Question

Find the image of the line $\dfrac{x-1}{3}=\dfrac{y-3}{1}=\dfrac{z-4}{-5}$ in the plane $2x-y+z+3=0$

Answer 1

We will use the concept of direction ratios and foot of perpendicular to solve this question. If a, b, c are three numbers proportional to the direction cosine l, m, n of a straight line then a, b, c are called its direction ratios. Also we need to find the dot product between line and plane to understand the relation between them.

Complete step-by-step answer:
To find the image of the given line all we need is a point on that line and the direction ratios of that line.
Now given line in the question is $\dfrac{x-1}{3}=\dfrac{y-3}{1}=\dfrac{z-4}{-5}......(1)$.
So the direction ratios of the line in equation (1) is (3, 1, -5) and the point is (1, 3, 4).
And now the given plane with respect to which we need to find the image is $2x-y+z+3=0......(2)$.
Now we know that the normal to the plane is the coefficients of x, y and z in the equation (2). So using this information we get the normal to the plane as (2, -1, 1).
Now dot product of normal to the plane and direction ratios of the line is,
$\Rightarrow 3\times 2+1\times (-1)+(-5)\times 1=6-1-5=0.......(3)$
Now from equation (3) we can see that the given line is parallel to the plane as the dot product turns out to be 0. So the image line will also be parallel to the plane and hence the direction ratios will be the same for the image line as it was for the given line.
Now direction ratios of the image line is (3, 1, -5). Now we need to find the point on this image line. For finding this we need to find the image of these points with respect to the plane.
Now for finding the image of point (3, 1, -5) on the plane $2x-y+z+3=0$ we will first find the foot of the perpendicular line.
We have the direction ratios of normal to the plane as (2, -1, 1) and also the equation of line in question. Now foot of the perpendicular line to the plane will be the line,
$\dfrac{x-1}{2}=\dfrac{y-3}{-1}=\dfrac{z-4}{1}=r......(3)$
Now solving equation (3) by equating each term to r one by one we get x, y and z as,

& \Rightarrow \dfrac{x-1}{2}=r \\\ & \Rightarrow x=2r+1.......(4) \\\ \end{aligned}$$ $$\begin{aligned} & \Rightarrow \dfrac{y-3}{-r}=r \\\ & \Rightarrow y=-r+3.......(5) \\\ \end{aligned}$$ $$\begin{aligned} & \Rightarrow \dfrac{z-4}{1}=r \\\ & \Rightarrow z=r+4......(6) \\\ \end{aligned}$$ So from equation (4), (5) and (6) the coordinates of the foot of perpendicular is (2r+1, -r+3, r+4). Now putting this point in the plane $$2x-y+z+3=0$$ we get, $$\Rightarrow 2(2r+1)-(-r+3)+(r+4)+3=0.....(7)$$ Now solving equation (7) for r we get, $$\begin{aligned} & \Rightarrow 4r+2+r-3+r+4+3=0 \\\ & \Rightarrow 6r=-6 \\\ & \Rightarrow r=-1......(8) \\\ \end{aligned}$$ So now putting the value of r from equation (8) in (2r+1, -r+3, r+4) we get, $$\Rightarrow x=2r+1=2\times -1+1=-1.......(9)$$ $$\Rightarrow y=-r+3=-(-1)+3=4........(10)$$ $$\Rightarrow z=r+4=-1+4=3........(11)$$ So from equation (9), (10) and (11) we get the foot of perpendicular as (-1, 4, 3). Now let the coordinate of the image be (a, b, c). So now the midpoint of (1, 3, 4) and (a, b, c) will be the foot of perpendicular (-1, 4, 3). So utilizing this information we can find (a, b, c). Hence we get, $$\begin{aligned} & \dfrac{a+1}{2}=-1,\,\dfrac{b+3}{2}=4,\,\dfrac{c+4}{2}=3 \\\ & \Rightarrow a=-2-1=-3,b=8-3=5,c=6-4=2 \\\ \end{aligned}$$ So we get (a, b, c) as (-3, 5, 2) and we know that the direction ratios are the same. Now the equation of the image of the line is $$\dfrac{x+3}{3}=\dfrac{y-5}{1}=\dfrac{z-2}{-5}$$. **Note:** We need to remember the midpoint formula and the concept of direction ratios and foot of the perpendicular to solve such types of questions. Also we need to know about the dot product. We can make a mistake in a hurry in solving equation (6) and hence we need to be careful while solving this step.