GSEB Class 12 Maths Solutions Chapter 11 Three Dimensional Geometry Exercise 11.1

Question 1. If a line makes angles 90°, 135° and 45° with x, y and z axes respectively, find its direction cosines.
Answer: The direction angles given are \( 90^\circ \), \( 135^\circ \), and \( 45^\circ \). To find the direction cosines, we take the cosine of each angle.
Let \( l \), \( m \), and \( n \) be the direction cosines.
\( l = \cos 90^\circ = 0 \)
\( m = \cos 135^\circ = -\frac{1}{\sqrt{2}} \)
\( n = \cos 45^\circ = \frac{1}{\sqrt{2}} \)
Therefore, the direction cosines for the line are \( 0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \).
In simple words: When a line makes certain angles with the x, y, and z axes, its direction cosines are simply the cosine of each of those angles.

Exam Tip: Remember that direction cosines are found by taking the cosine of the angles a line makes with the positive x, y, and z axes. Ensure you recall the cosine values for common angles like 90°, 135°, and 45°.

Question 2. Find the direction cosines of a line which makes equal angles with coordinate axes.
Answer: Let the line make an equal angle \( \alpha \) with the coordinate axes (x, y, and z axes).
The direction cosines of this line are \( \cos \alpha \), \( \cos \alpha \), and \( \cos \alpha \).
We know that for any line, the sum of the squares of its direction cosines is 1, i.e., \( l^2 + m^2 + n^2 = 1 \).
Substituting the direction cosines:
\( \cos^2 \alpha + \cos^2 \alpha + \cos^2 \alpha = 1 \)
\( 3\cos^2 \alpha = 1 \)

\( \implies \cos^2 \alpha = \frac{1}{3} \)

\( \implies \cos \alpha = \pm\frac{1}{\sqrt{3}} \)
Thus, the direction cosines of a line that makes equal angles with the coordinate axes are \( \pm\frac{1}{\sqrt{3}}, \pm\frac{1}{\sqrt{3}}, \pm\frac{1}{\sqrt{3}} \).
In simple words: If a line leans at the same angle to all three axes, then the cosine of that angle, squared and added three times, must equal one. This helps us calculate that cosine value.

Exam Tip: For lines making equal angles with the coordinate axes, the direction cosines are always \( \pm\frac{1}{\sqrt{3}} \). This is a standard result that can be quickly applied in multiple-choice questions.

Question 3. If a line has the direction ratios – 18, 12 and – 4, then what are its direction cosines?
Answer: The direction ratios of the line are \( a = -18 \), \( b = 12 \), and \( c = -4 \).
To find the direction cosines, we first calculate \( \sqrt{a^2 + b^2 + c^2} \).
\( \sqrt{(-18)^2 + (12)^2 + (-4)^2} \)
\( = \sqrt{324 + 144 + 16} \)
\( = \sqrt{484} \)
\( = 22 \)
The direction cosines are given by \( \frac{a}{\sqrt{a^2 + b^2 + c^2}}, \frac{b}{\sqrt{a^2 + b^2 + c^2}}, \frac{c}{\sqrt{a^2 + b^2 + c^2}} \).
So, the direction cosines are:
\( \frac{-18}{22}, \frac{12}{22}, \frac{-4}{22} \)
Simplifying these fractions, we get:
\( -\frac{9}{11}, \frac{6}{11}, -\frac{2}{11} \).
In simple words: To change direction ratios into direction cosines, first find the square root of the sum of the squares of the ratios. Then, divide each ratio by this number.

Exam Tip: Remember to simplify the fractions for direction cosines to their lowest terms. A common mistake is not simplifying or making sign errors with negative ratios.

Question 4. Show that the points (2, 3, 4), (- 1, – 2, 1) and (5, 8, 7) are collinear.
Answer: Let the given points be \( A(2, 3, 4) \), \( B(-1, -2, 1) \), and \( C(5, 8, 7) \).
We will find the direction ratios for the line segments AB and BC.
For AB, the direction ratios are \( (x_2 - x_1, y_2 - y_1, z_2 - z_1) \):
\( (-1 - 2), (-2 - 3), (1 - 4) \)
\( = (-3, -5, -3) \)
For BC, the direction ratios are \( (x_2 - x_1, y_2 - y_1, z_2 - z_1) \):
\( (5 - (-1)), (8 - (-2)), (7 - 1) \)
\( = (5 + 1), (8 + 2), (6) \)
\( = (6, 10, 6) \)
Now, we compare the direction ratios of AB and BC. We notice that the direction ratios of BC are \( -2 \) times the direction ratios of AB:
\( 6 = -2 \times (-3) \)
\( 10 = -2 \times (-5) \)
\( 6 = -2 \times (-3) \)
Since the direction ratios of AB and BC are proportional, it means that the line segments AB and BC are parallel to each other. Because point B is a common point to both segments AB and BC, all three points A, B, and C must lie on the same straight line.
Hence, A, B, and C are collinear.
In simple words: To prove points are collinear, find the direction ratios of two segments formed by the points (e.g., AB and BC). If these ratios are proportional, and the segments share a common point, the points are on the same line.

Exam Tip: Points are collinear if the direction ratios of the line segments formed by them are proportional and they share a common point. Ensure the common point is explicitly stated to justify collinearity.

Question 5. Find the direction cosines of the sides of the triangle whose vertices are (3, 5, – 4), (- 1, 1, 5, - 2).
Answer: The vertices of triangle ABC are given as \( A(3, 5, -4) \), \( B(-1, 1, 2) \), and \( C(-5, -5, -2) \). We need to find the direction cosines for each side: AB, BC, and CA.

(i) **For side AB**: Vertices are \( A(3, 5, -4) \) and \( B(-1, 1, 2) \).
The direction ratios \( (a, b, c) \) are \( (x_2 - x_1, y_2 - y_1, z_2 - z_1) \):
\( a = -1 - 3 = -4 \)
\( b = 1 - 5 = -4 \)
\( c = 2 - (-4) = 2 + 4 = 6 \)
Next, calculate \( \sqrt{a^2 + b^2 + c^2} = \sqrt{(-4)^2 + (-4)^2 + (6)^2} = \sqrt{16 + 16 + 36} = \sqrt{68} = 2\sqrt{17} \).
The direction cosines of AB are \( \left( \frac{-4}{2\sqrt{17}}, \frac{-4}{2\sqrt{17}}, \frac{6}{2\sqrt{17}} \right) = \left( \frac{-2}{\sqrt{17}}, \frac{-2}{\sqrt{17}}, \frac{3}{\sqrt{17}} \right) \).

(ii) **For side BC**: Vertices are \( B(-1, 1, 2) \) and \( C(-5, -5, -2) \).
The direction ratios \( (a, b, c) \) are \( (x_2 - x_1, y_2 - y_1, z_2 - z_1) \):
\( a = -5 - (-1) = -5 + 1 = -4 \)
\( b = -5 - 1 = -6 \)
\( c = -2 - 2 = -4 \)
These direction ratios \((-4, -6, -4)\) can be simplified to \((-2, -3, -2)\) by dividing by 2. We use the simplified ratios to calculate \( \sqrt{a^2 + b^2 + c^2} \).
\( \sqrt{(-2)^2 + (-3)^2 + (-2)^2} = \sqrt{4 + 9 + 4} = \sqrt{17} \).
The direction cosines of BC are \( \left( \frac{-2}{\sqrt{17}}, \frac{-3}{\sqrt{17}}, \frac{-2}{\sqrt{17}} \right) \).

(iii) **For side CA**: Vertices are \( C(-5, -5, -2) \) and \( A(3, 5, -4) \).
The direction ratios \( (a, b, c) \) are \( (x_2 - x_1, y_2 - y_1, z_2 - z_1) \):
\( a = 3 - (-5) = 3 + 5 = 8 \)
\( b = 5 - (-5) = 5 + 5 = 10 \)
\( c = -4 - (-2) = -4 + 2 = -2 \)
These direction ratios \((8, 10, -2)\) can be simplified to \((4, 5, -1)\) by dividing by 2. We use the simplified ratios to calculate \( \sqrt{a^2 + b^2 + c^2} \).
\( \sqrt{(4)^2 + (5)^2 + (-1)^2} = \sqrt{16 + 25 + 1} = \sqrt{42} \).
The direction cosines of CA are \( \left( \frac{4}{\sqrt{42}}, \frac{5}{\sqrt{42}}, \frac{-1}{\sqrt{42}} \right) \).
Thus, the direction cosines of the sides AB, BC, and CA are respectively:
\( \left( \frac{-2}{\sqrt{17}}, \frac{-2}{\sqrt{17}}, \frac{3}{\sqrt{17}} \right) \); \( \left( \frac{-2}{\sqrt{17}}, \frac{-3}{\sqrt{17}}, \frac{-2}{\sqrt{17}} \right) \); and \( \left( \frac{4}{\sqrt{42}}, \frac{5}{\sqrt{42}}, \frac{-1}{\sqrt{42}} \right) \).
In simple words: For each side of the triangle, calculate its direction ratios by subtracting coordinates. Then, find the magnitude of these ratios by squaring them, adding, and taking the square root. Finally, divide each ratio by its magnitude to get the direction cosines.

Exam Tip: When finding direction cosines for a triangle's sides, remember that the order of subtraction for coordinates matters (e.g., \( x_2 - x_1 \)). Always simplify the resulting fractions to their lowest terms and avoid calculation errors for square roots.