally whose coordinates are (4, -3, 1) and (2, 3, -5). And the midpoint of this line will lie in the plan.
Now, we will draw a figure accordingly,

Image: Plane perpendicular to a line
Now, we will have to find the direction ratio, for that purpose, we will find the difference between the coordinates between the M(4, -3, 1) and O(2, 3, -5). Therefore, the direction ratio is,
\[ \Rightarrow \](1, -3, 3).
Now we got the direction ratios that is.
\[\begin{array}{*{20}{c}}
{ \Rightarrow p}& = &1
\end{array}\], \[\begin{array}{*{20}{c}}
q& = &{ - 3}
\end{array}\] and \[\begin{array}{*{20}{c}}
r& = &3
\end{array}\]
Now, we know that the equation of the plane when the plane coordinates and the direction ratios are given,
Therefore, we Can write.
\[\begin{array}{*{20}{c}}
{p(x - {x_1}) + q(y - {y_1}) + r(z - {z_1})}& = &0
\end{array}\]
Now,
\[ \Rightarrow \begin{array}{*{20}{c}}
{1(x - 3) - 3(y - 0) + 3(z + 2)}& = &0
\end{array}\]
\[ \Rightarrow \begin{array}{*{20}{c}}
{x - 3y - 3z + 3}& = &0
\end{array}\] …………………. (b)
Now compare the equation (a) and (b). we will get,
\[ \Rightarrow \begin{array}{*{20}{c}}
a& = &1
\end{array}\], \[\begin{array}{*{20}{c}}
b& = &{ - 3}
\end{array}\] \[\begin{array}{*{20}{c}}
c& = &{ - 3}
\end{array}\] and \[\begin{array}{*{20}{c}}
d& = &3
\end{array}\]
Therefore, the minimum of \[{a^2} + {b^2} + {c^2} + {d^2}\]will be,
\[ \Rightarrow {(1)^2} + {( - 3)^2} + {( - 3)^2} + {(3)^2}\]
\[ \Rightarrow 28\]
Hence correct answer is 28.
Note: The first point to keep in mind is that the midpoint of the line lies in the plan. And with the help of the midpoint of the line and point M, we can determine the direction ratio. After that, we can find the equation of the plane with the help of the midpoint of the line which lies on the plane and the direction ratio.
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