Q.The vector equation of the line $\dfrac{x-5}{3} = \dfrac{y+4}{7} = \dfrac{z-6}{2}$ is __________.

The key idea is that the symmetric form $\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$ directly gives a point $(x_0, y_0, z_0)$ on the line and a direction vector $(a, b, c)$.

From the given equation, the point is $(5, -4, 6)$ and the direction vector is $(3, 7, 2)$.

The vector equation is $\vec{r} = \vec{a} + \lambda \vec{b}$, where $\vec{a}$ is the position vector of the point and $\vec{b}$ is the direction vector. …

The key idea here is that any line in space can be written in vector form once you know two things: a point it passes through, and its direction. The symmetric form of a line is just a tidy way to package that information.

When you see an equation like $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$, each denominator gives the direction ratios $(a, b, c)$, and each numerator's constant tells you the coordinates of a fixed point $(x_1, y_1, z_1)$. The equality of the three fractions means that as you move along the line, the changes in $x$, $y$, and $z$ are proportional to $a$, $b$, and $c$ respectively.

So the vector equation $\vec{r} = \vec{a} + \lambda \vec{b}$ is just saying: start at the fixed point $\vec{a}$, then add any scalar multiple $\lambda$ of the direction vector $\vec{b}$ to reach any point on the line.

Let's extract the numbers.

1. Identify the fixed point. From $\frac{x-5}{3}$, the numerator is $x - 5$, so $x_1 = 5$. From $\frac{y+4}{7}$, note that $y+4 = y - (-4)$, so $y_1 = -4$. From $\frac{z-6}{2}$, we get $z_1 = 6$. So the point is $(5, -4, 6)$. In vector form, this is $\vec{a} = 5\hat{i} - 4\hat{j} + 6\hat{k}$.

2. Identify the direction ratios. The denominators are $3$, $7$, and $2$. These are the direction ratios, so the direction vector is $\vec{b} = 3\hat{i} + 7\hat{j} + 2\hat{k}$. …

Method: Reading a Line's Point and Direction from Its Symmetric Form

This method converts any line written in symmetric (Cartesian) form $\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{z-z_1}{c}$ into vector form $\vec{r}=\vec{a}+\lambda\vec{b}$.

Steps

Step 1: Force every numerator into the strict pattern $\dfrac{x-x_1}{a}$.

The vector form needs a point and a direction, and the symmetric form hides both. A term such as $y+4$ must be read as $y-(-4)$ so the sign of each coordinate is unambiguous. This one habit prevents almost every sign error.

Step 2: Read the fixed point from the numerator constants.

The constants you subtract, $(x_1,y_1,z_1)$, are the coordinates of a point on the line, so $\vec{a}=x_1\hat{i}+y_1\hat{j}+z_1\hat{k}$. …

Common Mistakes

Mistake 1: Reading the point as $(5, 4, 6)$ instead of $(5, -4, 6)$.

Why it's wrong: the middle term is $y+4$, which is $y-(-4)$, so the $y$-coordinate is $-4$. Correct approach: rewrite every numerator as $x-x_1$ before reading it, so a "$+$" in a numerator always becomes a negative coordinate.

Mistake 2: Swapping the roles of numerator and denominator. …