RBSE Solutions Class 12 Maths Chapter 14 Three Dimensional Geometry Exercise 14.3

Question 1. Find the angle between the lines \( \vec{r} = 2\hat{i}-5\hat{j}+\hat{k}+\lambda(3\hat{i}+2\hat{j}+6\hat{k}) \) and \( \vec{r} = 7\hat{i}-6\hat{j}+\mu(\hat{i}+2\hat{j}+2\hat{k}) \).
Answer: First, we identify the direction vectors of the two lines. The direction vector for the first line is \( \vec{b_1} = 3\hat{i} + 2\hat{j} + 6\hat{k} \), and for the second line, it is \( \vec{b_2} = \hat{i} + 2\hat{j} + 2\hat{k} \).
The angle \( \theta \) between these two lines is found using the dot product formula:
\( \cos \theta = \frac { \vec{b_1} \cdot \vec{b_2} }{ |\vec{b_1}| |\vec{b_2}| } \)
Now, we calculate the dot product of the direction vectors:
\( \vec{b_1} \cdot \vec{b_2} = (3)(1) + (2)(2) + (6)(2) = 3 + 4 + 12 = 19 \)
Next, we find the magnitude (length) of each vector:
\( |\vec{b_1}| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7 \)
\( |\vec{b_2}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \)
We put these values back into the cosine formula:
\( \cos \theta = \frac{19}{7 \times 3} = \frac{19}{21} \)
To find the angle \( \theta \), we take the inverse cosine:
\( \theta = \cos^{-1} \left( \frac{19}{21} \right) \)
This calculation helps us determine how much the lines "turn" away from each other in three-dimensional space.
In simple words: We find the direction each line points. Then we use a special math rule (dot product and magnitudes) to figure out how far apart these two directions are, giving us the angle between the lines.

🎯 Exam Tip: Remember to correctly identify the direction vectors from the given line equations and apply the dot product formula carefully to avoid calculation errors.

Question 2. Find the angle between the lines \( \frac{x}{2} = \frac{y}{2} = \frac{z}{1} \) and \( \frac{x-5}{4} = \frac{y-2}{1} = \frac{z-3}{8} \).
Answer: First, we need to find the direction ratios for both lines. For the first line, which is in Cartesian form, the direction ratios are the denominators: \( a_1=2, b_1=2, c_1=1 \).
For the second line, the direction ratios are also the denominators: \( a_2=4, b_2=1, c_2=8 \).
The angle \( \theta \) between two lines with these direction ratios is given by the formula:
\( \cos \theta = \frac { a_1 a_2 + b_1 b_2 + c_1 c_2 }{ \sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2} } \)
Next, we calculate the top part of the fraction:
\( a_1 a_2 + b_1 b_2 + c_1 c_2 = (2)(4) + (2)(1) + (1)(8) = 8 + 2 + 8 = 18 \)
Now, we calculate the bottom part (magnitudes of the direction vectors):
\( \sqrt{a_1^2 + b_1^2 + c_1^2} = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \)
\( \sqrt{a_2^2 + b_2^2 + c_2^2} = \sqrt{4^2 + 1^2 + 8^2} = \sqrt{16 + 1 + 64} = \sqrt{81} = 9 \)
Substitute these values back into the formula for \( \cos \theta \):
\( \cos \theta = \frac{18}{3 \times 9} = \frac{18}{27} = \frac{2}{3} \)
Finally, to find the angle \( \theta \), we take the inverse cosine:
\( \theta = \cos^{-1} \left( \frac{2}{3} \right) \)
Understanding direction ratios helps visualize the orientation of lines in space.
In simple words: We get numbers (direction ratios) that show where each line is pointing. Then, we use a formula involving these numbers to find the angle between the two lines.

🎯 Exam Tip: When lines are given in Cartesian form, the denominators directly represent the direction ratios. Ensure you accurately calculate the square roots for the magnitudes.

Question 3. Show that the line passes through the points (1,-1,2) and (3,4,-2) is perpendicular to the line passes through the points (0, 3, 2) and (3, 5, 6).
Answer: To show that two lines are perpendicular, we first need to find their direction ratios.
For the first line passing through points \( P_1(1, -1, 2) \) and \( P_2(3, 4, -2) \), the direction ratios \( (a_1, b_1, c_1) \) are found by subtracting the coordinates:
\( (a_1, b_1, c_1) = (3-1, 4-(-1), -2-2) = (2, 5, -4) \)
For the second line passing through points \( Q_1(0, 3, 2) \) and \( Q_2(3, 5, 6) \), the direction ratios \( (a_2, b_2, c_2) \) are:
\( (a_2, b_2, c_2) = (3-0, 5-3, 6-2) = (3, 2, 4) \)
Two lines are perpendicular if the dot product of their direction ratios is zero:
\( a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 \)
Let's calculate this dot product:
\( (2)(3) + (5)(2) + (-4)(4) = 6 + 10 - 16 \)
\( \implies 16 - 16 = 0 \)
Since the result is 0, the two lines are indeed perpendicular to each other. This condition is a fundamental way to check orthogonality in 3D geometry.
In simple words: We find the direction numbers for both lines. If we multiply the matching direction numbers and add them up, and the answer is zero, then the lines cross each other at a perfect right angle.

🎯 Exam Tip: Remember that the direction ratios of a line passing through two points \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \) are \( (x_2-x_1, y_2-y_1, z_2-z_1) \). The dot product condition for perpendicularity is crucial.

Question 4. Find k, if lines \( \frac{x-1}{-3} = \frac{y-2}{2k} = \frac{z-3}{2} \) and \( \frac{x-1}{3k} = \frac{y-1}{1} = \frac{z-6}{-5} \) are perpendicular to each other.
Answer: We first identify the direction ratios for each line from their Cartesian equations.
For the first line, the direction ratios are \( a_1 = -3, b_1 = 2k, c_1 = 2 \).
For the second line, the direction ratios are \( a_2 = 3k, b_2 = 1, c_2 = -5 \).
Since the two lines are perpendicular, the dot product of their direction ratios must be equal to zero:
\( a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 \)
Now, we substitute the direction ratios into this equation:
\( (-3)(3k) + (2k)(1) + (2)(-5) = 0 \)
This simplifies to:
\( -9k + 2k - 10 = 0 \)
Combine the terms with \( k \):
\( -7k - 10 = 0 \)
Add 10 to both sides:
\( -7k = 10 \)
Finally, divide by -7 to find the value of \( k \):
\( k = -\frac{10}{7} \)
This value of k ensures the lines intersect at a right angle.
In simple words: We take the direction numbers of both lines. Because they are perpendicular, multiplying the matching numbers and adding them should give zero. We use this to solve for the unknown value 'k'.

🎯 Exam Tip: Always remember that for perpendicular lines, the sum of the products of their corresponding direction ratios is zero. This is a common property used to find unknown parameters.

Question 5. Find the vector equation of a line passing through point (1,2,-4) and perpendicular to the both lines \( \frac{x-8}{3} = \frac{y+19}{-16} = \frac{z-10}{7} \) and \( \frac{x-15}{3} = \frac{y-29}{8} = \frac{z-5}{-5} \).
Answer: Let the direction ratios of the required line be \( (b_1, b_2, b_3) \). The line passes through the point \( (1, 2, -4) \).
The vector equation of a line passing through a point \( \vec{a} \) and with direction vector \( \vec{b} \) is \( \vec{r} = \vec{a} + \lambda \vec{b} \).
The direction ratios of the first given line are \( (3, -16, 7) \). Since our required line is perpendicular to this line, their dot product is zero:
\( 3b_1 - 16b_2 + 7b_3 = 0 \) (Equation 1)
The direction ratios of the second given line are \( (3, 8, -5) \). Since our required line is also perpendicular to this line, their dot product is zero:
\( 3b_1 + 8b_2 - 5b_3 = 0 \) (Equation 2)
Now, we solve Equation 1 and Equation 2 for \( b_1, b_2, b_3 \) using the method of cross-multiplication:
\( \frac{b_1}{(-16)(-5) - (7)(8)} = \frac{b_2}{(7)(3) - (3)(-5)} = \frac{b_3}{(3)(8) - (-16)(3)} \)
\( \implies \frac{b_1}{80 - 56} = \frac{b_2}{21 - (-15)} = \frac{b_3}{24 - (-48)} \)
\( \implies \frac{b_1}{24} = \frac{b_2}{21 + 15} = \frac{b_3}{24 + 48} \)
\( \implies \frac{b_1}{24} = \frac{b_2}{36} = \frac{b_3}{72} \)
We can simplify these direction ratios by dividing each term by their greatest common divisor, which is 12:
\( \implies \frac{b_1}{2} = \frac{b_2}{3} = \frac{b_3}{6} \)
So, we can take the direction ratios of the required line as \( (2, 3, 6) \).
Now, we can write the vector equation of the line passing through \( (1, 2, -4) \) with direction vector \( 2\hat{i} + 3\hat{j} + 6\hat{k} \):
\( \vec{r} = \hat{i} + 2\hat{j} - 4\hat{k} + \lambda (2\hat{i} + 3\hat{j} + 6\hat{k}) \)
The Cartesian equation of this line is given by:
\( \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-(-4)}{6} \)
\( \implies \frac{x-1}{2} = \frac{y-2}{3} = \frac{z+4}{6} \)
This method is crucial for finding lines perpendicular to two other lines, often useful in geometry problems involving planes.
In simple words: We need a line that goes through a certain point and also makes a right angle with two other lines. We find the direction of this new line by solving two equations. Once we have the direction and the point, we can write the line's equation.

🎯 Exam Tip: When a line is perpendicular to two other lines, its direction vector is found by taking the cross product of the direction vectors of the two given lines. This problem uses a system of equations, which is an equivalent method.

Question 6. Find the cartesian equation of a line that passes through point (- 2, 4, -5) and is parallel to \( \frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6} \).
Answer: We need to find the Cartesian equation of a line. We know it passes through the point \( (-2, 4, -5) \).
The line is parallel to the given line \( \frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6} \).
If two lines are parallel, they have the same direction ratios. So, we can directly take the direction ratios from the given line's equation, which are the denominators: \( (3, 5, 6) \).
The Cartesian equation of a line passing through a point \( (x_1, y_1, z_1) \) with direction ratios \( (a, b, c) \) is given by:
\( \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} \)
Now, we substitute the given point \( (-2, 4, -5) \) and the direction ratios \( (3, 5, 6) \) into this formula:
\( \frac{x - (-2)}{3} = \frac{y - 4}{5} = \frac{z - (-5)}{6} \)
Simplifying the terms:
\( \frac{x + 2}{3} = \frac{y - 4}{5} = \frac{z + 5}{6} \)
This is the required Cartesian equation of the line. Understanding parallel lines helps predict their relative orientation in space.
In simple words: We want to find the equation for a new line. This new line goes through a specific point and runs in the same direction as another line. Since they are parallel, they share the same direction numbers. We just plug the point and these numbers into the standard line equation.

🎯 Exam Tip: Remember that parallel lines share the same direction ratios. The Cartesian form of a line equation is straightforward when you have a point and direction ratios.