Most Important 3D Questions: Image of a Point and Line Equation Conversions

This lesson covers two staple Class 12 three-dimensional geometry tasks: finding the image of a point in a line, and converting a line's equation between Cartesian and vector form.

Image of a point in a line

We want the image of the point $P(1, 6, 3)$ in the line

$\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}.$

A general point on the line

Set each ratio equal to a parameter $k$:

$\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3} = k.$

This gives $x = k$, $y = 2k + 1$, and $z = 3k + 2$, so a general point on the line is

$(k,\; 2k+1,\; 3k+2).$

Foot of the perpendicular

Let $M$ be the foot of the perpendicular from $P$ to the line, so $M = (k, 2k+1, 3k+2)$ for some $k$. The direction ratios of $PM$ are

$(k - 1,\; (2k+1) - 6,\; (3k+2) - 3) = (k - 1,\; 2k - 5,\; 3k - 1).$

The direction ratios of the line are the denominators in standard form, $(1, 2, 3)$. (If the line is not in standard form, convert it first, since that changes the direction ratios.)

Since $PM$ is perpendicular to the line, the sum of the products of corresponding direction ratios is zero:

$(k - 1)(1) + (2k - 5)(2) + (3k - 1)(3) = 0.$

Expanding,

$k - 1 + 4k - 10 + 9k - 3 = 0 \;\Rightarrow\; 14k - 14 = 0 \;\Rightarrow\; k = 1.$

Substituting $k = 1$ gives the foot of the perpendicular

$M = (1,\; 2(1)+1,\; 3(1)+2) = (1, 3, 5).$

Locating the image

The image $P'(\alpha, \beta, \gamma)$ is the reflection of $P$ in the line, so $M$ is the midpoint of $P$ and $P'$:

$\left(\frac{1 + \alpha}{2},\; \frac{6 + \beta}{2},\; \frac{3 + \gamma}{2}\right) = (1, 3, 5).$

Comparing components:

$1 + \alpha = 2 \Rightarrow \alpha = 1, \quad 6 + \beta = 6 \Rightarrow \beta = 0, \quad 3 + \gamma = 10 \Rightarrow \gamma = 7.$

So the image of $P$ is $(1, 0, 7)$.

Vector form to Cartesian form

Given the vector equation of a line

$\vec r = (2\,\hat\imath + 3\,\hat\jmath - 4\,\hat k) + \lambda\,(3\,\hat\imath - \hat\jmath + \hat k),$

read off the point $\vec a$ and the direction $\vec b$. The constant part gives the point $(2, 3, -4)$, and the coefficient of $\lambda$ gives direction ratios $(3, -1, 1)$. The Cartesian equation is therefore

$\frac{x - 2}{3} = \frac{y - 3}{-1} = \frac{z + 4}{1}.$

Cartesian form to vector form

Given the Cartesian equation

$\frac{x - 5}{2} = \frac{y - 3}{6} = \frac{z - 1}{5},$

take the $x - x_1$ form to read the point: $(5, 3, 1)$, so

$\vec a = 5\,\hat\imath + 3\,\hat\jmath + 1\,\hat k.$

The denominators are the direction ratios, giving

$\vec b = 2\,\hat\imath + 6\,\hat\jmath + 5\,\hat k.$

(First confirm the equation is in standard form; if not, convert it before reading the direction ratios.) The vector equation is

$\vec r = (5\,\hat\imath + 3\,\hat\jmath + 1\,\hat k) + \lambda\,(2\,\hat\imath + 6\,\hat\jmath + 5\,\hat k).$

Key takeaways

• The foot of the perpendicular from a point to a line is found by writing a general point on the line, then using the condition that corresponding direction ratios sum to zero.
• The image of a point in a line is its reflection: the foot of the perpendicular is the midpoint of the point and its image.
• To convert between forms, the point comes from the constant part (or the $x - x_1$ numerators) and the direction ratios come from the coefficient of the parameter (or the denominators).