Understanding the Cartesian Form of a Vector

In three-dimensional analytic geometry, the cartesian form of a vector expresses vectors, lines, and planes using coordinates in relation to orthogonal axes. This form provides unambiguous algebraic representation and manipulation of geometric objects within the cartesian coordinate system.

Algebraic Structure of a Vector in Cartesian Form

Let $O$ denote the origin of a right-handed rectangular coordinate system with axes $OX$, $OY$, and $OZ$. Any point $P(x, y, z)$ in this space determines a unique position vector $\vec{OP}$, written in terms of unit vectors as: \[ \vec{OP} = x\hat{i} + y\hat{j} + z\hat{k} \] where $\hat{i}$, $\hat{j}$, and $\hat{k}$ are unit vectors directed along the $x$-, $y$-, and $z$-axes, respectively.

The coordinates $x$, $y$, and $z$ are the scalar components of the vector in the respective directions. Thus, any vector $\vec{a}$ in three-dimensional space is represented as \[ \vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k} \] where $a_1$, $a_2$, $a_3$ are real numbers.

Interpretation of the Components in the Cartesian Form of a Vector

Given a vector $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$, the magnitude (length) of $\vec{a}$ is found using the distance formula: \[ |\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \] This is obtained by applying the Pythagorean theorem in three dimensions to the terminal point $(a_1, a_2, a_3)$ of the vector originating from the origin.

If $\vec{a}$ denotes the position vector of a point $A(a_1, a_2, a_3)$, then $a_1$, $a_2$, and $a_3$ are respectively the projections of $OA$ on the $x$-, $y$-, and $z$-axes. Conversion between vector form and point notation is straightforward using these coordinates.

Cartesian Form of the Equation of a Line Passing Through a Given Point with Specified Direction Ratios

The equation of a line that passes through a fixed point $A(x_1, y_1, z_1)$ and has direction ratios $a$, $b$, $c$ can be expressed in vector form as: \[ \vec{r} = \vec{a} + \lambda \vec{d} \] where $\vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$ is the position vector of $A$, $\vec{d} = a\hat{i} + b\hat{j} + c\hat{k}$ is any vector parallel to the line, and $\lambda$ is a scalar parameter.

To obtain the cartesian form, let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ represent an arbitrary point $P$ on the line. Equate corresponding components: \[ x\hat{i} + y\hat{j} + z\hat{k} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k} + \lambda(a\hat{i} + b\hat{j} + c\hat{k}) \] Separate the coefficients for each unit vector: \[ x = x_1 + \lambda a \] \[ y = y_1 + \lambda b \] \[ z = z_1 + \lambda c \] Eliminate $\lambda$: \[ x - x_1 = \lambda a \implies \frac{x - x_1}{a} = \lambda \] \[ y - y_1 = \lambda b \implies \frac{y - y_1}{b} = \lambda \] \[ z - z_1 = \lambda c \implies \frac{z - z_1}{c} = \lambda \] Since all expressions equal $\lambda$, set them equal to each other: \[ \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} \]

Result: This is the cartesian equation of a line through $(x_1, y_1, z_1)$ with direction ratios $a$, $b$, $c$.

Vector Algebra discusses the vector formulation underpinning such representations in greater detail.

Cartesian Form of the Equation of a Line Passing Through Two Points

If a line passes through two points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$, its direction ratios are \[ a = x_2 - x_1, \quad b = y_2 - y_1, \quad c = z_2 - z_1 \] Substitute these into the previous result to yield: \[ \frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1} \] This expresses every point $(x, y, z)$ on the line as a parametric function of the segment joining $A$ and $B$.

Cartesian Form of the Equation of a Plane with Known Normal

A plane perpendicular to the vector $\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}$ and at a distance $d$ from the origin can be written in vector form as \[ \vec{r} \cdot \vec{n} = d \] Let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$. Compute the scalar product: \[ (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (a\hat{i} + b\hat{j} + c\hat{k}) = d \] Expand this using properties of dot product: \[ ax + by + cz = d \]

Result: The general cartesian equation of a plane is $ax + by + cz = d$, where $a$, $b$, $c$ are direction ratios of the normal to the plane.

Cartesian Equation of a Plane Passing Through a Given Point and Perpendicular to a Given Vector

Suppose the plane passes through $A(x_1, y_1, z_1)$ and has normal vector $\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}$. The vector equation is \[ (\vec{r} - \vec{a}) \cdot \vec{n} = 0 \] Let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ and $\vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$. Write the difference: \[ (x\hat{i} + y\hat{j} + z\hat{k} - x_1\hat{i} - y_1\hat{j} - z_1\hat{k}) \cdot (a\hat{i} + b\hat{j} + c\hat{k}) = 0 \] \[ [(x - x_1)\hat{i} + (y - y_1)\hat{j} + (z - z_1)\hat{k}] \cdot (a\hat{i} + b\hat{j} + c\hat{k}) = 0 \] Perform the dot product: \[ a(x - x_1) + b(y - y_1) + c(z - z_1) = 0 \] Expand the terms: \[ ax - a x_1 + by - b y_1 + cz - c z_1 = 0 \] Or, \[ ax + by + cz = a x_1 + b y_1 + c z_1 \]

Result: The cartesian equation of the plane is $ax + by + cz = a x_1 + b y_1 + c z_1$.

Conversion: Vector Form to Cartesian Form of the Equation of a Line

Given the vector equation of a line: \[ \vec{r} = \vec{a} + \lambda \vec{b} \] where $\vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$ and $\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}$, the equivalent cartesian form is found by comparing each component: \[ x\hat{i} + y\hat{j} + z\hat{k} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k} + \lambda(a\hat{i} + b\hat{j} + c\hat{k}) \] Extract componentwise equations: \[ x = x_1 + \lambda a \] \[ y = y_1 + \lambda b \] \[ z = z_1 + \lambda c \] Eliminate the parameter $\lambda$ to obtain: \[ \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} \]

Result: The vector equation $\vec{r} = \vec{a} + \lambda\vec{b}$ is equivalent to the above cartesian equation.

Determinant Form: Plane Passing Through Three Non-Collinear Points

Let three non-collinear points be $A(x_1, y_1, z_1)$, $B(x_2, y_2, z_2)$, and $C(x_3, y_3, z_3)$. The cartesian equation of the plane passing through them is given by: \[ \begin{vmatrix} x - x_1 & y - y_1 & z - z_1 \\ x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ x_3 - x_1 & y_3 - y_1 & z_3 - z_1 \end{vmatrix} = 0 \] This determinant equals zero if and only if the vector from $A$ to a general point $(x, y, z)$, together with $\overrightarrow{AB}$ and $\overrightarrow{AC}$, are coplanar.

Worked Example: Cartesian Equation of a Line Through Two Points

Given: Points $A(3, 4, 2)$ and $B(5, -2, 4)$. Find the cartesian equation of the line passing through them.

Determine direction ratios: \[ a = 5 - 3 = 2,\quad b = -2 - 4 = -6,\quad c = 4 - 2 = 2 \] Write the cartesian equation: \[ \frac{x - 3}{2} = \frac{y - 4}{-6} = \frac{z - 2}{2} \]

Result: The required equation is $\dfrac{x - 3}{2} = \dfrac{y - 4}{-6} = \dfrac{z - 2}{2}$.

Graphical Methods of Vector Addition could aid in geometric verification of this result.

Worked Example: Cartesian Equation of a Plane Through a Point and Perpendicular to a Vector

Given: Plane passing through $(2, 3, 4)$ and perpendicular to vector $5\hat{i} - 3\hat{j} + 2\hat{k}$.

Let $a = 5$, $b = -3$, $c = 2$, $x_1 = 2$, $y_1 = 3$, $z_1 = 4$.

Substitute into general plane equation: \[ 5(x - 2) - 3(y - 3) + 2(z - 4) = 0 \] \[ 5x - 10 - 3y + 9 + 2z - 8 = 0 \] \[ 5x - 3y + 2z - 10 + 9 - 8 = 0 \] \[ 5x - 3y + 2z - 9 = 0 \]

Result: The required cartesian equation is $5x - 3y + 2z = 9$.

Distinction Between Cartesian and Vector Forms of Representation

The cartesian form expresses points, lines, and planes using coordinate relationships with respect to the axes, while the vector form utilizes directed quantities with magnitude and direction, using unit vectors and vector operations throughout.