Three Dimensional Geometry Class 12 NCERT Guide

Understanding Direction Cosines and Direction Ratios

To locate a directed line in space, we use its angles with the coordinate axes. If a directed line passing through the origin makes angles $\alpha$, $\beta$, and $\gamma$ with the positive x, y, and z-axes respectively, these angles are called direction angles. The cosines of these angles, denoted by $l = \cos\alpha$, $m = \cos\beta$, and $n = \cos\gamma$, are called the Direction Cosines (DCs) of the line. A fundamental relation governing them is $l^2 + m^2 + n^2 = 1$.

Any three numbers $a$, $b$, and $c$ that are proportional to the direction cosines $l$, $m$, and $n$ of a line are called its Direction Ratios (DRs). This means $l/a = m/b = n/c = k$. From this relation, we can find the direction cosines of a line if its direction ratios are known: $l = \pm \frac{a}{\sqrt{a^2+b^2+c^2}}$, $m = \pm \frac{b}{\sqrt{a^2+b^2+c^2}}$, and $n = \pm \frac{c}{\sqrt{a^2+b^2+c^2}}$. Knowing the difference between DCs (which are unique and normalized) and DRs (which can be any proportional set of infinite values) is crucial for solving 3D coordinate geometry problems.

Step-by-Step: Deriving and Writing Equations of a Line

1. Case 1: Line passing through a point and parallel to a vector — Let the line pass through a point $A$ with position vector $\vec{a}$ and be parallel to vector $\vec{b}$. The Vector Equation is given by $\vec{r} = \vec{a} + \lambda\vec{b}$, where $\lambda$ is a scalar parameter. To write the Cartesian Equation, let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$, $\vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$, and $\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}$. Equating coefficients yields the Cartesian form: $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$.
2. Case 2: Line passing through two given points — Let the line pass through two points $A$ and $B$ with position vectors $\vec{a}$ and $\vec{b}$. The Vector Equation is given by $\vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a})$. In Cartesian coordinates, if the points are $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$, the direction ratios are $(x_2 - x_1)$, $(y_2 - y_1)$, and $(z_2 - z_1)$. The Cartesian equation becomes: $\frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1}$.

Worked Examples: Angle and Shortest Distance

• Example 1: Find the shortest distance between the skew lines whose vector equations are: $\vec{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k})$ and $\vec{r} = (2\hat{i} - \hat{j} - \hat{k}) + \mu(2\hat{i} + \hat{j} + 2\hat{k})$. Step 1: Identify parameters. $\vec{a}_1 = \hat{i} + 2\hat{j} + \hat{k}$, $\vec{b}_1 = \hat{i} - \hat{j} + \hat{k}$ $\vec{a}_2 = 2\hat{i} - \hat{j} - \hat{k}$, $\vec{b}_2 = 2\hat{i} + \hat{j} + 2\hat{k}$ Step 2: Find $\vec{a}_2 - \vec{a}_1$: $\vec{a}_2 - \vec{a}_1 = (2-1)\hat{i} + (-1-2)\hat{j} + (-1-1)\hat{k} = \hat{i} - 3\hat{j} - 2\hat{k}$ Step 3: Calculate cross product $\vec{b}_1 \times \vec{b}_2$: $\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 2 & 1 & 2 \end{vmatrix} = \hat{i}(-2 - 1) - \hat{j}(2 - 2) + \hat{k}(1 - (-2)) = -3\hat{i} + 3\hat{k}$ Step 4: Find magnitude $|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-3)^2 + 0^2 + 3^2} = \sqrt{18} = 3\sqrt{2}$ Step 5: Compute scalar dot product $(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1)$: $(-3\hat{i} + 3\hat{k}) \cdot (\hat{i} - 3\hat{j} - 2\hat{k}) = (-3)(1) + (0)(-3) + (3)(-2) = -3 + 0 - 6 = -9$ Step 6: Use Shortest Distance formula $d = \left| \frac{(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1)}{|\vec{b}_1 \times \vec{b}_2|} \right|$ $d = \left| \frac{-9}{3\sqrt{2}} \right| = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}$ units.
• Example 2: Find the value of $p$ so that the lines $\frac{1-x}{3} = \frac{7y-14}{2p} = \frac{z-3}{2}$ and $\frac{7-7x}{3p} = \frac{y-5}{1} = \frac{6-z}{5}$ are perpendicular. Step 1: Rewrite both equations in the standard Cartesian form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$: First line: $\frac{x-1}{-3} = \frac{y-2}{2p/7} = \frac{z-3}{2}$. Direction ratios $\vec{b}_1 = (-3, \frac{2p}{7}, 2)$. Second line: $\frac{x-1}{-3p/7} = \frac{y-5}{1} = \frac{z-6}{-5}$. Direction ratios $\vec{b}_2 = (-\frac{3p}{7}, 1, -5)$. Step 2: Apply the perpendicular condition $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$: $(-3)(-\frac{3p}{7}) + (\frac{2p}{7})(1) + (2)(-5) = 0$ $\frac{9p}{7} + \frac{2p}{7} - 10 = 0 \implies \frac{11p}{7} = 10 \implies p = \frac{70}{11}$.

Board Exam Trap: Standard Cartesian Form

One of the most common places CBSE students lose marks in Class 12 Three Dimensional Geometry is failing to convert the given Cartesian equations of lines into their standard form before pulling out direction ratios.

Always ensure that the coefficients of $x$, $y$, and $z$ in the numerators of the equations are exactly $+1$. For example, if you are given $\frac{3-x}{2} = \frac{2y+3}{4} = z$, you must rewrite it as:
$\frac{x-3}{-2} = \frac{y + 3/2}{2} = \frac{z-0}{1}$
Only then can you correctly state that the direction ratios of the line are $-2, 2, 1$. Applying formulas using raw numbers from non-standard equations will result in incorrect vectors and zero credit on board evaluations.

Practice Questions with Solutions

• Q: Find the coordinates of the foot of the perpendicular drawn from the point $P(1, 2, 1)$ to the line $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$. A: Step 1: Write a general point on the given line. Let $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} = \lambda$. This gives $x = 2\lambda + 1$, $y = 3\lambda + 2$, $z = 4\lambda + 3$. Let this general point be $Q(2\lambda+1, 3\lambda+2, 4\lambda+3)$ represent the foot of the perpendicular. Step 2: Find the direction ratios of the line segment $PQ$: $DRs = (2\lambda + 1 - 1, 3\lambda + 2 - 2, 4\lambda + 3 - 1) = (2\lambda, 3\lambda, 4\lambda + 2)$. Step 3: Since $PQ$ is perpendicular to the given line (whose DRs are $2, 3, 4$), the dot product of their DRs must be zero: $2(2\lambda) + 3(3\lambda) + 4(4\lambda + 2) = 0$ $4\lambda + 9\lambda + 16\lambda + 8 = 0$ $29\lambda + 8 = 0 \implies \lambda = -\frac{8}{29}$. Step 4: Substitute $\lambda$ back into the coordinates of $Q$: $x = 2(-\frac{8}{29}) + 1 = \frac{13}{29}$ $y = 3(-\frac{8}{29}) + 2 = \frac{34}{29}$ $z = 4(-\frac{8}{29}) + 3 = \frac{55}{29}$. Final answer: The coordinates of the foot of the perpendicular are $\left(\frac{13}{29}, \frac{34}{29}, \frac{55}{29}\right)$.
• Q: Show that the line joining the points $(4, 7, 8)$ and $(2, 3, 4)$ is parallel to the line joining the points $(-1, -2, 1)$ and $(1, 2, 5)$. A: Step 1: Find the direction ratios of the first line, $L_1$, joining $A(4, 7, 8)$ and $B(2, 3, 4)$: $a_1 = 2 - 4 = -2$ $b_1 = 3 - 7 = -4$ $c_1 = 4 - 8 = -4$ So, the DRs of $L_1$ are $(-2, -4, -4)$, which simplifies to $(1, 2, 2)$ by dividing by $-2$. Step 2: Find the direction ratios of the second line, $L_2$, joining $C(-1, -2, 1)$ and $D(1, 2, 5)$: $a_2 = 1 - (-1) = 2$ $b_2 = 2 - (-2) = 4$ $c_2 = 5 - 1 = 4$ So, the DRs of $L_2$ are $(2, 4, 4)$, which simplifies to $(1, 2, 2)$ by dividing by $2$. Step 3: Compare the direction ratios. Two lines are parallel if their direction ratios are proportional: $\frac{a_1}{a_2} = \frac{-2}{2} = -1$ $\frac{b_1}{b_2} = \frac{-4}{4} = -1$ $\frac{c_1}{c_2} = \frac{-4}{4} = -1$ Final answer: Since the direction ratios are proportional (ratio is $-1$), the two lines are parallel.
• Q: Find the shortest distance between the lines $L_1$ and $L_2$ given by Cartesian equations $\frac{x-1}{1} = \frac{y-2}{-1} = \frac{z-1}{1}$ and $\frac{x-2}{2} = \frac{y-1}{1} = \frac{z-3}{2}$. A: Step 1: Extract vectors $\vec{a}_1$, $\vec{a}_2$, $\vec{b}_1$, and $\vec{b}_2$ from the standard forms: $\vec{a}_1 = \hat{i} + 2\hat{j} + \hat{k}$, $\vec{b}_1 = \hat{i} - \hat{j} + \hat{k}$ $\vec{a}_2 = 2\hat{i} + \hat{j} + 3\hat{k}$, $\vec{b}_2 = 2\hat{i} + \hat{j} + 2\hat{k}$ Step 2: Find $\vec{a}_2 - \vec{a}_1 = \hat{i} - \hat{j} + 2\hat{k}$. Step 3: Calculate the cross product $\vec{b}_1 \times \vec{b}_2$: $\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 2 & 1 & 2 \end{vmatrix} = \hat{i}(-2 - 1) - \hat{j}(2 - 2) + \hat{k}(1 - (-2)) = -3\hat{i} + 3\hat{k}$. Step 4: Find magnitude $|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-3)^2 + 3^2} = 3\sqrt{2}$. Step 5: Calculate the dot product $(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1)$: $(-3\hat{i} + 3\hat{k}) \cdot (\hat{i} - \hat{j} + 2\hat{k}) = -3(1) + 0(-1) + 3(2) = -3 + 6 = 3$. Step 6: Compute the shortest distance: $d = \left| \frac{3}{3\sqrt{2}} \right| = \frac{1}{\sqrt{2}}$ units. Final answer: The shortest distance is $\frac{1}{\sqrt{2}}$ units.
• Q: Find the vector equation of a line passing through the point $(2, -1, 4)$ and parallel to the line joining the points $(1, 2, -1)$ and $(3, -2, 5)$. A: Step 1: Find the direction ratios of the line joining $(1, 2, -1)$ and $(3, -2, 5)$. Vector along the parallel line $\vec{b} = (3-1)\hat{i} + (-2-2)\hat{j} + (5 - (-1))\hat{k} = 2\hat{i} - 4\hat{j} + 6\hat{k}$. Step 2: Simplify the direction ratios (optional but standard practice). We can use $\vec{b} = \hat{i} - 2\hat{j} + 3\hat{k}$ as the direction vector. Step 3: Write the vector equation of the line passing through point $A(2, -1, 4)$ with position vector $\vec{a} = 2\hat{i} - \hat{j} + 4\hat{k}$: $\vec{r} = \vec{a} + \lambda\vec{b}$ $\vec{r} = (2\hat{i} - \hat{j} + 4\hat{k}) + \lambda(\hat{i} - 2\hat{j} + 3\hat{k})$. Final answer: The vector equation is $\vec{r} = (2\hat{i} - \hat{j} + 4\hat{k}) + \lambda(\hat{i} - 2\hat{j} + 3\hat{k})$.

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