RBSE Solutions Class 12 Maths Chapter 14 Three Dimensional Geometry Exercise 14.1

Question 1. Find the direction cosines of a line which makes equal angles with the coordinate axis.
Answer: Let the line make an equal angle \( \theta \) with each coordinate axis (X, Y, and Z). The direction cosines of the line are \( l = \cos \alpha \), \( m = \cos \beta \), and \( n = \cos \gamma \). Since the angles are equal, \( \alpha = \beta = \gamma = \theta \). Therefore, the direction cosines are \( l = \cos \theta \), \( m = \cos \theta \), and \( n = \cos \theta \). We know that the sum of the squares of the direction cosines is always 1: \( l^2 + m^2 + n^2 = 1 \) Substitute the direction cosines: \( (\cos \theta)^2 + (\cos \theta)^2 + (\cos \theta)^2 = 1 \) \( 3 \cos^2 \theta = 1 \)
\( \implies \cos^2 \theta = \frac{1}{3} \)
\( \implies \cos \theta = \pm \sqrt{\frac{1}{3}} \)
\( \implies \cos \theta = \pm \frac{1}{\sqrt{3}} \) So, the direction cosines of the line are \( \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}} \). This means there are two possible sets of direction cosines, corresponding to lines making equal angles in different directions.In simple words: When a line makes the same angle with all three axes, its direction cosines are all equal. Because the square of these cosines must add up to one, each cosine has to be plus or minus one divided by the square root of three.

🎯 Exam Tip: Remember the fundamental property that \( \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \) for direction cosines. This is crucial for solving problems involving angles with axes.

Question 2. Find the direction of the line joining two points (4, 2, 3) and (4, 5, 7).
Answer: Let the two given points be \( P(x_1, y_1, z_1) = (4, 2, 3) \) and \( Q(x_2, y_2, z_2) = (4, 5, 7) \). First, we find the direction ratios of the line segment PQ. The direction ratios are given by \( (x_2 - x_1), (y_2 - y_1), (z_2 - z_1) \). Direction ratios \( = (4 - 4), (5 - 2), (7 - 3) \) Direction ratios \( = (0, 3, 4) \) Next, we find the magnitude of the vector PQ, which is the distance between the two points. \( PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \) \( PQ = \sqrt{(0)^2 + (3)^2 + (4)^2} \) \( PQ = \sqrt{0 + 9 + 16} \) \( PQ = \sqrt{25} \) \( PQ = 5 \) Finally, the direction cosines (l, m, n) are found by dividing each direction ratio by the magnitude. \( l = \frac{x_2 - x_1}{PQ} = \frac{0}{5} = 0 \) \( m = \frac{y_2 - y_1}{PQ} = \frac{3}{5} \) \( n = \frac{z_2 - z_1}{PQ} = \frac{4}{5} \) So, the direction cosines of the line joining the points (4, 2, 3) and (4, 5, 7) are \( 0, \frac{3}{5}, \frac{4}{5} \). These cosines describe the exact orientation of the line in 3D space.In simple words: To find the line's direction, first subtract the coordinates of the two points to get its direction ratios. Then, divide these ratios by the total length of the line segment to get the direction cosines.

🎯 Exam Tip: Remember that direction ratios can be any set of numbers proportional to the direction cosines. However, direction cosines are unique for a given direction and their squares sum to 1.

Question 3. If a line has direction ratios 2, -1, -2 determine its direction cosines.
Answer: Let the given direction ratios of the line be \( a = 2, b = -1, c = -2 \). To find the direction cosines (l, m, n), we first need to calculate the magnitude (or the normalizing factor) \( r \), which is \( \sqrt{a^2 + b^2 + c^2} \). \( r = \sqrt{(2)^2 + (-1)^2 + (-2)^2} \) \( r = \sqrt{4 + 1 + 4} \) \( r = \sqrt{9} \) \( r = 3 \) Now, we can find the direction cosines by dividing each direction ratio by \( r \): \( l = \frac{a}{r} = \frac{2}{3} \) \( m = \frac{b}{r} = \frac{-1}{3} \) \( n = \frac{c}{r} = \frac{-2}{3} \) Thus, the direction cosines of the line are \( \frac{2}{3}, -\frac{1}{3}, -\frac{2}{3} \). These values help define the line's orientation and relationship with the coordinate axes.In simple words: First, find the "length" of the direction ratios by squaring them, adding them, and taking the square root. Then, divide each original direction ratio by this "length" to get the direction cosines.

🎯 Exam Tip: Always calculate the magnitude \( r \) accurately, as it is used to normalize the direction ratios into direction cosines. Double-check your squaring and square root operations.

Question 4. A vector \( \overrightarrow {r} \) makes angles of \( 45^\circ \), \( 60^\circ \), \( 120^\circ \) with X, Y and Z-axis respectively. If magnitude of \( \overrightarrow {r} \) is 2 units then find \( \overrightarrow {r} \)
Answer: We are given the angles that the vector \( \overrightarrow{r} \) makes with the X, Y, and Z axes: \( \alpha = 45^\circ \) (with X-axis) \( \beta = 60^\circ \) (with Y-axis) \( \gamma = 120^\circ \) (with Z-axis) The magnitude of the vector is given as \( |\overrightarrow{r}| = 2 \) units. First, we find the direction cosines (l, m, n) using these angles: \( l = \cos \alpha = \cos 45^\circ = \frac{1}{\sqrt{2}} \) \( m = \cos \beta = \cos 60^\circ = \frac{1}{2} \) \( n = \cos \gamma = \cos 120^\circ = -\frac{1}{2} \) A vector \( \overrightarrow{r} \) can be expressed in terms of its magnitude and direction cosines as: \( \overrightarrow{r} = |\overrightarrow{r}| (l \hat{i} + m \hat{j} + n \hat{k}) \) Substitute the given magnitude and the calculated direction cosines into this formula: \( \overrightarrow{r} = 2 \left( \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{2} \hat{j} - \frac{1}{2} \hat{k} \right) \) Now, distribute the magnitude (2) to each component: \( \overrightarrow{r} = \left( 2 \times \frac{1}{\sqrt{2}} \right) \hat{i} + \left( 2 \times \frac{1}{2} \right) \hat{j} - \left( 2 \times \frac{1}{2} \right) \hat{k} \) \( \overrightarrow{r} = \sqrt{2} \hat{i} + 1 \hat{j} - 1 \hat{k} \) So, the vector \( \overrightarrow{r} \) is \( \sqrt{2} \hat{i} + \hat{j} - \hat{k} \). This is the component form of the vector, showing its direction and length in a structured way.In simple words: We are given how much the vector "leans" towards each axis and how long it is. We use the cosine of these angles to find its direction parts. Then, we multiply each direction part by the vector's total length to get the final vector components.

🎯 Exam Tip: Be careful with the signs of cosine values, especially for angles in the second quadrant (like \( \cos 120^\circ \)). A negative cosine indicates a component in the negative direction of that axis.