Question

Find the image of point \[\left( {3, - 2,1} \right)\] in the plane \[3x - y + 4z = 2\].

Answer 1

Hint:
We will find the equation of the line joining the point and its image. We will assume the midpoint of the line and its image as \[T\]. We will find the midpoint of the given point and its image using the midpoint formula and equate it with \[T\]. We will substitute the coordinates of point \[T\] in the equation of the plane as that point will lie on the plane. We will find the image by using basic algebra on the linear equation.

Formulas used:
We will use the following formulas:
1. The equation of a line passing through the point \[\left( {a,b,c} \right)\] and having direction ratios \[p,q\] and \[r\] is \[\dfrac{{x - a}}{p} = \dfrac{{y - b}}{q} = \dfrac{{z - c}}{r}\] .
2. The midpoint of 2 points \[\left( {a,b,c} \right)\] and \[\left( {p,q,r} \right)\] is \[\left( {\dfrac{{a + p}}{2},\dfrac{{b + q}}{2},\dfrac{{c + r}}{2}} \right)\]

Complete step by step solution:
We will first draw the diagram.

We will assume that the image of point \[A\left( {3, - 2,1} \right)\] in the plane \[3x - y + 4z = 2\] is point \[B\] . The line joining \[A\] and \[B\] will be normal to the plane \[3x - y + 4z = 2\]. So, the direction ratios of the line \[AB\] will be 3, \[ - 4\] and 1.
We will formulate the equation of the line \[AB\].
Substituting 3 for \[a\] , \[ - 2\] for \[b\] , 1 for \[c\] , 3 for \[p\] , \[ - 4\] for \[q\] and 1 for \[r\] in the formula \[\dfrac{{x - a}}{p} = \dfrac{{y - b}}{q} = \dfrac{{z - c}}{r}\] , we get
\[\dfrac{{x - 3}}{3} = \dfrac{{y - \left( { - 2} \right)}}{{ - 1}} = \dfrac{{z - 1}}{4}\]
We will assume that the 3 ratios are equal to \[k\], Therefore,
 \[ \Rightarrow \dfrac{{x - 3}}{3} = \dfrac{{y + 2}}{{ - 1}} = \dfrac{{z - 1}}{4} = k\]
Now we will find the values of \[x,y\] and \[z\] in terms of \[k\].
So,
\[ \Rightarrow \dfrac{{x - 3}}{3} = k\]
On cross multiplication, we get
\[\begin{array}{l} \Rightarrow x - 3 = 3k\\ \Rightarrow x = 3k + 3\end{array}\]
Now equating the \[y\] term with \[k\], we get
\[ \Rightarrow \dfrac{{y + 2}}{{ - 1}} = k\]
On cross multiplication, we get
\[\begin{array}{l} \Rightarrow y + 2 = - 1 \cdot k\\ \Rightarrow y = - k - 2\end{array}\]
Now equating the \[y\] term with \[k\], we get
\[ \Rightarrow \dfrac{{z - 1}}{4} = k\]
On cross multiplication, we get
\[\begin{array}{l} \Rightarrow z - 1 = 4k\\ \Rightarrow z = 4k + 1\end{array}\]
Any point lying on line \[AB\] will be of the form \[\left( {3k + 3, - k - 2,4k + 1} \right)\].
We will take point \[B \equiv \left( {3k + 3, - k - 2,4k + 1} \right)\].
We will assume that point \[T\] is the mid-point of line \[AB\].
We will find the coordinates of \[T\] using the mid-point formula, \[\left( {\dfrac{{a + p}}{2},\dfrac{{b + q}}{2},\dfrac{{c + r}}{2}} \right)\]. So,
\[\begin{array}{l}T \equiv \left( {\dfrac{{3 + 3k + 3}}{2},\dfrac{{ - 2 - k - 2}}{2},\dfrac{{1 + 4k + 1}}{2}} \right)\\ \Rightarrow T \equiv \left( {\dfrac{3}{2}k + 3, - \dfrac{k}{2} - 2,2k + 1} \right)\end{array}\]
We know that point \[T\] will lie on the plane \[3x - y + 4z = 2\] . It will satisfy the equation of the plane:
\[ \Rightarrow 3x - y + 4z = 2\]
Now substituting the values of \[x\], \[y\] and \[z\], we get
\[ \Rightarrow 3\left( {\dfrac{3}{2}k + 3} \right) - \left( { - \dfrac{k}{2} - 2} \right) + 4\left( {2k + 1} \right) = 2\]
Multiplying the terms, we get
\[ \Rightarrow \dfrac{9}{2}k + 9 + \dfrac{k}{2} + 2 + 8k + 4 = 2\]
Adding the terms, we get
\[\begin{array}{l} \Rightarrow 13k = - 13\\ \Rightarrow k = - 1\end{array}\]
We have calculated the value of \[k\] .
Now substituting \[k = - 1\] in the equation of point \[B\], we get
\[\begin{array}{l}B = \left( {3\left( { - 1} \right) + 3, - \left( { - 1} \right) - 2,4\left( { - 1} \right) + 1} \right)\\ \Rightarrow B = \left( {0, - 1, - 3} \right)\end{array}\]
$\therefore $ The image of \[\left( {3, - 2,1} \right)\] in the plane \[3x - y + 4z = 2\] is \[\left( {0, - 1, - 3} \right)\] .

Note:
The coefficient of \[x,y\] and \[z\] in the equation of a plane are the direction ratios of the normal to the plane. That is why we have taken the direction ratios of line \[AB\] as 3, \[ - 1\] and 4.
Direction ratios are also known as direction component or direction numbers. It can be said that direction ratios are the numbers proportional to direction cosines.