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CBSE - Class 12 Mathematics Three Dimensional Geometry Worksheet
EXERCISE 11.1
Worksheet Answers
Solution:
The direction cosines of a line in three-dimensional space are defined as the cosines of the angles ($\alpha, \beta, \gamma$) that the line makes with the positive directions of the $x$, $y$, and $z$-axes, respectively. These are denoted by $l, m,$ and $n$.
According to the definition:
From the problem statement, the angles are given as:
We calculate each direction cosine individually using trigonometric evaluations:
For $l$ (x-axis):
$l = \cos(90^{\circ}) = 0$
For $m$ (y-axis):
$m = \cos(135^{\circ}) = \cos(180^{\circ} - 45^{\circ})$
Using the trigonometric identity $\cos(180^{\circ} - \theta) = -\cos(\theta)$ [By the Supplementary Angle Theorem]:
$m = -\cos(45^{\circ}) = -\frac{1}{\sqrt{2}}$
For $n$ (z-axis):
$n = \cos(45^{\circ}) = \frac{1}{\sqrt{2}}$
The direction cosines must satisfy the fundamental identity $l^2 + m^2 + n^2 = 1$.
$(0)^2 + (-\frac{1}{\sqrt{2}})^2 + (\frac{1}{\sqrt{2}})^2 = 0 + \frac{1}{2} + \frac{1}{2} = 1$.
[The identity holds, confirming the values are correct.]
Final Solution: The direction cosines of the line are $0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}$.
Solution:
The direction ratios of a line are given as:
[Per the definition of direction ratios, any triplet $(a, b, c)$ proportional to the direction cosines $(l, m, n)$ represents the orientation of the line.]
To find the direction cosines, we must first determine the magnitude of the vector representing the direction ratios, denoted by $r = \sqrt{a^2 + b^2 + c^2}$.
$r = \sqrt{(-18)^2 + (12)^2 + (-4)^2}$
$r = \sqrt{324 + 144 + 16}$
$r = \sqrt{484}$
$r = 22$
The direction cosines $(l, m, n)$ are calculated by dividing each direction ratio by the magnitude $r$:
$l = \frac{a}{r} = \frac{-18}{22} = -\frac{9}{11}$
$m = \frac{b}{r} = \frac{12}{22} = \frac{6}{11}$
$n = \frac{c}{r} = \frac{-4}{22} = -\frac{2}{11}$
[By the relation $l^2 + m^2 + n^2 = 1$, we verify: $(-\frac{9}{11})^2 + (\frac{6}{11})^2 + (-\frac{2}{11})^2 = \frac{81 + 36 + 4}{121} = \frac{121}{121} = 1$.]
Since the direction cosines are defined by the vector divided by its magnitude, taking the negative of the magnitude also yields a valid set of direction cosines for the same line (representing the opposite orientation):
$(l, m, n) = (\frac{9}{11}, -\frac{6}{11}, \frac{2}{11})$
Final Solution: The direction cosines are $\left( -\frac{9}{11}, \frac{6}{11}, -\frac{2}{11} \right)$ or $\left( \frac{9}{11}, -\frac{6}{11}, \frac{2}{11} \right)$.
Solution:
Let the line make an angle $\alpha$ with each of the $x$, $y$, and $z$ axes. Consequently, the direction angles are $\alpha = \beta = \gamma = \alpha$.
The direction cosines of a line are defined as $l = \cos \alpha$, $m = \cos \beta$, and $n = \cos \gamma$. [Per the definition of direction cosines in 3D Euclidean space].
For any line, the sum of the squares of its direction cosines is unity [By the fundamental identity for direction cosines]:
$l^2 + m^2 + n^2 = 1$
Substituting the condition of equal angles ($\cos \alpha = \cos \beta = \cos \gamma$):
$\cos^2 \alpha + \cos^2 \alpha + \cos^2 \alpha = 1$
$3\cos^2 \alpha = 1$
Solving for $\cos \alpha$:
$\cos^2 \alpha = \frac{1}{3}$
$\cos \alpha = \pm \frac{1}{\sqrt{3}}$
Since $\cos \alpha = \cos \beta = \cos \gamma$, the direction cosines are:
$(l, m, n) = \left( \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}} \right)$
The set of direction cosines identifies the orientation of the line in space. Given the symmetry of the problem relative to the axes, two cases emerge based on the direction of the vector:
| Orientation | Direction Cosines $(l, m, n)$ |
|---|---|
| Case 1 | $\left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right)$ |
| Case 2 | $\left( -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}} \right)$ |
Final Solution: The direction cosines are $\pm \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right)$.
Solution:
Let the vertices of the triangle be defined as:
The direction cosines of a line segment connecting two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ are proportional to the direction ratios $(a, b, c) = (x_2-x_1, y_2-y_1, z_2-z_1)$. If $L$ is the length of the segment, where $L = \sqrt{a^2 + b^2 + c^2}$, the direction cosines $(l, m, n)$ are given by $\left( \frac{a}{L}, \frac{b}{L}, \frac{c}{L} \right)$ [Per the definition of direction cosines in $\mathbb{R}^3$].
Direction ratios for $\vec{AB}$: $a_1 = -1 - 3 = -4$, $b_1 = 1 - 5 = -4$, $c_1 = 2 - (-4) = 6$.
Length $L_{AB} = \sqrt{(-4)^2 + (-4)^2 + 6^2} = \sqrt{16 + 16 + 36} = \sqrt{68} = 2\sqrt{17}$.
Direction cosines $(l_1, m_1, n_1) = \left( \frac{-4}{2\sqrt{17}}, \frac{-4}{2\sqrt{17}}, \frac{6}{2\sqrt{17}} \right) = \mathbf{\left( -\frac{2}{\sqrt{17}}, -\frac{2}{\sqrt{17}}, \frac{3}{\sqrt{17}} \right)}$.
Direction ratios for $\vec{BC}$: $a_2 = -5 - (-1) = -4$, $b_2 = -5 - 1 = -6$, $c_2 = -2 - 2 = -4$.
Length $L_{BC} = \sqrt{(-4)^2 + (-6)^2 + (-4)^2} = \sqrt{16 + 36 + 16} = \sqrt{68} = 2\sqrt{17}$.
Direction cosines $(l_2, m_2, n_2) = \left( \frac{-4}{2\sqrt{17}}, \frac{-6}{2\sqrt{17}}, \frac{-4}{2\sqrt{17}} \right) = \mathbf{\left( -\frac{2}{\sqrt{17}}, -\frac{3}{\sqrt{17}}, -\frac{2}{\sqrt{17}} \right)}$.
Direction ratios for $\vec{CA}$: $a_3 = 3 - (-5) = 8$, $b_3 = 5 - (-5) = 10$, $c_3 = -4 - (-2) = -2$.
Length $L_{CA} = \sqrt{8^2 + 10^2 + (-2)^2} = \sqrt{64 + 100 + 4} = \sqrt{168} = 2\sqrt{42}$.
Direction cosines $(l_3, m_3, n_3) = \left( \frac{8}{2\sqrt{42}}, \frac{10}{2\sqrt{42}}, \frac{-2}{2\sqrt{42}} \right) = \mathbf{\left( \frac{4}{\sqrt{42}}, \frac{5}{\sqrt{42}}, -\frac{1}{\sqrt{42}} \right)}$.
| Side | Direction Cosines $(l, m, n)$ |
|---|---|
| $AB$ | $\left( -\frac{2}{\sqrt{17}}, -\frac{2}{\sqrt{17}}, \frac{3}{\sqrt{17}} \right)$ |
| $BC$ | $\left( -\frac{2}{\sqrt{17}}, -\frac{3}{\sqrt{17}}, -\frac{2}{\sqrt{17}} \right)$ |
| $CA$ | $\left( \frac{4}{\sqrt{42}}, \frac{5}{\sqrt{42}}, -\frac{1}{\sqrt{42}} \right)$ |
Final Solution: The direction cosines for the sides are $AB: (-\frac{2}{\sqrt{17}}, -\frac{2}{\sqrt{17}}, \frac{3}{\sqrt{17}})$, $BC: (-\frac{2}{\sqrt{17}}, -\frac{3}{\sqrt{17}}, -\frac{2}{\sqrt{17}})$, and $CA: (\frac{4}{\sqrt{42}}, \frac{5}{\sqrt{42}}, -\frac{1}{\sqrt{42}})$.
Solution:
To prove that three points $A$, $B$, and $C$ are collinear, we must demonstrate that they lie on the same straight line. In three-dimensional coordinate geometry, three points are collinear if the direction ratios of the line segments joining them are proportional. Alternatively, one can show that the distance between the two furthest points is equal to the sum of the distances between the adjacent segments [By the Triangle Inequality Theorem, if $d(A, B) + d(B, C) = d(A, C)$, the points are collinear].
Let the given points be:
The direction ratios $(a, b, c)$ of a line segment connecting $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ are given by $(x_2 - x_1, y_2 - y_1, z_2 - z_1)$.
For vector $\vec{AB}$:
$a_1 = -1 - 2 = -3$
$b_1 = -2 - 3 = -5$
$c_1 = 1 - 4 = -3$
Direction ratios of $AB = (-3, -5, -3)$.
For vector $\vec{BC}$:
$a_2 = 5 - (-1) = 6$
$b_2 = 8 - (-2) = 10$
$c_2 = 7 - 1 = 6$
Direction ratios of $BC = (6, 10, 6)$.
Two lines are parallel (or collinear if they share a common point) if their direction ratios are proportional [Condition of Collinearity: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$].
Taking the ratio of components of $\vec{AB}$ and $\vec{BC}$:
$\frac{a_1}{a_2} = \frac{-3}{6} = -\frac{1}{2}$
$\frac{b_1}{b_2} = \frac{-5}{10} = -\frac{1}{2}$
$\frac{c_1}{c_2} = \frac{-3}{6} = -\frac{1}{2}$
Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = -\frac{1}{2}$, the direction ratios are proportional. Because the line segments $AB$ and $BC$ share a common point $B$, the points $A$, $B$, and $C$ must lie on the same straight line.
Final Solution: Since the direction ratios of the segments $AB$ and $BC$ are proportional and the segments share the common point $B$, the points $(2, 3, 4)$, $(-1, -2, 1)$, and $(5, 8, 7)$ are proved to be collinear.
