Q.The cartesian equation of a line is $\frac{x-5}{3} = \frac{y+4}{7} = \frac{z-6}{2}$. Write its vector form.

The key idea is to convert the symmetric form into the vector form by identifying a point on the line and the direction vector.

Step 1: From the given symmetric equation $\frac{x-5}{3} = \frac{y+4}{7} = \frac{z-6}{2}$, the line passes through the point $(5, -4, 6)$ and has direction ratios $(3, 7, 2)$.

Step 2: The direction vector is $\vec{d} = 3\hat{i} + 7\hat{j} + 2\hat{k}$, and the position vector of the fixed point is $\vec{a} = 5\hat{i} - 4\hat{j} + 6\hat{k}$. …

The symmetric (or cartesian) form of a line is a compact way to say: "start at a known point, then move only along a fixed direction." Each fraction tells you how much the $x$, $y$, and $z$ coordinates change relative to a common parameter. The key is that the denominators are the components of the direction vector, and the numerators (after shifting by the constants) are the coordinates of a point on the line.

Let’s extract that information cleanly.

1. Identify a point on the line

   The symmetric form $\frac{x-5}{3} = \frac{y+4}{7} = \frac{z-6}{2}$ means that when the common value is zero, we get $x-5=0$, $y+4=0$, $z-6=0$. So the point $(5, -4, 6)$ lies on the line. In vector terms, this is the position vector $\vec{a} = 5\hat{i} - 4\hat{j} + 6\hat{k}$.

2. Identify the direction vector

   The denominators $3$, $7$, and $2$ are the direction ratios. They tell us that for every $3$ units moved along $x$, we move $7$ along $y$ and $2$ along $z$. So the direction vector is $\vec{b} = 3\hat{i} + 7\hat{j} + 2\hat{k}$.

3. Write the vector equation

   The vector form of a line is $\vec{r} = \vec{a} + \lambda \vec{b}$, where $\lambda$ is a scalar parameter. Substituting what we have: …

Method: Convert symmetric (Cartesian) form to vector form

The symmetric form and the vector form of a line carry exactly the same two pieces of data — a point and a direction. Converting is just reading them off and re-packaging.

Steps

Step 1: Read the point from the numerators. In $\dfrac{x-x_0}{a}=\dfrac{y-y_0}{b}=\dfrac{z-z_0}{c}$, setting each fraction to $0$ gives the point $(x_0,y_0,z_0)$. Mind the signs: $y+4$ means $y-(-4)$, so that coordinate is $-4$, not $+4$.

Step 2: Read the direction from the denominators. The direction vector is $\vec b=a\hat i+b\hat j+c\hat k$.

Step 3: Assemble the vector form. …

Common Mistakes

Mistake 1: Misreading the point's sign from $y+4$.

Why it's wrong: $y+4=y-(-4)$, so the $y$-coordinate of the point is $-4$, not $+4$. Correct approach: rewrite each numerator as $x-x_0$ etc. before reading the point, giving $(5,-4,6)$.

Mistake 2: Swapping the roles of numerators and denominators. …