OP Malhotra Class 12 Maths Solutions Chapter 23 Three Dimensional Geometry Exercise 23 (A)

Question 1. The direction ratios of a line are 1,-2,-2. What are their direction cosines?
Answer: The direction ratios of the line are \( < 1, -2, -2 > \).
Let the direction cosines of the line be \( l, m, n \).
We know that if \( a, b, c \) are direction ratios, then direction cosines are \( \frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}} \).
First, calculate the magnitude: \( \sqrt{1^2 + (-2)^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \).
Therefore, the direction cosines of the line are:
\( \frac{1}{3}, \frac{-2}{3}, \frac{-2}{3} \).
So, the direction cosines are \( < \frac{1}{3}, \frac{-2}{3}, \frac{-2}{3} > \).
In simple words: Direction cosines are found by dividing each direction ratio by the total length (magnitude) of the vector they represent. This gives you unit vectors that point in the same direction.

🎯 Exam Tip: Remember that direction cosines are unique for a line, whereas direction ratios are proportional to the direction cosines and can have infinitely many values. Always simplify the ratios to their simplest form before finding cosines.

Question 2. If \( \alpha, \beta, \gamma \) are angles which a line makes with the axes, prove that \( \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 2 \)
Answer: Given that \( \alpha, \beta \) and \( \gamma \) are the angles which a line makes with the x, y, and z axes, respectively.
The direction cosines of the line are \( \cos \alpha, \cos \beta, \cos \gamma \).
We know a fundamental identity that the sum of the squares of the direction cosines of any line is equal to 1:
\( \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \). This is a key property in 3D geometry.
We also know that \( \cos^2 x = 1 - \sin^2 x \). Applying this to each term:
\( (1 - \sin^2 \alpha) + (1 - \sin^2 \beta) + (1 - \sin^2 \gamma) = 1 \)

\( \implies 3 - (\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma) = 1 \)

\( \implies \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 3 - 1 \)

\( \implies \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 2 \).
Hence Proved.
In simple words: The squares of the cosines of the angles a line makes with the axes always add up to one. Using this, we can show that the squares of the sines of these same angles will always add up to two.

🎯 Exam Tip: This proof relies on the identity \( \cos^2 \theta + \sin^2 \theta = 1 \). Be sure to clearly state this identity and the basic direction cosine property \( l^2 + m^2 + n^2 = 1 \) at the beginning of your proof for full marks.

Question 3. Can a line have direction angles 45°, 60°, 120° ?
Answer: For a line to have these as its direction angles, the sum of the squares of its direction cosines must be equal to 1.
The given direction angles are \( \alpha = 45^\circ, \beta = 60^\circ, \gamma = 120^\circ \).
First, find the direction cosines:
\( l = \cos \alpha = \cos 45^\circ = \frac{1}{\sqrt{2}} \)
\( m = \cos \beta = \cos 60^\circ = \frac{1}{2} \)
\( n = \cos \gamma = \cos 120^\circ = \cos (180^\circ - 60^\circ) = -\cos 60^\circ = -\frac{1}{2} \).
Now, check if \( l^2 + m^2 + n^2 = 1 \):
\( l^2 + m^2 + n^2 = \left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)^2 \)
\( = \frac{1}{2} + \frac{1}{4} + \frac{1}{4} \)
\( = \frac{2}{4} + \frac{1}{4} + \frac{1}{4} \)
\( = \frac{4}{4} = 1 \).
Since the sum of the squares of the direction cosines is 1, a line can indeed have these direction angles.
In simple words: To check if a set of angles can be direction angles for a line, we calculate the cosine of each angle, square them, and then add them up. If the total is exactly 1, then those angles are valid direction angles.

🎯 Exam Tip: Be careful with the signs of cosine values for angles in different quadrants, especially for angles like 120°, 135°, or 150°. A common mistake is to forget the negative sign for cosines of obtuse angles.

Question 4. Prove that 1,1,1 cannot be direction cosines of a straight line.
Answer: Let the given numbers be \( l=1, m=1, n=1 \).
For any set of numbers to be the direction cosines of a straight line, the sum of their squares must be equal to 1. This is a fundamental property of direction cosines in three-dimensional space.
We need to check if \( l^2 + m^2 + n^2 = 1 \).
Substitute the given values:
\( 1^2 + 1^2 + 1^2 \)
\( = 1 + 1 + 1 \)
\( = 3 \).
Since \( 3 \neq 1 \), the numbers \( < 1, 1, 1 > \) cannot be the direction cosines of a straight line.
In simple words: Direction cosines tell us how a line points in space. For any valid set, if you square each number and add them, the answer must be 1. Since 1 squared plus 1 squared plus 1 squared equals 3, and not 1, these numbers cannot be direction cosines.

🎯 Exam Tip: Always remember the core condition \( l^2 + m^2 + n^2 = 1 \) for direction cosines. This condition is frequently tested and is the key to solving many problems related to direction cosines and angles.

Question 5. Find the direction cosines and direction ratios of the line joining the points
(i) A(0, 0, 0), B(4, 8, -8)
(ii) A(1, 3, 5), B(-1, 0, -1)
(iii) A(5, 6, -3), B(1, -6, 3)
(iv) A(4, 2, -6), B(-2, 1, 3).
Answer: We know that if two points are \( A(x_1, y_1, z_1) \) and \( B(x_2, y_2, z_2) \), then the direction ratios of the line joining them are \( < x_2 - x_1, y_2 - y_1, z_2 - z_1 > \).

(i) For A(0, 0, 0) and B(4, 8, -8):
Direction ratios of line AB are \( < 4 - 0, 8 - 0, -8 - 0 > \)
i.e., \( < 4, 8, -8 > \). These can be simplified by dividing by 4, so they are proportional to \( < 1, 2, -2 > \).
Now, calculate the magnitude for direction cosines: \( \sqrt{4^2 + 8^2 + (-8)^2} = \sqrt{16 + 64 + 64} = \sqrt{144} = 12 \).
Direction cosines of line AB are \( < \frac{4}{12}, \frac{8}{12}, \frac{-8}{12} > \)
i.e., \( < \frac{1}{3}, \frac{2}{3}, \frac{-2}{3} > \).

(ii) For A(1, 3, 5) and B(-1, 0, -1):
Direction ratios of line AB are \( < -1 - 1, 0 - 3, -1 - 5 > \)
i.e., \( < -2, -3, -6 > \). These are proportional to \( < 2, 3, 6 > \) by multiplying by -1.
Now, calculate the magnitude: \( \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \).
Direction cosines of line AB are \( < \frac{2}{7}, \frac{3}{7}, \frac{6}{7} > \).

(iii) For A(5, 6, -3) and B(1, -6, 3):
Direction ratios of line AB are \( < 1 - 5, -6 - 6, 3 - (-3) > \)
i.e., \( < -4, -12, 6 > \). These are proportional to \( < 2, 6, -3 > \) by dividing by -2.
Now, calculate the magnitude: \( \sqrt{2^2 + 6^2 + (-3)^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7 \).
Direction cosines of line AB are \( < \frac{2}{7}, \frac{6}{7}, \frac{-3}{7} > \).

(iv) For A(4, 2, -6) and B(-2, 1, 3):
Direction ratios of line AB are \( < -2 - 4, 1 - 2, 3 - (-6) > \)
i.e., \( < -6, -1, 9 > \). These are proportional to \( < 6, 1, -9 > \) by multiplying by -1.
Now, calculate the magnitude: \( \sqrt{6^2 + 1^2 + (-9)^2} = \sqrt{36 + 1 + 81} = \sqrt{118} \).
Direction cosines of line AB are \( < \frac{6}{\sqrt{118}}, \frac{1}{\sqrt{118}}, \frac{-9}{\sqrt{118}} > \).
In simple words: First, find the "direction ratios" by subtracting the coordinates of the first point from the second. Then, to get "direction cosines", divide each of these ratios by the square root of the sum of their squares. This normalization gives the actual cosine values.

🎯 Exam Tip: When finding direction ratios, you can use \( (x_2-x_1, y_2-y_1, z_2-z_1) \) or \( (x_1-x_2, y_1-y_2, z_1-z_2) \). The set of direction ratios will be proportional to each other. Always simplify the direction ratios before calculating the magnitude to make computations easier. A common error is not simplifying them.

Question 6. By using direction ratios method, show that the following set of points are collinear:
(i) A(1, 2, 3), B(4, 0, 4) and C(-2, 4, 2)
(ii) (-2, 4, 7), (3, -6, -8), (1 -2, -2)
Answer: Two lines are parallel if their direction ratios are proportional. If two lines are parallel and share a common point, then the three points forming those lines are collinear (lie on the same straight line).

(i) For points A(1, 2, 3), B(4, 0, 4), C(-2, 4, 2):
Direction ratios of line AB: \( < 4 - 1, 0 - 2, 4 - 3 > = < 3, -2, 1 > \).
Direction ratios of line BC: \( < -2 - 4, 4 - 0, 2 - 4 > = < -6, 4, -2 > \).
Now, let's check if the direction ratios are proportional:
\( \frac{3}{-6} = -\frac{1}{2} \)
\( \frac{-2}{4} = -\frac{1}{2} \)
\( \frac{1}{-2} = -\frac{1}{2} \).
Since \( \frac{3}{-6} = \frac{-2}{4} = \frac{1}{-2} = -\frac{1}{2} \), the direction ratios of line AB and line BC are proportional. This means line AB is parallel to line BC. Also, point B is common to both lines AB and BC. Therefore, points A, B, and C lie on the same line, which means they are collinear.

(ii) For points P(-2, 4, 7), Q(3, -6, -8), R(1, -2, -2) (renaming for clarity):
Direction ratios of line PQ: \( < 3 - (-2), -6 - 4, -8 - 7 > = < 5, -10, -15 > \). These are proportional to \( < 1, -2, -3 > \) (by dividing by 5).
Direction ratios of line QR: \( < 1 - 3, -2 - (-6), -2 - (-8) > = < -2, 4, 6 > \). These are proportional to \( < 1, -2, -3 > \) (by dividing by -2).
Let's check for proportionality between PQ and QR:
\( \frac{5}{-2} \)
\( \frac{-10}{4} = -\frac{5}{2} \)
\( \frac{-15}{6} = -\frac{5}{2} \).
Since \( \frac{5}{-2} = \frac{-10}{4} = \frac{-15}{6} \), the direction ratios of line PQ and line QR are proportional. This means line PQ is parallel to line QR. Also, point Q is common to both lines PQ and QR. Therefore, points P, Q, and R lie on the same line, which means they are collinear.
In simple words: To prove points are in a straight line, we find the "direction ratios" between the first two points and then between the second and third points. If these sets of ratios are simply multiples of each other, it means the lines are parallel. Since they share a middle point, all three must lie on the same line.

🎯 Exam Tip: When proving collinearity using direction ratios, always check two conditions: (1) Proportionality of direction ratios for two segments (e.g., AB and BC), and (2) A common point between those two segments (e.g., B). Both are essential for concluding collinearity.

Question 7. A line makes an angle of \( \frac{\pi}{4} \) with each of the x-axis and the y-axis. Find the angle made by it with the z-axis.
Answer: Let \( \alpha, \beta, \gamma \) be the angles the line makes with the x, y, and z axes, respectively.
Given that the line makes an angle of \( \frac{\pi}{4} \) with the x-axis and the y-axis.
So, \( \alpha = \frac{\pi}{4} \) and \( \beta = \frac{\pi}{4} \).
Let \( \theta \) be the angle the line makes with the z-axis, so \( \gamma = \theta \).
The direction cosines of the line are \( \cos \alpha, \cos \beta, \cos \theta \).
So, \( \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} \) and \( \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} \).
We know that the sum of the squares of the direction cosines is 1:
\( \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \)
\( \left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 + \cos^2 \theta = 1 \)
\( \frac{1}{2} + \frac{1}{2} + \cos^2 \theta = 1 \)
\( 1 + \cos^2 \theta = 1 \)

\( \implies \cos^2 \theta = 1 - 1 \)

\( \implies \cos^2 \theta = 0 \)

\( \implies \cos \theta = 0 \).
For \( \cos \theta = 0 \), the angle \( \theta \) is \( \frac{\pi}{2} \) (or 90 degrees).
Thus, the line makes an angle of \( \frac{\pi}{2} \) with the z-axis.
In simple words: We are given how a line slants compared to the x and y axes. We use a rule that says if you square the cosine of each slant angle and add them up, you always get 1. By putting in the known angles, we can find the cosine of the angle with the z-axis, which turns out to be 0, meaning the line is perpendicular to the z-axis.

🎯 Exam Tip: This question directly tests the fundamental property of direction cosines, \( l^2 + m^2 + n^2 = 1 \). Ensure you correctly recall the cosine values for standard angles like \( \frac{\pi}{4} \) (45°) and know how to solve for the unknown angle.

Question 8. If the line O P makes with the x-axis an angle of measure 120° and withy x-axis an angle of measure 60°. Find the angle made by the line with the z-axis.
Answer: This question seems to have a typo ("withy x-axis"). Assuming it means "with the y-axis," we proceed.
Let the angles the line OP makes with the x, y, and z axes be \( \alpha, \beta, \theta \) respectively.
Given: \( \alpha = 120^\circ \) (with x-axis) and \( \beta = 60^\circ \) (with y-axis).
Let \( \theta \) be the angle the line OP makes with the z-axis.
The direction cosines of the line OP are \( \cos \alpha, \cos \beta, \cos \theta \).
\( \cos \alpha = \cos 120^\circ = -\frac{1}{2} \)
\( \cos \beta = \cos 60^\circ = \frac{1}{2} \)
We know the identity: \( \cos^2 \alpha + \cos^2 \beta + \cos^2 \theta = 1 \). This rule is crucial for problems involving direction angles.
Substitute the known values:
\( \left(-\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 + \cos^2 \theta = 1 \)
\( \frac{1}{4} + \frac{1}{4} + \cos^2 \theta = 1 \)
\( \frac{2}{4} + \cos^2 \theta = 1 \)
\( \frac{1}{2} + \cos^2 \theta = 1 \)

\( \implies \cos^2 \theta = 1 - \frac{1}{2} \)

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

\( \implies \cos \theta = \pm \frac{1}{\sqrt{2}} \).
This means \( \theta \) can be \( \frac{\pi}{4} \) (45 degrees) or \( \frac{3\pi}{4} \) (135 degrees), since \( 0 < \theta < \pi \). Both are valid possibilities.
In simple words: If we know how much a line slants with the first two main directions (x and y axes), we can use a special rule to find how much it slants with the third direction (z-axis). We use the cosine values of the known angles, square them, and then use the rule that all three squared cosines must add up to one. This helps us find the missing angle.

🎯 Exam Tip: When \( \cos^2 \theta \) yields a value, remember to consider both positive and negative square roots for \( \cos \theta \), as angles between 0° and 180° can have both positive and negative cosines. Both \( \frac{\pi}{4} \) and \( \frac{3\pi}{4} \) are valid angles for a line with an axis.

Question 9. Find the angle between the vectors whose direction cosines are proportional to 2, 3, -6 and 3, -4. 5.
Answer: Let the direction ratios of the first vector be \( a_1 = 2, b_1 = 3, c_1 = -6 \).
Let the direction ratios of the second vector be \( a_2 = 3, b_2 = -4, c_2 = 5 \).
The angle \( \theta \) between two lines with direction ratios \( < a_1, b_1, c_1 > \) and \( < a_2, b_2, c_2 > \) is given by the formula:
\( \cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} \). The absolute value is used since we are usually interested in the acute angle.
First, calculate the dot product of the direction ratios:
\( a_1 a_2 + b_1 b_2 + c_1 c_2 = (2)(3) + (3)(-4) + (-6)(5) = 6 - 12 - 30 = -36 \).
Next, calculate the magnitudes of the direction ratios vectors:
\( \sqrt{a_1^2 + b_1^2 + c_1^2} = \sqrt{2^2 + 3^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \).
\( \sqrt{a_2^2 + b_2^2 + c_2^2} = \sqrt{3^2 + (-4)^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2} \).
Now, substitute these values into the cosine formula:
\( \cos \theta = \frac{|-36|}{(7)(5\sqrt{2})} = \frac{36}{35\sqrt{2}} \).
To rationalize the denominator, multiply the numerator and denominator by \( \sqrt{2} \):
\( \cos \theta = \frac{36\sqrt{2}}{35 \times 2} = \frac{18\sqrt{2}}{35} \).
So, the angle \( \theta \) between the vectors is \( \theta = \cos^{-1}\left(\frac{18\sqrt{2}}{35}\right) \).
In simple words: To find the angle between two lines (or vectors), we use their direction ratios. We multiply corresponding ratios and add them up, then divide this by the product of the lengths of the two ratio sets. The result is the cosine of the angle. We take the inverse cosine to get the angle itself.

🎯 Exam Tip: Remember to use the formula \( \cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} \) for the acute angle. If the problem asks for *an* angle and not specifically the acute one, you might omit the absolute value and get \( \cos \theta = -\frac{18\sqrt{2}}{35} \), leading to an obtuse angle.

Question 10. If \( \alpha, \beta, \gamma \) are the angles that a line makes with the axes, then find \( \cos \gamma \) if
(i) \( \cos \alpha = \frac{14}{15}, \cos \beta = -\frac{1}{3} \)
(ii) \( \alpha = 60^\circ, \beta = 135^\circ \).
Answer: We use the fundamental property of direction cosines: \( \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \).

(i) Given \( \cos \alpha = \frac{14}{15} \) and \( \cos \beta = -\frac{1}{3} \). We need to find \( \cos \gamma \).
Substitute the given values into the identity:
\( \left(\frac{14}{15}\right)^2 + \left(-\frac{1}{3}\right)^2 + \cos^2 \gamma = 1 \)
\( \frac{196}{225} + \frac{1}{9} + \cos^2 \gamma = 1 \)
To add the fractions, find a common denominator, which is 225:
\( \frac{196}{225} + \frac{25}{225} + \cos^2 \gamma = 1 \)
\( \frac{196 + 25}{225} + \cos^2 \gamma = 1 \)
\( \frac{221}{225} + \cos^2 \gamma = 1 \)

\( \implies \cos^2 \gamma = 1 - \frac{221}{225} \)

\( \implies \cos^2 \gamma = \frac{225 - 221}{225} \)

\( \implies \cos^2 \gamma = \frac{4}{225} \)

\( \implies \cos \gamma = \pm \sqrt{\frac{4}{225}} = \pm \frac{2}{15} \).

(ii) Given \( \alpha = 60^\circ \) and \( \beta = 135^\circ \). We need to find \( \cos \gamma \).
First, find \( \cos \alpha \) and \( \cos \beta \):
\( \cos \alpha = \cos 60^\circ = \frac{1}{2} \)
\( \cos \beta = \cos 135^\circ = -\frac{1}{\sqrt{2}} \). Remember that \( \cos 135^\circ = \cos(180^\circ - 45^\circ) = -\cos 45^\circ \).
Substitute these values into the identity:
\( \left(\frac{1}{2}\right)^2 + \left(-\frac{1}{\sqrt{2}}\right)^2 + \cos^2 \gamma = 1 \)
\( \frac{1}{4} + \frac{1}{2} + \cos^2 \gamma = 1 \)
\( \frac{1}{4} + \frac{2}{4} + \cos^2 \gamma = 1 \)
\( \frac{3}{4} + \cos^2 \gamma = 1 \)

\( \implies \cos^2 \gamma = 1 - \frac{3}{4} \)

\( \implies \cos^2 \gamma = \frac{1}{4} \)

\( \implies \cos \gamma = \pm \sqrt{\frac{1}{4}} = \pm \frac{1}{2} \).
In simple words: A line's direction is described by three angles with the x, y, and z axes. We use a key rule: if you square the cosine of each of these three angles and add them up, the total will always be one. We plug in the known cosine values or angles, and then do some simple math to find the missing cosine value for the z-axis angle.

🎯 Exam Tip: Pay close attention to the sign of cosine values for angles greater than 90 degrees (obtuse angles). Also, remember to take both positive and negative square roots when solving for \( \cos \gamma \), as both are mathematically valid possibilities for the direction cosine.

Question 11. If the coordinates of A and B be (2, 3, 4) and (1, -2, 1) respectively, prove that O A is perpendicular to O B, where O is the origin.
Answer: The origin O has coordinates (0, 0, 0).
Point A has coordinates (2, 3, 4).
Point B has coordinates (1, -2, 1).
First, find the direction ratios of line OA:
Direction ratios of OA are \( < 2 - 0, 3 - 0, 4 - 0 > = < 2, 3, 4 > \). Let these be \( a_1, b_1, c_1 \).
Next, find the direction ratios of line OB:
Direction ratios of OB are \( < 1 - 0, -2 - 0, 1 - 0 > = < 1, -2, 1 > \). Let these be \( a_2, b_2, c_2 \).
Two lines are perpendicular if the sum of the products of their corresponding direction ratios is zero. That is, \( a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 \). This is the condition for perpendicularity.
Let's calculate the sum of the products:
\( (2)(1) + (3)(-2) + (4)(1) \)
\( = 2 - 6 + 4 \)
\( = 6 - 6 \)
\( = 0 \).
Since \( a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 \), line OA is perpendicular to line OB.
Hence Proved.
In simple words: To show two lines from the origin are at a right angle, we first find their direction ratios by simply using the coordinates of the given points. Then, we multiply the first numbers from each set, then the second numbers, then the third numbers, and add these three results. If the final sum is zero, the lines are perpendicular.

🎯 Exam Tip: Clearly state the coordinates of the origin (0,0,0) and the formula for perpendicularity of lines using direction ratios: \( a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 \). Show each step of the calculation to ensure full marks.

Question 12. Show that therein of the points (1, 2, 3), (4, 5, 7) is parallel to the join of the points (-4, 3, -6) and (2, 9, 2).
Answer: Let the first line be L1, joining points A(1, 2, 3) and B(4, 5, 7).
Let the second line be L2, joining points C(-4, 3, -6) and D(2, 9, 2).
First, find the direction ratios of line L1 (AB):
Direction ratios of AB are \( < 4 - 1, 5 - 2, 7 - 3 > = < 3, 3, 4 > \). Let these be \( a_1, b_1, c_1 \).
Next, find the direction ratios of line L2 (CD):
Direction ratios of CD are \( < 2 - (-4), 9 - 3, 2 - (-6) > = < 2 + 4, 6, 2 + 6 > = < 6, 6, 8 > \). Let these be \( a_2, b_2, c_2 \).
Two lines are parallel if their direction ratios are proportional. That is, \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \). This is the condition for parallel lines.
Let's check the 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 \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{1}{2} \), the direction ratios are proportional. This demonstrates that line AB is parallel to line CD.
Hence Proved.
In simple words: To show two lines are parallel, we first find their direction ratios by subtracting the coordinates of their end points. Then, we check if the ratios of corresponding numbers are the same. If they are, it means the lines are pointing in the same direction and are therefore parallel.

🎯 Exam Tip: Clearly define the points and lines. The condition for parallel lines, \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \), is essential and must be shown explicitly with the calculated ratios. Make sure to simplify the ratios to confirm proportionality.

Question 13. Find the angles between the lines whose direction ratios are
(i) 5, -12, 13; -3, 4, 5;
(ii) 1, 1, 2 ; \( \sqrt{3} - 1, -\sqrt{3} - 1, 4 \).
Answer: The angle \( \theta \) between two lines with direction ratios \( < a_1, b_1, c_1 > \) and \( < a_2, b_2, c_2 > \) is given by the formula:
\( \cos \theta = \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} \). For the angle between lines, it can be acute or obtuse, so the absolute value is usually omitted unless specified.

(i) Given direction ratios for the first line: \( < 5, -12, 13 > \). So, \( a_1 = 5, b_1 = -12, c_1 = 13 \).
Given direction ratios for the second line: \( < -3, 4, 5 > \). So, \( a_2 = -3, b_2 = 4, c_2 = 5 \).
Calculate the numerator (dot product):
\( a_1 a_2 + b_1 b_2 + c_1 c_2 = (5)(-3) + (-12)(4) + (13)(5) = -15 - 48 + 65 = 2 \).
Calculate the magnitude of the first vector:
\( \sqrt{a_1^2 + b_1^2 + c_1^2} = \sqrt{5^2 + (-12)^2 + 13^2} = \sqrt{25 + 144 + 169} = \sqrt{338} = \sqrt{2 \times 169} = 13\sqrt{2} \).
Calculate the magnitude of the second vector:
\( \sqrt{a_2^2 + b_2^2 + c_2^2} = \sqrt{(-3)^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2} \).
Now, substitute these into the formula for \( \cos \theta \):
\( \cos \theta = \frac{2}{(13\sqrt{2})(5\sqrt{2})} = \frac{2}{13 \times 5 \times 2} = \frac{2}{130} = \frac{1}{65} \).
So, the angle is \( \theta = \cos^{-1}\left(\frac{1}{65}\right) \).

(ii) Given direction ratios for the first line: \( < 1, 1, 2 > \). So, \( a_1 = 1, b_1 = 1, c_1 = 2 \).
Given direction ratios for the second line: \( < \sqrt{3} - 1, -\sqrt{3} - 1, 4 > \). So, \( a_2 = \sqrt{3} - 1, b_2 = -\sqrt{3} - 1, c_2 = 4 \).
Calculate the numerator (dot product):
\( a_1 a_2 + b_1 b_2 + c_1 c_2 = (1)(\sqrt{3} - 1) + (1)(-\sqrt{3} - 1) + (2)(4) \)
\( = \sqrt{3} - 1 - \sqrt{3} - 1 + 8 \)
\( = -2 + 8 = 6 \).
Calculate the magnitude of the first vector:
\( \sqrt{a_1^2 + b_1^2 + c_1^2} = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6} \).
Calculate the magnitude of the second vector:
\( \sqrt{a_2^2 + b_2^2 + c_2^2} = \sqrt{(\sqrt{3} - 1)^2 + (-\sqrt{3} - 1)^2 + 4^2} \)
\( = \sqrt{(3 - 2\sqrt{3} + 1) + (3 + 2\sqrt{3} + 1) + 16} \)
\( = \sqrt{4 - 2\sqrt{3} + 4 + 2\sqrt{3} + 16} \)
\( = \sqrt{8 + 16} = \sqrt{24} = 2\sqrt{6} \).
Now, substitute these into the formula for \( \cos \theta \):
\( \cos \theta = \frac{6}{(\sqrt{6})(2\sqrt{6})} = \frac{6}{2 \times 6} = \frac{6}{12} = \frac{1}{2} \).
So, the angle is \( \theta = \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} \) (or 60 degrees).
In simple words: To find the angle between two lines using their direction ratios, we use a specific formula. We multiply corresponding direction ratios and add them up for the top part of the fraction. For the bottom part, we find the "length" (magnitude) of each set of direction ratios by squaring and adding them, then taking the square root, and then multiply these two lengths. The final fraction is the cosine of the angle, so we use inverse cosine to get the angle itself.

🎯 Exam Tip: When dealing with complex direction ratios like those in part (ii), be careful with algebraic expansions, especially when squaring terms like \( (\sqrt{3}-1)^2 \). Simplify the square roots in the denominator carefully to avoid calculation errors.

Question 14. If P, Q, R are respectively (2, 3, 5),(-1, 3, 2) and (3, 5, -2), find the direction cosines of the sides of the triangle P Q R.
Answer: Let the coordinates of the vertices of the triangle be \( P(2, 3, 5), Q(-1, 3, 2), \) and \( R(3, 5, -2) \). We need to find the direction cosines for each side: PQ, QR, and PR.

**For side PQ:**
Direction ratios of PQ are \( < -1 - 2, 3 - 3, 2 - 5 > = < -3, 0, -3 > \). These can be simplified to \( < 1, 0, 1 > \) by dividing by -3.
Magnitude of PQ direction ratios: \( \sqrt{1^2 + 0^2 + 1^2} = \sqrt{1 + 0 + 1} = \sqrt{2} \).
Direction cosines of PQ are \( < \frac{1}{\sqrt{2}}, \frac{0}{\sqrt{2}}, \frac{1}{\sqrt{2}} > \)
i.e., \( < \frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}} > \).

**For side QR:**
Direction ratios of QR are \( < 3 - (-1), 5 - 3, -2 - 2 > = < 4, 2, -4 > \). These can be simplified to \( < 2, 1, -2 > \) by dividing by 2.
Magnitude of QR direction ratios: \( \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \).
Direction cosines of QR are \( < \frac{2}{3}, \frac{1}{3}, \frac{-2}{3} > \).

**For side PR:**
Direction ratios of PR are \( < 3 - 2, 5 - 3, -2 - 5 > = < 1, 2, -7 > \).
Magnitude of PR direction ratios: \( \sqrt{1^2 + 2^2 + (-7)^2} = \sqrt{1 + 4 + 49} = \sqrt{54} = 3\sqrt{6} \).
Direction cosines of PR are \( < \frac{1}{\sqrt{54}}, \frac{2}{\sqrt{54}}, \frac{-7}{\sqrt{54}} > \)
i.e., \( < \frac{1}{3\sqrt{6}}, \frac{2}{3\sqrt{6}}, \frac{-7}{3\sqrt{6}} > \).
In simple words: For each side of the triangle, first find the "direction ratios" by subtracting the coordinates of its two end points. Then, to get the "direction cosines", divide each direction ratio by the length of that side (which is found by squaring the direction ratios, adding them, and taking the square root). Repeat this for all three sides.

🎯 Exam Tip: Clearly label the direction ratios and direction cosines for each side (PQ, QR, PR). Simplify the direction ratios before calculating the magnitude to make computations easier. Remember to rationalize denominators if required, although \( \frac{1}{\sqrt{54}} \) is often acceptable unless specified otherwise.

Question 15. Prove that the three points P, Q, R, whose coordinates are respectively (3, 2, -4), (5, 4, -6) and (9, 8, -10) are collinear and find the ratio in which Q divides P R.
Answer: Given points: \( P(3, 2, -4), Q(5, 4, -6), \) and \( R(9, 8, -10) \).

**Part 1: Prove collinearity**
First, find the direction ratios of line segment PQ:
Direction ratios of PQ are \( < 5 - 3, 4 - 2, -6 - (-4) > = < 2, 2, -2 > \). These can be simplified to \( < 1, 1, -1 > \).
Next, find the direction ratios of line segment QR:
Direction ratios of QR are \( < 9 - 5, 8 - 4, -10 - (-6) > = < 4, 4, -4 > \). These can be simplified to \( < 1, 1, -1 > \).
Comparing the direction ratios of PQ and QR: \( \frac{2}{4} = \frac{1}{2}, \frac{2}{4} = \frac{1}{2}, \frac{-2}{-4} = \frac{1}{2} \).
Since \( \frac{2}{4} = \frac{2}{4} = \frac{-2}{-4} \), the direction ratios of PQ and QR are proportional. This means line PQ is parallel to line QR. Also, point Q is common to both segments. Therefore, the points P, Q, and R are collinear.

**Part 2: Find the ratio in which Q divides PR**
Let point Q divide PR in the ratio \( K:1 \). Using the section formula, the coordinates of Q are:
\( \left(\frac{K x_2 + 1 x_1}{K+1}, \frac{K y_2 + 1 y_1}{K+1}, \frac{K z_2 + 1 z_1}{K+1}\right) \)
Substitute the coordinates of P(3, 2, -4) as \( (x_1, y_1, z_1) \) and R(9, 8, -10) as \( (x_2, y_2, z_2) \):
\( Q = \left(\frac{K(9) + 1(3)}{K+1}, \frac{K(8) + 1(2)}{K+1}, \frac{K(-10) + 1(-4)}{K+1}\right) \)
\( Q = \left(\frac{9K + 3}{K+1}, \frac{8K + 2}{K+1}, \frac{-10K - 4}{K+1}\right) \).
We know the coordinates of Q are (5, 4, -6). Equate the corresponding coordinates:
For the x-coordinate:
\( \frac{9K + 3}{K+1} = 5 \)
\( 9K + 3 = 5(K+1) \)
\( 9K + 3 = 5K + 5 \)
\( 4K = 2 \)
\( K = \frac{2}{4} = \frac{1}{2} \).
Let's verify with the y-coordinate:
\( \frac{8K + 2}{K+1} = 4 \)
\( 8K + 2 = 4(K+1) \)
\( 8K + 2 = 4K + 4 \)
\( 4K = 2 \)
\( K = \frac{1}{2} \).
And with the z-coordinate:
\( \frac{-10K - 4}{K+1} = -6 \)
\( -10K - 4 = -6(K+1) \)
\( -10K - 4 = -6K - 6 \)
\( -4K = -2 \)
\( K = \frac{-2}{-4} = \frac{1}{2} \).
Since all coordinates yield \( K = \frac{1}{2} \), the point Q divides PR in the ratio \( \frac{1}{2}:1 \), which simplifies to \( 1:2 \).
In simple words: First, we prove the points are in a straight line by checking if the direction ratios between P and Q are proportional to those between Q and R. Since Q is a common point, they must be collinear. Next, we assume Q divides PR in some ratio, say K:1. Using a formula for dividing a line segment, we set up equations for the coordinates and solve for K. This gives us the exact ratio.

🎯 Exam Tip: For collinearity, clearly show the proportionality of direction ratios for two segments sharing a common point. For finding the ratio, consistently use the section formula and verify the ratio using all three coordinates to catch any calculation errors.

Question 16. Find the angle not greater than 90° between the lines joining the following pairs of points:
(i) (8, 2, 0), (4, 6, -7), and (-3, 1, 2), (-9, -2, 4);
(ii) (4, -2, 3), (6, 1, 7), and (4, -2, 3),(5, 4, -2);
(iii) (3, 1, -2), (4, 0, -4), and (4, -3, 3), (6, -2, 2).
Answer: To find the angle between two lines, we first determine their direction ratios. If the lines are defined by two points \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \), their direction ratios are \( < x_2 - x_1, y_2 - y_1, z_2 - z_1 > \). The angle \( \theta \) between two lines with direction ratios \( < a_1, b_1, c_1 > \) and \( < a_2, b_2, c_2 > \) is given by \( \cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} \) for the acute angle (not greater than 90°).

(i) Line 1: Joining A(8, 2, 0) and B(4, 6, -7).
Direction ratios of AB: \( < 4 - 8, 6 - 2, -7 - 0 > = < -4, 4, -7 > \).
Line 2: Joining C(-3, 1, 2) and D(-9, -2, 4).
Direction ratios of CD: \( < -9 - (-3), -2 - 1, 4 - 2 > = < -6, -3, 2 > \).
Numerator: \( a_1 a_2 + b_1 b_2 + c_1 c_2 = (-4)(-6) + (4)(-3) + (-7)(2) = 24 - 12 - 14 = -2 \).
Magnitude 1: \( \sqrt{(-4)^2 + 4^2 + (-7)^2} = \sqrt{16 + 16 + 49} = \sqrt{81} = 9 \).
Magnitude 2: \( \sqrt{(-6)^2 + (-3)^2 + 2^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7 \).
\( \cos \theta = \frac{|-2|}{9 \times 7} = \frac{2}{63} \).
So, \( \theta = \cos^{-1}\left(\frac{2}{63}\right) \).

(ii) Line 1: Joining A(4, -2, 3) and B(6, 1, 7).
Direction ratios of AB: \( < 6 - 4, 1 - (-2), 7 - 3 > = < 2, 3, 4 > \).
Line 2: Joining C(4, -2, 3) and D(5, 4, -2).
Direction ratios of CD: \( < 5 - 4, 4 - (-2), -2 - 3 > = < 1, 6, -5 > \).
Numerator: \( a_1 a_2 + b_1 b_2 + c_1 c_2 = (2)(1) + (3)(6) + (4)(-5) = 2 + 18 - 20 = 0 \).
Since the numerator is 0, \( \cos \theta = 0 \).
This implies \( \theta = 90^\circ \) or \( \frac{\pi}{2} \). The lines are perpendicular.

(iii) Line 1: Joining A(3, 1, -2) and B(4, 0, -4).
Direction ratios of AB: \( < 4 - 3, 0 - 1, -4 - (-2) > = < 1, -1, -2 > \).
Line 2: Joining C(4, -3, 3) and D(6, -2, 2).
Direction ratios of CD: \( < 6 - 4, -2 - (-3), 2 - 3 > = < 2, 1, -1 > \).
Numerator: \( a_1 a_2 + b_1 b_2 + c_1 c_2 = (1)(2) + (-1)(1) + (-2)(-1) = 2 - 1 + 2 = 3 \).
Magnitude 1: \( \sqrt{1^2 + (-1)^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6} \).
Magnitude 2: \( \sqrt{2^2 + 1^2 + (-1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \).
\( \cos \theta = \frac{3}{\sqrt{6} \times \sqrt{6}} = \frac{3}{6} = \frac{1}{2} \).
So, \( \theta = \cos^{-1}\left(\frac{1}{2}\right) = 60^\circ \) or \( \frac{\pi}{3} \).
In simple words: For each pair of lines, first find their "direction ratios" by subtracting the coordinates of the given points. Then, use a formula: multiply the matching direction ratios from both lines and add them up. Divide this total by the product of the "lengths" of each set of direction ratios. This gives you the cosine of the angle. Finally, use the inverse cosine to find the angle itself. Make sure to use the absolute value in the numerator to get an angle not greater than 90 degrees.

🎯 Exam Tip: Always make sure to calculate the direction ratios correctly by subtracting the coordinates. Remember to use the absolute value in the numerator of the cosine formula when asked for the angle "not greater than 90°" (acute angle). A common mistake is not using the absolute value, which can lead to an obtuse angle.

Question 17. Find the direction cosines of the line which is perpendicular to the lines with direction cosines proportional to 1, – 2, –2 ; 0, 2, 1.
Answer: Let the direction ratios of the required line be \( < a, b, c > \).
The first given line has direction ratios proportional to \( < 1, -2, -2 > \).
The second given line has direction ratios proportional to \( < 0, 2, 1 > \).
Since the required line is perpendicular to both given lines, its direction ratios \( < a, b, c > \) must satisfy the perpendicularity condition for both lines:
\( a(1) + b(-2) + c(-2) = 0 \implies a - 2b - 2c = 0 \) (Equation 1)
\( a(0) + b(2) + c(1) = 0 \implies 2b + c = 0 \) (Equation 2)
We can solve these two equations for \( a, b, c \) using the cross-multiplication method:
\( \frac{a}{(-2)(1) - (-2)(2)} = \frac{b}{(-2)(0) - (1)(1)} = \frac{c}{(1)(2) - (-2)(0)} \)
\( \frac{a}{-2 - (-4)} = \frac{b}{0 - 1} = \frac{c}{2 - 0} \)
\( \frac{a}{2} = \frac{b}{-1} = \frac{c}{2} \).
So, the direction ratios of the required line are \( < 2, -1, 2 > \).
Now, we need to find the direction cosines. First, calculate the magnitude of these direction ratios:
\( \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \).
The direction cosines of the required line are \( < \frac{2}{3}, \frac{-1}{3}, \frac{2}{3} > \).
In simple words: We are looking for a line that is at a right angle to two other lines. We find the "direction ratios" of this new line by setting up two equations based on the perpendicularity rule (which means their dot product is zero). We solve these equations using a method called cross-multiplication to get the direction ratios. Finally, we convert these ratios into "direction cosines" by dividing each by the line's total length.

🎯 Exam Tip: When finding a line perpendicular to two given lines, remember that its direction ratios can be found using the cross product of the direction ratios of the two given lines. This is equivalent to solving the system of linear equations using cross-multiplication. Ensure correct signs in the cross-multiplication steps.

Question 18. Find the direction ratios of a line perpendicular to the two lines determined by the pairs of points (2, 3, -4), (-3, 3, -2) and (-1, 4, 2), (3, 5, 1).
Answer: Let the first line L1 be determined by points \( A(2, 3, -4) \) and \( B(-3, 3, -2) \).
Direction ratios of L1: \( < -3 - 2, 3 - 3, -2 - (-4) > = < -5, 0, 2 > \). Let these be \( a_1, b_1, c_1 \).
Let the second line L2 be determined by points \( C(-1, 4, 2) \) and \( D(3, 5, 1) \).
Direction ratios of L2: \( < 3 - (-1), 5 - 4, 1 - 2 > = < 4, 1, -1 > \). Let these be \( a_2, b_2, c_2 \).
Let the direction ratios of the line perpendicular to both L1 and L2 be \( < a, b, c > \).
Since this line is perpendicular to L1, their dot product of direction ratios is zero:
\( a(-5) + b(0) + c(2) = 0 \implies -5a + 2c = 0 \) (Equation 1)
Since this line is also perpendicular to L2, their dot product of direction ratios is zero:
\( a(4) + b(1) + c(-1) = 0 \implies 4a + b - c = 0 \) (Equation 2)
We can solve these two equations for \( a, b, c \) using the cross-multiplication method:
From Equation 1, \( -5a + 0b + 2c = 0 \)
From Equation 2, \( 4a + 1b - 1c = 0 \)
\( \frac{a}{(0)(-1) - (2)(1)} = \frac{b}{(2)(4) - (-5)(-1)} = \frac{c}{(-5)(1) - (0)(4)} \)
\( \frac{a}{0 - 2} = \frac{b}{8 - 5} = \frac{c}{-5 - 0} \)
\( \frac{a}{-2} = \frac{b}{3} = \frac{c}{-5} \).
So, the direction ratios of the required line are \( < -2, 3, -5 > \). These represent a vector that is mutually orthogonal to both given lines.
In simple words: We first find the "direction ratios" for each of the two given lines. Then, we look for a third line that is at a right angle to both of them. We do this by setting up two equations using the rule that perpendicular lines have a dot product of zero for their direction ratios. Solving these equations by cross-multiplication gives us the direction ratios for the third line.

🎯 Exam Tip: Be careful with signs when calculating direction ratios and applying the cross-multiplication method. A common mistake is sign errors during subtraction or multiplication. Ensure your final direction ratios are consistent with the cross product of the two given direction ratio vectors.

Question 19. For what value of x will the line through (4, 1, 2) and (5, x, 0) be parallel to the line through (2, 1, 1) and (3, 3, -1).
Answer: Let the first line be L1, passing through \( A(4, 1, 2) \) and \( B(5, x, 0) \).
Direction ratios of L1: \( < 5 - 4, x - 1, 0 - 2 > = < 1, x - 1, -2 > \). Let these be \( a_1, b_1, c_1 \).
Let the second line be L2, passing through \( C(2, 1, 1) \) and \( D(3, 3, -1) \).
Direction ratios of L2: \( < 3 - 2, 3 - 1, -1 - 1 > = < 1, 2, -2 > \). Let these be \( a_2, b_2, c_2 \).
For two lines to be parallel, their direction ratios must be proportional. That means \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \). This proportional relationship ensures the lines have the same orientation.
Set up the proportionality:
\( \frac{1}{1} = \frac{x-1}{2} = \frac{-2}{-2} \).
From \( \frac{1}{1} = \frac{-2}{-2} \), we get \( 1 = 1 \), which is consistent.
Now, use the middle part to solve for x:
\( \frac{x-1}{2} = 1 \)
\( x - 1 = 2 \)
\( x = 2 + 1 \)
\( x = 3 \).
Thus, for the lines to be parallel, the value of x must be 3.
In simple words: To make two lines point in the same direction (be parallel), their "direction ratios" must be simple multiples of each other. We find these ratios for both lines. Then, we set up an equation where the ratios of corresponding numbers are equal and solve for the unknown 'x'.

🎯 Exam Tip: The condition for parallel lines, \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \), is crucial. Write down the direction ratios clearly for both lines and then equate the ratios to solve for the unknown variable. Double-check your algebraic steps.

Question 20. For what value of x will the lines in Problem 19 be perpendicular?
Answer: From Problem 19, the direction ratios of the first line (L1) are \( < 1, x - 1, -2 > \). Let these be \( a_1, b_1, c_1 \).
And the direction ratios of the second line (L2) are \( < 1, 2, -2 > \). Let these be \( a_2, b_2, c_2 \).
For two lines to be perpendicular, the sum of the products of their corresponding direction ratios must be zero. That is, \( a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 \). This is a fundamental condition for orthogonal lines.
Substitute the direction ratios into the perpendicularity condition:
\( (1)(1) + (x - 1)(2) + (-2)(-2) = 0 \)
\( 1 + 2(x - 1) + 4 = 0 \)
\( 1 + 2x - 2 + 4 = 0 \)
Combine the constant terms:
\( 2x + (1 - 2 + 4) = 0 \)
\( 2x + 3 = 0 \)
\( 2x = -3 \)
\( x = -\frac{3}{2} \).
Thus, for the lines to be perpendicular, the value of x must be \( -\frac{3}{2} \).
In simple words: To make two lines form a right angle, we use a different rule. We find the "direction ratios" for both lines, multiply their matching numbers, and add these results. If the lines are perpendicular, this sum must be zero. We set up an equation using this rule and solve for 'x'.

🎯 Exam Tip: The condition for perpendicular lines, \( a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 \), is key. Clearly write down the direction ratios and perform the dot product calculation carefully. Avoid algebraic errors when solving the resulting linear equation for x.

Question 21. Show that the points (4, 7, 8), (2, 3, 4), (-1, -2, 1) and (1, 2, 5) are the vertices of a parallelogram.
Answer: Let the given points be \( A(4, 7, 8), B(2, 3, 4), C(-1, -2, 1), \) and \( D(1, 2, 5) \).
A quadrilateral is a parallelogram if its diagonals bisect each other. This means the midpoint of one diagonal must be the same as the midpoint of the other diagonal.
Let's find the midpoint of diagonal AC:
Midpoint of AC \( = \left(\frac{4 + (-1)}{2}, \frac{7 + (-2)}{2}, \frac{8 + 1}{2}\right) = \left(\frac{3}{2}, \frac{5}{2}, \frac{9}{2}\right) \).
Let's find the midpoint of diagonal BD:
Midpoint of BD \( = \left(\frac{2 + 1}{2}, \frac{3 + 2}{2}, \frac{4 + 5}{2}\right) = \left(\frac{3}{2}, \frac{5}{2}, \frac{9}{2}\right) \).
Since the midpoint of AC is the same as the midpoint of BD, the diagonals bisect each other. This is a property of parallelograms. Therefore, the points A, B, C, and D are the vertices of a parallelogram.
Alternatively, we can show that opposite sides are parallel by comparing their direction ratios.
Direction ratios of side AB: \( < 2 - 4, 3 - 7, 4 - 8 > = < -2, -4, -4 > \).
Direction ratios of side DC: \( < -1 - 1, -2 - 2, 1 - 5 > = < -2, -4, -4 > \).
Since the direction ratios of AB and DC are identical, AB is parallel to DC.
Direction ratios of side AD: \( < 1 - 4, 2 - 7, 5 - 8 > = < -3, -5, -3 > \).
Direction ratios of side BC: \( < -1 - 2, -2 - 3, 1 - 4 > = < -3, -5, -3 > \).
Since the direction ratios of AD and BC are identical, AD is parallel to BC.
Since both pairs of opposite sides are parallel, the quadrilateral ABCD is a parallelogram.
In simple words: To prove that four points form a parallelogram, we can check if their diagonal lines cut each other exactly in half. We find the middle point of each diagonal. If both middle points are the same, then the shape is a parallelogram. Another way is to show that opposite sides have the same direction, meaning they are parallel.

🎯 Exam Tip: You can prove a parallelogram using either the midpoint property of diagonals or by showing that both pairs of opposite sides are parallel (using direction ratios). Choosing the midpoint method is often quicker. Clearly state the coordinates of the midpoints or the direction ratios for each pair of sides.

Question 22. Show that the points (5, -1, 1), (7, -4, 7), (1, -6, 10), and (-1, -3, 4) are the vertices of a rhombus.
Answer: Let the given points be \( A(5, -1, 1), B(7, -4, 7), C(1, -6, 10), \) and \( D(-1, -3, 4) \).
A rhombus is a parallelogram with all four sides equal in length. Also, its diagonals bisect each other at right angles.

**Part 1: Check if diagonals bisect each other at right angles.**
Midpoint of diagonal AC: \( \left(\frac{5 + 1}{2}, \frac{-1 + (-6)}{2}, \frac{1 + 10}{2}\right) = \left(\frac{6}{2}, \frac{-7}{2}, \frac{11}{2}\right) = \left(3, -\frac{7}{2}, \frac{11}{2}\right) \).
Midpoint of diagonal BD: \( \left(\frac{7 + (-1)}{2}, \frac{-4 + (-3)}{2}, \frac{7 + 4}{2}\right) = \left(\frac{6}{2}, \frac{-7}{2}, \frac{11}{2}\right) = \left(3, -\frac{7}{2}, \frac{11}{2}\right) \).
Since the midpoints are the same, the diagonals bisect each other, confirming it's a parallelogram.

Now, check if the diagonals are perpendicular. Find their direction ratios.
Direction ratios of AC: \( < 1 - 5, -6 - (-1), 10 - 1 > = < -4, -5, 9 > \). Let these be \( a_1, b_1, c_1 \).
Direction ratios of BD: \( < -1 - 7, -3 - (-4), 4 - 7 > = < -8, 1, -3 > \). Let these be \( a_2, b_2, c_2 \).
Check perpendicularity condition: \( a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 \).
\( (-4)(-8) + (-5)(1) + (9)(-3) = 32 - 5 - 27 = 0 \).
Since the dot product is 0, the diagonals AC and BD are perpendicular. This confirms it is a rhombus.

**Part 2: Check side lengths (optional, but good for verification if not asked to use diagonals)**
Length of AB: \( \sqrt{(7-5)^2 + (-4-(-1))^2 + (7-1)^2} = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4+9+36} = \sqrt{49} = 7 \).
Length of BC: \( \sqrt{(1-7)^2 + (-6-(-4))^2 + (10-7)^2} = \sqrt{(-6)^2 + (-2)^2 + 3^2} = \sqrt{36+4+9} = \sqrt{49} = 7 \).
Length of CD: \( \sqrt{(-1-1)^2 + (-3-(-6))^2 + (4-10)^2} = \sqrt{(-2)^2 + 3^2 + (-6)^2} = \sqrt{4+9+36} = \sqrt{49} = 7 \).
Length of DA: \( \sqrt{(5-(-1))^2 + (-1-(-3))^2 + (1-4)^2} = \sqrt{6^2 + 2^2 + (-3)^2} = \sqrt{36+4+9} = \sqrt{49} = 7 \).
Since all sides are equal, it is indeed a rhombus.

**Part 3: Check diagonal lengths (optional, to differentiate from square)**
Length of AC: \( \sqrt{(1-5)^2 + (-6-(-1))^2 + (10-1)^2} = \sqrt{(-4)^2 + (-5)^2 + 9^2} = \sqrt{16+25+81} = \sqrt{122} \).
Length of BD: \( \sqrt{(-1-7)^2 + (-3-(-4))^2 + (4-7)^2} = \sqrt{(-8)^2 + 1^2 + (-3)^2} = \sqrt{64+1+9} = \sqrt{74} \).
Since \( \text{AC} \neq \text{BD} \), the diagonals are not equal, confirming it's a rhombus and not a square.
Clearly, A, B, C and D are the vertices of a rhombus.
In simple words: To prove that four points form a rhombus, we first show that it's a parallelogram by checking if its diagonals cut each other in half. Then, we check if these diagonals cross at a perfect right angle. If both conditions are met, the shape is a rhombus. We can also verify by calculating all four side lengths and making sure they are equal.

🎯 Exam Tip: The most efficient way to prove a rhombus is to show that its diagonals bisect each other (same midpoint) and are perpendicular (dot product of direction ratios is zero). Optionally, you can also prove all four side lengths are equal, which implies it's a rhombus. Clearly state the criteria you are using.

Question 23. Find the foot of the perpendicular drawn from the point A(1, 0, 3) to the join of the points B(4, 7, 1) and C(3, 5, 3).
Answer: Let D be the foot of the perpendicular drawn from point \( A(1, 0, 3) \) to the line segment BC. Point D lies on the line segment BC.
Let D divide the line segment BC in the ratio \( K:1 \). Using the section formula, the coordinates of D are:
\( D = \left(\frac{K(x_2) + 1(x_1)}{K+1}, \frac{K(y_2) + 1(y_1)}{K+1}, \frac{K(z_2) + 1(z_1)}{K+1}\right) \)
Substitute coordinates of B(4, 7, 1) as \( (x_1, y_1, z_1) \) and C(3, 5, 3) as \( (x_2, y_2, z_2) \):
\( D = \left(\frac{3K + 4}{K+1}, \frac{5K + 7}{K+1}, \frac{3K + 1}{K+1}\right) \).
Now, find the direction ratios of the line AD:
Direction ratios of AD \( = \left< \frac{3K + 4}{K+1} - 1, \frac{5K + 7}{K+1} - 0, \frac{3K + 1}{K+1} - 3 \right> \)
\( = \left< \frac{3K + 4 - (K+1)}{K+1}, \frac{5K + 7}{K+1}, \frac{3K + 1 - 3(K+1)}{K+1} \right> \)
\( = \left< \frac{2K + 3}{K+1}, \frac{5K + 7}{K+1}, \frac{3K + 1 - 3K - 3}{K+1} \right> \)
\( = \left< \frac{2K + 3}{K+1}, \frac{5K + 7}{K+1}, \frac{-2}{K+1} \right> \).
Next, find the direction ratios of the line BC:
Direction ratios of BC \( = < 3 - 4, 5 - 7, 3 - 1 > = < -1, -2, 2 > \).
Since AD is perpendicular to BC, the dot product of their direction ratios must be zero:
\( \left(\frac{2K + 3}{K+1}\right)(-1) + \left(\frac{5K + 7}{K+1}\right)(-2) + \left(\frac{-2}{K+1}\right)(2) = 0 \).
Since \( K+1 \) is in the denominator of each term, we can multiply the entire equation by \( (K+1) \) (assuming \( K \neq -1 \)):
\( -(2K + 3) - 2(5K + 7) - 2(2) = 0 \)
\( -2K - 3 - 10K - 14 - 4 = 0 \)
\( -12K - 21 = 0 \)
\( -12K = 21 \)
\( K = -\frac{21}{12} = -\frac{7}{4} \).
Now, substitute the value of \( K = -\frac{7}{4} \) back into the coordinates of D to find the foot of the perpendicular:
\( D_x = \frac{3(-\frac{7}{4}) + 4}{-\frac{7}{4} + 1} = \frac{-\frac{21}{4} + \frac{16}{4}}{\frac{-7+4}{4}} = \frac{-\frac{5}{4}}{-\frac{3}{4}} = \frac{5}{3} \).
\( D_y = \frac{5(-\frac{7}{4}) + 7}{-\frac{7}{4} + 1} = \frac{-\frac{35}{4} + \frac{28}{4}}{-\frac{3}{4}} = \frac{-\frac{7}{4}}{-\frac{3}{4}} = \frac{7}{3} \).
\( D_z = \frac{3(-\frac{7}{4}) + 1}{-\frac{7}{4} + 1} = \frac{-\frac{21}{4} + \frac{4}{4}}{-\frac{3}{4}} = \frac{-\frac{17}{4}}{-\frac{3}{4}} = \frac{17}{3} \).
So, the coordinates of the foot of the perpendicular D are \( \left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right) \).
A point is dividing the line segment externally since K is negative.
In simple words: To find where a perpendicular line from point A hits line BC, we first assume this meeting point (D) divides BC in some ratio (K:1). We use a formula to write D's coordinates in terms of K. Then, we find the direction ratios for the line AD and for the line BC. Since AD is perpendicular to BC, the dot product of their direction ratios must be zero. This gives us an equation to solve for K. Once K is found, we plug it back into D's coordinate formula to get the exact location of point D.

🎯 Exam Tip: This problem combines the section formula and the condition for perpendicularity. Clearly define the variable K and express the coordinates of D using it. Remember that if K is negative, the division is external. Pay meticulous attention to algebraic calculations when solving for K and substituting it back into the coordinates.

Question 24. A(1, 0, 4) and B(0, -11, 3), C(2, -3, 1) are three points and D is the foot of the perpendicular from A on B C. Find the coordinates of D.
Answer: Let D be a point on the line segment BC. We can assume that D divides the line BC in the ratio \( K:1 \).
The coordinates of point D are given by the section formula:
\( D = \left(\frac{K x_C + 1 x_B}{K+1}, \frac{K y_C + 1 y_B}{K+1}, \frac{K z_C + 1 z_B}{K+1}\right) \)
\( D = \left(\frac{2K+0}{K+1}, \frac{-3K-11}{K+1}, \frac{K+3}{K+1}\right) \)
Now, we find the direction ratios of the line AD, using A(1, 0, 4) and D's coordinates:
Direction ratios of AD are \( \left(x_D-x_A, y_D-y_A, z_D-z_A\right) \):
\( \left(\frac{2K}{K+1}-1, \frac{-3K-11}{K+1}-0, \frac{K+3}{K+1}-4\right) \)
\( \implies \left(\frac{2K-(K+1)}{K+1}, \frac{-3K-11}{K+1}, \frac{K+3-4(K+1)}{K+1}\right) \)
\( \implies \left(\frac{K-1}{K+1}, \frac{-3K-11}{K+1}, \frac{K+3-4K-4}{K+1}\right) \)
\( \implies \left(\frac{K-1}{K+1}, \frac{-3K-11}{K+1}, \frac{-3K-1}{K+1}\right) \)
Next, we find the direction ratios of the line BC, using B(0, -11, 3) and C(2, -3, 1):
Direction ratios of BC are \( \left(x_C-x_B, y_C-y_B, z_C-z_B\right) \):
\( (2-0, -3-(-11), 1-3) = (2, 8, -2) \)
Since the line AD is perpendicular to the line BC, the dot product of their direction ratios must be zero:
\( \left(\frac{K-1}{K+1}\right)(2) + \left(\frac{-3K-11}{K+1}\right)(8) + \left(\frac{-3K-1}{K+1}\right)(-2) = 0 \)
Multiplying the entire equation by \( (K+1) \) (assuming \( K \neq -1 \)):
\( 2(K-1) + 8(-3K-11) - 2(-3K-1) = 0 \)
\( 2K - 2 - 24K - 88 + 6K + 2 = 0 \)
Combine the K terms and constant terms:
\( (2 - 24 + 6)K + (-2 - 88 + 2) = 0 \)
\( -16K - 88 = 0 \)
\( -16K = 88 \)
\( K = \frac{88}{-16} = -\frac{11}{2} \)
Now, substitute the value of \( K \) back into the coordinates of point D:
\( x_D = \frac{2K}{K+1} = \frac{2(-\frac{11}{2})}{-\frac{11}{2}+1} = \frac{-11}{\frac{-11+2}{2}} = \frac{-11}{-\frac{9}{2}} = \frac{22}{9} \)
\( y_D = \frac{-3K-11}{K+1} = \frac{-3(-\frac{11}{2})-11}{-\frac{11}{2}+1} = \frac{\frac{33}{2}-11}{-\frac{9}{2}} = \frac{\frac{33-22}{2}}{-\frac{9}{2}} = \frac{\frac{11}{2}}{-\frac{9}{2}} = -\frac{11}{9} \)
\( z_D = \frac{K+3}{K+1} = \frac{-\frac{11}{2}+3}{-\frac{11}{2}+1} = \frac{\frac{-11+6}{2}}{-\frac{9}{2}} = \frac{-\frac{5}{2}}{-\frac{9}{2}} = \frac{5}{9} \)
Thus, the coordinates of the foot of the perpendicular D are \( \left(\frac{22}{9}, -\frac{11}{9}, \frac{5}{9}\right) \). This point is where the perpendicular from A meets the line BC.
In simple words: To find point D, we assume it divides the line BC in some ratio K. We use this ratio to write D's coordinates. Then, we find the direction of lines AD and BC. Since they are perpendicular, multiplying their direction numbers and adding them up must equal zero. Solving this equation gives us K. Finally, we plug K back into D's coordinates to get the answer.

🎯 Exam Tip: Remember to use the section formula to find coordinates of D. The key to solving this problem is understanding that the dot product of the direction ratios of two perpendicular lines is zero.

Question 25. Calculate the cosine of the angle A of the triangle with vertices A(1,-1, 2), B(6, 11, 2),
Answer: The vertices of the triangle are A(1, -1, 2), B(6, 11, 2), and C(1, 2, 6). To find the cosine of angle A, we need to consider the vectors formed by sides AB and AC.
First, find the direction ratios of the vector \(\vec{AB}\):
\( (x_B-x_A, y_B-y_A, z_B-z_A) = (6-1, 11-(-1), 2-2) \)
\( \implies (5, 12, 0) \)
Let \( a_1 = 5, b_1 = 12, c_1 = 0 \).
Next, find the direction ratios of the vector \(\vec{AC}\):
\( (x_C-x_A, y_C-y_A, z_C-z_A) = (1-1, 2-(-1), 6-2) \)
\( \implies (0, 3, 4) \)
Let \( a_2 = 0, b_2 = 3, c_2 = 4 \).
The cosine of the angle A between two lines with direction ratios \( (a_1, b_1, c_1) \) and \( (a_2, b_2, c_2) \) is given by the formula:
\( \cos A = \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}} \)
Substitute the values:
\( \cos A = \frac{(5)(0) + (12)(3) + (0)(4)}{\sqrt{5^2+12^2+0^2} \sqrt{0^2+3^2+4^2}} \)
\( \implies \cos A = \frac{0 + 36 + 0}{\sqrt{25+144+0} \sqrt{0+9+16}} \)
\( \implies \cos A = \frac{36}{\sqrt{169} \sqrt{25}} \)
\( \implies \cos A = \frac{36}{13 \times 5} \)
\( \implies \cos A = \frac{36}{65} \)
Therefore, the cosine of angle A is \( \frac{36}{65} \). This calculation helps determine the angle at vertex A without finding all three angles of the triangle.
In simple words: To find the cosine of angle A, we first figure out the "direction numbers" for the lines from A to B and from A to C. Then, we use a special formula that combines these direction numbers. This formula helps us calculate the cosine value, which tells us how wide or narrow the angle A is.

🎯 Exam Tip: Always define the vectors originating from the common vertex (in this case, A) when calculating an angle within a triangle. Make sure to accurately compute the magnitudes of the vectors in the denominator to avoid errors.

Question 26. If A, B, C, D are the points (6, -6, 0), (-1, -7, 6), (3, -4, 4), (2, -9, 2) respectively, prove that A B is perpendicular to C D.
Answer: We are given the coordinates of four points: A(6, -6, 0), B(-1, -7, 6), C(3, -4, 4), and D(2, -9, 2). To prove that line AB is perpendicular to line CD, we need to find their direction ratios and then check if their dot product is zero.
First, find the direction ratios of line AB:
\( (x_B-x_A, y_B-y_A, z_B-z_A) = (-1-6, -7-(-6), 6-0) \)
\( \implies (-7, -1, 6) \)
Let these be \( a_1 = -7, b_1 = -1, c_1 = 6 \).
Next, find the direction ratios of line CD:
\( (x_D-x_C, y_D-y_C, z_D-z_C) = (2-3, -9-(-4), 2-4) \)
\( \implies (-1, -5, -2) \)
Let these be \( a_2 = -1, b_2 = -5, c_2 = -2 \).
For two lines to be perpendicular, the sum of the products of their corresponding direction ratios must be zero: \( a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 \).
Calculate the dot product:
\( (-7)(-1) + (-1)(-5) + (6)(-2) \)
\( = 7 + 5 - 12 \)
\( = 12 - 12 \)
\( = 0 \)
Since the dot product of the direction ratios of line AB and line CD is 0, this proves that line AB is perpendicular to line CD. This means they meet at a right angle.
In simple words: We find the direction of line AB and line CD using their given points. If two lines are at a 90-degree angle to each other, a special calculation involving their directions (called a dot product) will always result in zero. Since our calculation for these two lines gives zero, we know they are perpendicular.

🎯 Exam Tip: Remember the condition for perpendicular lines: \( a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 \). Ensure you correctly calculate the difference in coordinates for each component to get accurate direction ratios.

Question 27. Find the angle between any two diagonals of a cube.
Answer: Let's assume the side length of the cube is 'a'. We can place one corner of the cube at the origin O(0,0,0) of a three-dimensional coordinate system. This helps us define the coordinates of all other vertices easily.
Consider two main body diagonals of the cube:
1. Diagonal OP: This diagonal connects the origin O(0,0,0) to the farthest vertex P(a,a,a).
The direction ratios of OP are \( (a-0, a-0, a-0) = (a, a, a) \).
2. Diagonal AR: This diagonal connects vertex A(a,0,0) (on the x-axis) to vertex R(0,a,a) (on the plane yz).
The direction ratios of AR are \( (0-a, a-0, a-0) = (-a, a, a) \).
Let \( \theta \) be the angle between these two diagonals. The cosine of the angle between two lines with direction ratios \( (a_1, b_1, c_1) \) and \( (a_2, b_2, c_2) \) is given by the formula:
\( \cos \theta = \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}} \)
For diagonal OP, \( a_1 = a, b_1 = a, c_1 = a \).
For diagonal AR, \( a_2 = -a, b_2 = a, c_2 = a \).
Substitute these values into the formula:
\( \cos \theta = \frac{(a)(-a) + (a)(a) + (a)(a)}{\sqrt{a^2+a^2+a^2} \sqrt{(-a)^2+a^2+a^2}} \)
\( \implies \cos \theta = \frac{-a^2 + a^2 + a^2}{\sqrt{3a^2} \sqrt{3a^2}} \)
\( \implies \cos \theta = \frac{a^2}{(\sqrt{3}a)(\sqrt{3}a)} \)
\( \implies \cos \theta = \frac{a^2}{3a^2} \)
\( \implies \cos \theta = \frac{1}{3} \)
Therefore, the angle \( \theta = \cos^{-1}\left(\frac{1}{3}\right) \). This angle is the same for any pair of diagonals in a cube, demonstrating a consistent geometric property.
In simple words: Imagine a cube and pick any two long lines that go through the middle of the cube from corner to opposite corner. If you want to know the angle where these two lines cross, it will always be the same. By using math with coordinates and a special formula for angles between lines, we find that the cosine of this angle is always one-third.

🎯 Exam Tip: When finding angles in 3D geometry problems, set one vertex at the origin (0,0,0) for simpler coordinate assignment. This allows for easier calculation of direction ratios for lines and diagonals. Remember that the angle between any two body diagonals of a cube is always constant, \( \cos^{-1}\left(\frac{1}{3}\right) \).