NCERT Class 12 Maths: 3 Dimensional Geometry Exercise 11.1

Understanding Lines in Three-Dimensional Space

Before we dive into specifics, let's briefly recall our understanding of the 3D coordinate system. We use three mutually perpendicular axes – X, Y, and Z – intersecting at the origin (0,0,0) to define any point in space as (x, y, z). While a point has a position, a line in 3D space has both a position and a direction. Describing this direction accurately is crucial, and that's where direction cosines and direction ratios come into play.

Imagine a line passing through the origin. Its direction can be uniquely determined by the angles it makes with the positive directions of the X, Y, and Z axes. If the line does not pass through the origin, we can draw a line parallel to it through the origin. Since parallel lines have the same direction, the angles made by this parallel line with the coordinate axes will represent the direction of the original line. This concept forms the basis for defining direction cosines.

Direction Cosines: The Angles of Orientation

Direction Cosines

The direction cosines of a directed line are the cosines of the angles which the line makes with the positive directions of the coordinate axes. If a line makes angles α, β, and γ with the positive X, Y, and Z axes respectively, then its direction cosines are $\cos\alpha$, $\cos\beta$, and $\cos\gamma$. These are usually denoted by $l$, $m$, and $n$ respectively. So, $l = \cos\alpha$, $m = \cos\beta$, $n = \cos\gamma$.

Fundamental Relation

For any set of direction cosines $(l, m, n)$ of a line, the following fundamental relation always holds true: $l^2 + m^2 + n^2 = 1$. This identity is extremely important and is derived from the Pythagorean theorem in 3D space. It implies that the direction cosines are not independent of each other.

Unique Direction

A directed line has a unique set of direction cosines. If the line is undirected, then it can be traversed in two opposite directions. If $(l, m, n)$ are the direction cosines for one direction, then $(-l, -m, -n)$ will be the direction cosines for the opposite direction.

Direction Ratios: Proportional Representation of Direction

While direction cosines uniquely define the direction of a line, they can sometimes be cumbersome to work with directly, especially when dealing with lines connecting two points. This is where direction ratios become incredibly useful. Direction ratios are simpler to calculate and provide a proportional representation of the direction of a line.

Any three numbers $a, b, c$ which are proportional to the direction cosines $l, m, n$ of a line are called its direction ratios. This means that $l = ka$, $m = kb$, $n = kc$ for some non-zero constant $k$. Since $l^2 + m^2 + n^2 = 1$, we can substitute these values:

$(ka)^2 + (kb)^2 + (kc)^2 = 1$
$k^2(a^2 + b^2 + c^2) = 1$
$k = \frac{1}{\pm\sqrt{a^2 + b^2 + c^2}}$

Therefore, the direction cosines can be expressed in terms of direction ratios as:
$l = \frac{a}{\pm\sqrt{a^2 + b^2 + c^2}}$
$m = \frac{b}{\pm\sqrt{a^2 + b^2 + c^2}}$
$n = \frac{c}{\pm\sqrt{a^2 + b^2 + c^2}}$

It's important to note that a line can have infinitely many sets of direction ratios, all proportional to each other. For example, if $(1, 2, 3)$ are direction ratios, then $(2, 4, 6)$ or $(-1, -2, -3)$ are also valid sets of direction ratios for the same line. However, it will have only two sets of direction cosines $(l, m, n)$ and $(-l, -m, -n)$ corresponding to the two opposite directions of the line.

Process: Finding Direction Ratios and Cosines from Two Points

1. Step 1: Identify the Coordinates of the Two Points — Let the two given points be $P_1(x_1, y_1, z_1)$ and $P_2(x_2, y_2, z_2)$. These points define the line segment for which we want to find the direction.
2. Step 2: Calculate the Direction Ratios (DRs) — The direction ratios of the line segment joining $P_1$ to $P_2$ are given by the differences in their coordinates: $a = x_2 - x_1$, $b = y_2 - y_1$, and $c = z_2 - z_1$. These values $(a, b, c)$ form a set of direction ratios for the line.
3. Step 3: Calculate the Magnitude (or Modulus) of the DRs — Find the value $\sqrt{a^2 + b^2 + c^2}$. This is the magnitude of the vector representing the direction, and it's also the distance between the two points, $P_1P_2$.
4. Step 4: Calculate the Direction Cosines (DCs) — Using the direction ratios $(a, b, c)$ and the magnitude calculated in Step 3, the direction cosines $(l, m, n)$ are: $l = \frac{a}{\sqrt{a^2 + b^2 + c^2}}$ $m = \frac{b}{\sqrt{a^2 + b^2 + c^2}}$ $n = \frac{c}{\sqrt{a^2 + b^2 + c^2}}$ Remember, if the line is undirected, you can also have $(-l, -m, -n)$ as the direction cosines.
5. Step 5: Verify (Optional but Recommended) — To ensure your calculations are correct, check if $l^2 + m^2 + n^2 = 1$. If this identity holds, your direction cosines are correctly calculated.

Worked Examples

• Example 1: If a line makes angles 90°, 135°, 45° with the positive X, Y, and Z axes respectively, find its direction cosines. Solution: Step 1: Identify the given angles. We are given $\alpha = 90°$, $\beta = 135°$, $\gamma = 45°$. Step 2: Calculate the cosines of these angles. $l = \cos\alpha = \cos 90° = 0$ $m = \cos\beta = \cos 135° = -\frac{1}{\sqrt{2}}$ $n = \cos\gamma = \cos 45° = \frac{1}{\sqrt{2}}$ Step 3: State the direction cosines. The direction cosines of the line are $(0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$. Step 4: Verify (Optional). $l^2 + m^2 + n^2 = (0)^2 + (-\frac{1}{\sqrt{2}})^2 + (\frac{1}{\sqrt{2}})^2 = 0 + \frac{1}{2} + \frac{1}{2} = 1$. The relation holds. Final Answer: The direction cosines are $(0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$.
• Example 2: Find the direction ratios and direction cosines of the line joining the points A(1, 2, -3) and B(-1, -2, 1). Solution: Step 1: Identify the coordinates of the two points. $P_1 = (x_1, y_1, z_1) = (1, 2, -3)$ $P_2 = (x_2, y_2, z_2) = (-1, -2, 1)$ Step 2: Calculate the direction ratios (DRs). $a = x_2 - x_1 = -1 - 1 = -2$ $b = y_2 - y_1 = -2 - 2 = -4$ $c = z_2 - z_1 = 1 - (-3) = 1 + 3 = 4$ So, the direction ratios are $(-2, -4, 4)$. Step 3: Calculate the magnitude of the DRs. $\sqrt{a^2 + b^2 + c^2} = \sqrt{(-2)^2 + (-4)^2 + (4)^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$. Step 4: Calculate the direction cosines (DCs). $l = \frac{a}{\sqrt{a^2 + b^2 + c^2}} = \frac{-2}{6} = -\frac{1}{3}$ $m = \frac{b}{\sqrt{a^2 + b^2 + c^2}} = \frac{-4}{6} = -\frac{2}{3}$ $n = \frac{c}{\sqrt{a^2 + b^2 + c^2}} = \frac{4}{6} = \frac{2}{3}$ Final Answer: The direction ratios are $(-2, -4, 4)$ and the direction cosines are $(-\frac{1}{3}, -\frac{2}{3}, \frac{2}{3})$. (For the opposite direction, they would be $(\frac{1}{3}, \frac{2}{3}, -\frac{2}{3})$).
• Example 3: Show that the points A(2, 3, 4), B(-1, -2, 1), and C(5, 8, 7) are collinear. Solution: Step 1: Find the direction ratios of line segment AB. DRs of AB: $a_1 = -1 - 2 = -3$, $b_1 = -2 - 3 = -5$, $c_1 = 1 - 4 = -3$. So, DRs of AB are $(-3, -5, -3)$. Step 2: Find the direction ratios of line segment BC. DRs of BC: $a_2 = 5 - (-1) = 6$, $b_2 = 8 - (-2) = 10$, $c_2 = 7 - 1 = 6$. So, DRs of BC are $(6, 10, 6)$. Step 3: Check for proportionality of direction ratios. For points to be collinear, the direction ratios of AB and BC must be proportional. That is, $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$. $\frac{-3}{6} = -\frac{1}{2}$ $\frac{-5}{10} = -\frac{1}{2}$ $\frac{-3}{6} = -\frac{1}{2}$ Since the ratios are equal, DRs of AB are proportional to DRs of BC. Step 4: Conclude collinearity. As the direction ratios of AB and BC are proportional, and point B is common to both segments, the points A, B, and C are collinear. Final Answer: The points A, B, and C are collinear.

Exam Tips and Common Mistakes to Avoid

When solving problems related to direction cosines and direction ratios, students often make a few common errors. Be mindful of these to score well:

1. Confusing Direction Cosines with Direction Ratios: Remember, direction cosines $(l, m, n)$ always satisfy $l^2 + m^2 + n^2 = 1$. Direction ratios $(a, b, c)$ are only proportional to direction cosines and can be any set of three numbers, not necessarily satisfying $a^2 + b^2 + c^2 = 1$. There's only one set of direction cosines for a directed line (and its negative for the opposite direction), but infinite sets of direction ratios.
2. Sign Convention: Pay close attention to the signs when calculating direction ratios $(x_2 - x_1, y_2 - y_1, z_2 - z_1)$. A wrong sign can completely change the direction and lead to incorrect answers. Also, when finding direction cosines from ratios, ensure the sign of $\sqrt{a^2 + b^2 + c^2}$ is consistent (usually taken as positive unless a specific direction is implied).
3. Angles with Axes: Ensure the angles $\alpha, \beta, \gamma$ are measured with the positive directions of the X, Y, and Z axes respectively. An angle greater than 90° will result in a negative cosine value.
4. Collinearity Check: For collinearity, not only must the direction ratios be proportional, but there must also be a common point among the line segments. Just checking proportionality isn't enough; for example, two parallel lines have proportional DRs but are not collinear if they don't share a point.

Practice Questions with Solutions

• Q: A line makes equal angles with the coordinate axes. What are its direction cosines? A: Step 1: Let the equal angles be $\alpha$. So, $\alpha = \beta = \gamma$. Step 2: Use the fundamental relation for direction cosines. We know that $l^2 + m^2 + n^2 = 1$. Since $l = \cos\alpha$, $m = \cos\beta$, $n = \cos\gamma$, we have $\cos^2\alpha + \cos^2\alpha + \cos^2\alpha = 1$. Step 3: Solve for $\cos\alpha$. $3\cos^2\alpha = 1 \implies \cos^2\alpha = \frac{1}{3} \implies \cos\alpha = \pm\frac{1}{\sqrt{3}}$. Final answer: The direction cosines are $(\pm\frac{1}{\sqrt{3}}, \pm\frac{1}{\sqrt{3}}, \pm\frac{1}{\sqrt{3}})$. (Note: All signs must be the same for a single direction, so it's either all positive or all negative).
• Q: Find the direction cosines of a line whose direction ratios are 2, -1, -2. A: Step 1: Identify the given direction ratios. Let $a = 2$, $b = -1$, $c = -2$. Step 2: Calculate $\sqrt{a^2 + b^2 + c^2}$. $\sqrt{2^2 + (-1)^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$. Step 3: Calculate the direction cosines. $l = \frac{a}{\sqrt{a^2 + b^2 + c^2}} = \frac{2}{3}$ $m = \frac{b}{\sqrt{a^2 + b^2 + c^2}} = \frac{-1}{3}$ $n = \frac{c}{\sqrt{a^2 + b^2 + c^2}} = \frac{-2}{3}$ Final answer: The direction cosines are $(\frac{2}{3}, -\frac{1}{3}, -\frac{2}{3})$.
• Q: Determine the direction ratios and direction cosines of the line joining the origin (0, 0, 0) to the point P(6, -2, 3). A: Step 1: Identify the two points. $P_1 = (x_1, y_1, z_1) = (0, 0, 0)$ $P_2 = (x_2, y_2, z_2) = (6, -2, 3)$ Step 2: Calculate the direction ratios (DRs). $a = x_2 - x_1 = 6 - 0 = 6$ $b = y_2 - y_1 = -2 - 0 = -2$ $c = z_2 - z_1 = 3 - 0 = 3$ So, the direction ratios are $(6, -2, 3)$. Step 3: Calculate the magnitude of the DRs. $\sqrt{a^2 + b^2 + c^2} = \sqrt{6^2 + (-2)^2 + 3^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7$. Step 4: Calculate the direction cosines (DCs). $l = \frac{6}{7}$, $m = \frac{-2}{7}$, $n = \frac{3}{7}$. Final answer: The direction ratios are $(6, -2, 3)$ and the direction cosines are $(\frac{6}{7}, -\frac{2}{7}, \frac{3}{7})$.
• Q: Show that the points P(1, 2, 3), Q(4, 5, 7), and R(-2, -1, -1) are not collinear. A: Step 1: Find the direction ratios of line segment PQ. DRs of PQ: $a_1 = 4 - 1 = 3$, $b_1 = 5 - 2 = 3$, $c_1 = 7 - 3 = 4$. So, DRs of PQ are $(3, 3, 4)$. Step 2: Find the direction ratios of line segment QR. DRs of QR: $a_2 = -2 - 4 = -6$, $b_2 = -1 - 5 = -6$, $c_2 = -1 - 7 = -8$. So, DRs of QR are $(-6, -6, -8)$. Step 3: Check for proportionality of direction ratios. $\frac{a_1}{a_2} = \frac{3}{-6} = -\frac{1}{2}$ $\frac{b_1}{b_2} = \frac{3}{-6} = -\frac{1}{2}$ $\frac{c_1}{c_2} = \frac{4}{-8} = -\frac{1}{2}$ Since the ratios are equal, the direction ratios are proportional. This means PQ is parallel to QR. Step 4: Conclude collinearity. Since the direction ratios are proportional AND Q is a common point to both segments PQ and QR, the points P, Q, and R are collinear. (My initial thought was 'not collinear', but the calculation shows they are.) Final answer: The points P, Q, and R are collinear.

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