GSEB Class 12 Maths Solutions Chapter 11 Three Dimensional Geometry Exercise 11.2

Question 1. Show that the three lines with direction cosines \( \frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}; \frac{4}{13}, \frac{12}{13}, \frac{3}{13} \) and \( \frac{3}{13}, \frac{-4}{13}, \frac{12}{13} \) are mutually perpendicular.
Answer: The line with direction cosines \( l_1, m_1, n_1 \) and \( l_2, m_2, n_2 \) are perpendicular if \( l_1l_2 + m_1m_2 + n_1n_2 = 0 \).
(i) Lines with direction cosines \( \frac{12}{13}, \frac{-3}{13}, \frac{-4}{13} \) and \( \frac{4}{13}, \frac{12}{13}, \frac{3}{13} \) are perpendicular if:
\( \frac{12}{13} \times \frac{4}{13} + (\frac{-3}{13}) \times (\frac{12}{13}) + (\frac{-4}{13}) \times (\frac{3}{13}) = 0 \)
or, if \( \frac{48 - 36 - 12}{169} = 0 \), which is true.
Therefore, these lines are perpendicular.
(ii) The lines with direction cosines \( \frac{4}{13}, \frac{12}{13}, \frac{3}{13} \) and \( \frac{3}{13}, \frac{-4}{13}, \frac{12}{13} \) are perpendicular if:
\( \frac{4}{13} \times \frac{3}{13} + \frac{12}{13} \times (\frac{-4}{13}) + \frac{3}{13} \times \frac{12}{13} = 0 \)
or, if \( \frac{12 - 48 + 36}{169} = 0 \), which is true.
Therefore, these lines are perpendicular.
(iii) The lines with direction cosines \( \frac{3}{13}, \frac{-4}{13}, \frac{12}{13} \) and \( \frac{12}{13}, \frac{-3}{13}, \frac{-4}{13} \) are perpendicular if:
\( \frac{3}{13} \times \frac{12}{13} + (\frac{-4}{13}) \times (\frac{-3}{13}) + \frac{12}{13} \times (\frac{-4}{13}) = 0 \)
or, if \( \frac{36 + 12 - 48}{169} = 0 \), which is true.
Therefore, these lines are perpendicular.
Since all pairs of lines satisfy the perpendicularity condition, the given lines are mutually perpendicular.
In simple words: To show that two lines are perpendicular, we multiply their matching direction cosines and add them up. If the total is zero, they are perpendicular. We check this for all three pairs of lines, and since each pair gives zero, all the lines are perpendicular to each other.

Exam Tip: Remember that two lines with direction cosines \( (l_1, m_1, n_1) \) and \( (l_2, m_2, n_2) \) are perpendicular if and only if \( l_1l_2 + m_1m_2 + n_1n_2 = 0 \).

Question 2. Show that the line through the points (1, – 1, 2) and (3, 4, – 2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).
Answer: Let A be (1, -1, 2) and B be (3, 4, -2). The direction ratios of AB are \( (3-1), (4-(-1)), (-2-2) \), which simplifies to \( (2, 5, -4) \).
Let C be (0, 3, 2) and D be (3, 5, 6). The direction ratios of CD are \( (3-0), (5-3), (6-2) \), which simplifies to \( (3, 2, 4) \).
Two lines with direction ratios \( (a_1, b_1, c_1) \) and \( (a_2, b_2, c_2) \) are perpendicular if \( a_1a_2 + b_1b_2 + c_1c_2 = 0 \).
For AB and CD, we check: \( 2 \times 3 + 5 \times 2 + (-4) \times 4 \)
\( = 6 + 10 - 16 \)
\( = 0 \), which is true.
Therefore, AB is perpendicular to CD.
In simple words: First, we find the direction ratios for each line by subtracting the coordinates of their points. Then, we multiply the corresponding direction ratios and add them up. If the final sum is zero, the two lines are perpendicular.

Exam Tip: When finding direction ratios of a line passing through two points \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \), simply calculate \( (x_2-x_1, y_2-y_1, z_2-z_1) \).

Question 3. Show that the line through the points (4, 7, 8) and (2, 3, 4) is parallel to the line through the points (- 1, – 2, 1) and (1, 2, 5).
Answer: Let A be (4, 7, 8) and B be (2, 3, 4). The direction ratios of AB are \( (2-4), (3-7), (4-8) \), which simplifies to \( (-2, -4, -4) \). We can also express these as \( (1, 2, 2) \) by dividing by -2.
Let C be (-1, -2, 1) and D be (1, 2, 5). The direction ratios of CD are \( (1-(-1)), (2-(-2)), (5-1) \), which simplifies to \( (2, 4, 4) \).
Two lines with direction ratios \( (a_1, b_1, c_1) \) and \( (a_2, b_2, c_2) \) are parallel if \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \).
Comparing the simplified direction ratios of AB \( (1, 2, 2) \) and CD \( (2, 4, 4) \):
\( \frac{1}{2} = \frac{2}{4} = \frac{2}{4} \), which simplifies to \( \frac{1}{2} = \frac{1}{2} = \frac{1}{2} \). This is true.
Therefore, AB is parallel to CD.
In simple words: We find the direction ratios for both lines. If these ratios are proportional (meaning you can multiply one set by a constant to get the other), then the lines are parallel.

Exam Tip: For lines to be parallel, their direction ratios must be proportional. Always simplify direction ratios to their simplest form before comparing them.

Question 4. Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector \( 3\hat {i} + 2\hat {j} – 2\hat {k} \).
Answer: The equation of a line passing through a point with position vector \( \vec{a} \) and parallel to vector \( \vec{b} \) is given by \( \vec{r} = \vec{a} + \lambda\vec{b} \).
Here, the position vector of the point is \( \vec{a} = \hat {i} + 2\hat {j} + 3\hat {k} \).
The vector parallel to the line is \( \vec{b} = 3\hat {i} + 2\hat {j} – 2\hat {k} \).
Therefore, the equation of the required line is \( \vec{r} = (\hat {i} + 2\hat {j} + 3\hat {k} ) + \lambda(3\hat {i} + 2\hat {j} – 2\hat {k} ) \), where \( \lambda \in R \).
In simple words: The line's equation is found by starting from the point it goes through (given by vector 'a') and then adding any multiple of the direction it points in (given by vector 'b').

Exam Tip: Always identify the position vector of the point the line passes through and the direction vector it is parallel to. These two components are key to forming the line's equation.

Question 5. Find the equation of the line in vector and in cartesian form that passes through the point with position vector \( 2\hat {i} – \hat {j} + 4\hat {k} \) and is in the direction \( \hat {i} + 2\hat {j} – \hat {k} \).
Answer: We know that the vector equation of a line passing through a point with position vector \( \vec{a} \) and parallel to the vector \( \vec{b} \) is \( \vec{r} = \vec{a} + \lambda\vec{b} \).
Here, the position vector is \( \vec{a} = 2\hat {i} – \hat {j} + 4\hat {k} \).
The direction vector is \( \vec{b} = \hat {i} + 2\hat {j} – \hat {k} \).
So, the vector equation of the required line is:
\( \vec{r} = (2\hat {i} – \hat {j} + 4\hat {k} ) + \lambda(\hat {i} + 2\hat {j} – \hat {k} ) \). Let this be equation (1), where \( \lambda \) is a parameter.
To find the Cartesian form, we substitute \( \vec{r} = x\hat {i} + y\hat {j} + z\hat {k} \) into equation (1):
\( x\hat {i} + y\hat {j} + z\hat {k} = (2\hat {i} – \hat {j} + 4\hat {k} ) + \lambda(\hat {i} + 2\hat {j} – \hat {k}) \)
\( \implies x\hat {i} + y\hat {j} + z\hat {k} = (2 + \lambda)\hat {i} + (-1 + 2\lambda)\hat {j} + (4 – \lambda)\hat {k} \)
Equating coefficients of \( \hat {i}, \hat {j} \) and \( \hat {k} \), we get:
\( x = 2 + \lambda \implies \lambda = x - 2 \)
\( y = -1 + 2\lambda \implies \lambda = \frac{y+1}{2} \)
\( z = 4 – \lambda \implies \lambda = \frac{z-4}{-1} \)
Since all these expressions equal \( \lambda \), we can set them equal to each other to get the Cartesian form:
\( \frac{x-2}{1} = \frac{y+1}{2} = \frac{z-4}{-1} \).
In simple words: We first write the equation in vector form using the starting point and the direction. To get the Cartesian form, we replace the general position vector with (x, y, z) and then match the coefficients of i, j, and k. This gives us individual equations for x, y, and z in terms of a variable, which we then combine.

Exam Tip: When converting from vector to Cartesian form, remember that the coefficients of \( \hat{i}, \hat{j}, \hat{k} \) in the direction vector become the denominators, and the coordinates of the passing point are subtracted from \( x, y, z \) in the numerators.

Question 6. Find the cartesian equation of the line which passes through the point (- 2, 4, – 5) and parallel to the line given by \( \frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6} \).
Answer: The equation of a line passing through a point \( (x_1, y_1, z_1) \) and parallel to another line with direction ratios \( (a, b, c) \) is given by \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \).
Here, the line passes through the point \( (-2, 4, -5) \). So, \( x_1 = -2, y_1 = 4, z_1 = -5 \).
The given parallel line is \( \frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6} \). From this, we can get the direction ratios as \( (3, 5, 6) \). So, \( a=3, b=5, c=6 \).
Substituting these values into the Cartesian equation formula:
\( \frac{x-(-2)}{3} = \frac{y-4}{5} = \frac{z-(-5)}{6} \)
\( \implies \frac{x+2}{3} = \frac{y-4}{5} = \frac{z+5}{6} \).
This is the Cartesian equation of the required line.
In simple words: If a new line is parallel to an existing line, it shares the same direction ratios. We use these ratios, along with the coordinates of the point the new line goes through, to write its Cartesian equation.

Exam Tip: Remember that parallel lines have identical or proportional direction ratios. The denominators in the Cartesian equation represent these direction ratios.

Question 7. The cartesian equation of a line \( \frac{x-5}{3} = \frac{y+4}{7} = \frac{z-6}{2} \). Write its vector form.
Answer: The Cartesian equation of the line is \( \frac{x-5}{3} = \frac{y+4}{7} = \frac{z-6}{2} \).
By comparing this with the standard Cartesian form \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \), we can identify:
The line passes through the point A with coordinates \( (5, -4, 6) \). So, the position vector \( \vec{a} = 5\hat {i} – 4\hat {j} + 6\hat {k} \).
The direction ratios of the line are \( (3, 7, 2) \). So, the direction vector \( \vec{b} = 3\hat {i} + 7\hat {j} + 2\hat {k} \).
The vector equation of a line passing through \( \vec{a} \) and in the direction of \( \vec{b} \) is \( \vec{r} = \vec{a} + \lambda\vec{b} \).
Substituting the identified vectors:
\( \vec{r} = (5\hat {i} – 4\hat {j} + 6\hat {k} ) + \lambda(3\hat {i} + 7\hat {j} + 2\hat {k} ) \).
In simple words: From the Cartesian form, the numbers being subtracted from x, y, and z (or added, then we change the sign) give us the point the line passes through. The denominators give us the direction vector. We then combine these into the standard vector equation.

Exam Tip: When \( x+a \) appears in the numerator, the coordinate is \( -a \). For example, \( y+4 \) means the y-coordinate is \( -4 \). Be careful with signs.

Question 8. Find the vector and cartesian equations of the line that passes through the origin and the point B(5, – 2, 3).
Answer: The line passes through the origin O \( (0, 0, 0) \) and point B \( (5, -2, 3) \).
The position vector of the point through which the line passes (origin) is \( \vec{a} = \vec{0} = 0\hat{i} + 0\hat{j} + 0\hat{k} \).
The direction ratios of the line joining two points \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \) are \( (x_2-x_1, y_2-y_1, z_2-z_1) \).
So, the direction ratios of the line joining O(0, 0, 0) and B(5, -2, 3) are \( (5-0, -2-0, 3-0) \), which simplifies to \( (5, -2, 3) \).
The direction vector \( \vec{b} = 5\hat {i} – 2\hat {j} + 3\hat {k} \).
(i) Equation of line OB in vector form:
\( \vec{r} = \vec{a} + \lambda\vec{b} \)
\( = \vec{0} + \lambda(5\hat {i} – 2\hat {j} + 3\hat {k} ) \)
\( = \lambda(5\hat {i} – 2\hat {j} + 3\hat {k} ) \).
(ii) Equation of line OB in Cartesian form:
The standard Cartesian equation is \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \), where \( (x_1, y_1, z_1) \) is a point on the line and \( (a, b, c) \) are the direction ratios.
Using the origin \( (0, 0, 0) \) as the point and \( (5, -2, 3) \) as the direction ratios:
\( \frac{x-0}{5} = \frac{y-0}{-2} = \frac{z-0}{3} \)
\( \implies \frac{x}{5} = \frac{y}{-2} = \frac{z}{3} \).
In simple words: For a line going through two points, we use one point as the starting point and the vector between the two points as the direction. Then, we just write both the vector and Cartesian equations based on these details. Since one point is the origin, its coordinates are all zeros.

Exam Tip: When a line passes through the origin, its position vector is \( \vec{0} \), simplifying the vector equation to just \( \lambda\vec{b} \).

Question 9. Find the vector and cartesian equations of the line that passes through the points P(3, – 2, – 5) and Q(3, – 2, 6).
Answer: The line PQ passes through the point P(3, -2, -5). So, the position vector \( \vec{a} = 3\hat {i} – 2\hat {j} – 5\hat {k} \).
The direction ratios of PQ, where P is \( (3, -2, -5) \) and Q is \( (3, -2, 6) \), are \( (x_2-x_1, y_2-y_1, z_2-z_1) \).
So, the direction ratios are \( (3-3, -2-(-2), 6-(-5)) \), which simplifies to \( (0, 0, 11) \).
The direction vector \( \vec{b} = 0\hat {i} + 0\hat {j} + 11\hat {k} = 11\hat {k} \).
(i) Equation of line PQ in vector form:
\( \vec{r} = \vec{a} + \lambda\vec{b} \)
\( = (3\hat {i} – 2\hat {j} – 5\hat {k} ) + \lambda(11\hat {k} ) \).
(ii) Equation of line PQ in Cartesian form:
The standard Cartesian equation is \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \).
Using point P \( (3, -2, -5) \) and direction ratios \( (0, 0, 11) \):
\( \frac{x-3}{0} = \frac{y-(-2)}{0} = \frac{z-(-5)}{11} \)
\( \implies \frac{x-3}{0} = \frac{y+2}{0} = \frac{z+5}{11} \).
In simple words: To find the equations, we pick one of the points as the starting point and calculate the vector between the two given points to get the direction. Then we write down the vector form and the Cartesian form. Note that a zero in the denominator for the Cartesian form implies the line is parallel to that axis in a special way.

Exam Tip: When a direction ratio is zero, it means the line is parallel to the corresponding coordinate plane. For example, if 'a' is 0, the line is parallel to the YZ-plane.

Question 10. Find the angle between the following pairs of lines:
(i) \( \vec{r} = 2\hat {i} – 5\hat {j} + \hat {k} + \lambda(3\hat {i} + 2\hat {j} + 6\hat {k}) \) and \( \vec{r} = 7\hat {i} – 6\hat {k} + \lambda(\hat {i} + 2\hat {j} + 2\hat {k}) \)
(ii) \( \vec{r} = 3\hat {i} + \hat {j} – 2\hat {k} + \lambda(\hat {i} – \hat {j} – 2\hat {k} ) \) and \( \vec{r} = 2\hat {i} – \hat {j} – 56\hat {k} + \mu(3\hat {i} – 5\hat {j} – 4\hat {k}) \)
Answer: The angle \( \theta \) between two lines with direction vectors \( \vec{b}_1 \) and \( \vec{b}_2 \) is given by \( \cos \theta = \frac{|\vec{b}_1 \cdot \vec{b}_2|}{|\vec{b}_1| |\vec{b}_2|} \).
(i) For the first pair of lines:
\( \vec{b}_1 = 3\hat {i} + 2\hat {j} + 6\hat {k} \)
\( \vec{b}_2 = \hat {i} + 2\hat {j} + 2\hat {k} \)
\( \vec{b}_1 \cdot \vec{b}_2 = (3)(1) + (2)(2) + (6)(2) = 3 + 4 + 12 = 19 \)
\( |\vec{b}_1| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7 \)
\( |\vec{b}_2| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \)
\( \cos \theta = \frac{|19|}{(7)(3)} = \frac{19}{21} \)
\( \implies \theta = \cos^{-1}\left(\frac{19}{21}\right) \).
(ii) For the second pair of lines:
\( \vec{b}_1 = \hat {i} – \hat {j} – 2\hat {k} \)
\( \vec{b}_2 = 3\hat {i} – 5\hat {j} – 4\hat {k} \)
\( \vec{b}_1 \cdot \vec{b}_2 = (1)(3) + (-1)(-5) + (-2)(-4) = 3 + 5 + 8 = 16 \)
\( |\vec{b}_1| = \sqrt{1^2 + (-1)^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6} \)
\( |\vec{b}_2| = \sqrt{3^2 + (-5)^2 + (-4)^2} = \sqrt{9 + 25 + 16} = \sqrt{50} = 5\sqrt{2} \)
\( \cos \theta = \frac{|16|}{\sqrt{6} \times \sqrt{50}} = \frac{16}{\sqrt{300}} = \frac{16}{10\sqrt{3}} = \frac{8}{5\sqrt{3}} = \frac{8\sqrt{3}}{15} \)
\( \implies \theta = \cos^{-1}\left(\frac{8\sqrt{3}}{15}\right) \).
In simple words: To find the angle between two lines, we use their direction vectors. We calculate the dot product of these vectors and divide it by the product of their magnitudes. The inverse cosine of this value gives us the angle.

Exam Tip: Remember to extract only the direction vectors \( \vec{b}_1 \) and \( \vec{b}_2 \) from the line equations \( \vec{r} = \vec{a} + \lambda\vec{b} \). The position vectors \( \vec{a} \) are not used in calculating the angle between lines.

Question 11. Find the angle between the following pair of lines:
(i) \( \frac{x-2}{2} = \frac{y-1}{5} = \frac{z+3}{-3} \) and \( \frac{x+2}{-1} = \frac{y-4}{8} = \frac{z-5}{4} \)
(ii) \( \frac{x}{2} = \frac{y}{2} = \frac{z}{1} \) and \( \frac{x-5}{4} = \frac{y-2}{1} = \frac{z-3}{8} \)
Answer: The angle \( \theta \) between two lines with direction vectors \( \vec{b}_1 \) and \( \vec{b}_2 \) is given by \( \cos \theta = \frac{|\vec{b}_1 \cdot \vec{b}_2|}{|\vec{b}_1| |\vec{b}_2|} \).
(i) For the first pair of lines, we extract the direction vectors from the denominators of their Cartesian equations:
\( \vec{b}_1 = 2\hat {i} + 5\hat {j} – 3\hat {k} \)
\( \vec{b}_2 = -\hat {i} + 8\hat {j} + 4\hat {k} \)
\( \vec{b}_1 \cdot \vec{b}_2 = (2)(-1) + (5)(8) + (-3)(4) = -2 + 40 - 12 = 26 \)
\( |\vec{b}_1| = \sqrt{2^2 + 5^2 + (-3)^2} = \sqrt{4 + 25 + 9} = \sqrt{38} \)
\( |\vec{b}_2| = \sqrt{(-1)^2 + 8^2 + 4^2} = \sqrt{1 + 64 + 16} = \sqrt{81} = 9 \)
\( \cos \theta = \frac{|26|}{\sqrt{38} \times 9} = \frac{26}{9\sqrt{38}} \)
\( \implies \theta = \cos^{-1}\left(\frac{26}{9\sqrt{38}}\right) \).
(ii) For the second pair of lines:
\( \vec{b}_1 = 2\hat {i} + 2\hat {j} + \hat {k} \)
\( \vec{b}_2 = 4\hat {i} + \hat {j} + 8\hat {k} \)
\( \vec{b}_1 \cdot \vec{b}_2 = (2)(4) + (2)(1) + (1)(8) = 8 + 2 + 8 = 18 \)
\( |\vec{b}_1| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \)
\( |\vec{b}_2| = \sqrt{4^2 + 1^2 + 8^2} = \sqrt{16 + 1 + 64} = \sqrt{81} = 9 \)
\( \cos \theta = \frac{|18|}{3 \times 9} = \frac{18}{27} = \frac{2}{3} \)
\( \implies \theta = \cos^{-1}\left(\frac{2}{3}\right) \).
In simple words: First, we find the direction vectors for each line from their Cartesian equations (these are the denominators). Then we use the formula for the angle between two vectors: the dot product of the vectors divided by the product of their lengths. The result is then used with inverse cosine to find the angle.

Exam Tip: When given Cartesian equations, the direction ratios are directly the denominators. Be sure to correctly form the direction vectors before calculating the dot product and magnitudes.

Question 12. Find the value of p so that the lines \( \frac{x-1}{-3} = \frac{7y-14}{2p} = \frac{z-3}{2} \) and \( \frac{7-7x}{3p} = \frac{y-5}{1} = \frac{6-z}{5} \) are at right angles.
Answer: The given equations are not in the standard Cartesian form \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \). We must rewrite them.
For the first line:
\( \frac{x-1}{-3} = \frac{7(y-2)}{2p} = \frac{z-3}{2} \)
\( \implies \frac{x-1}{-3} = \frac{y-2}{\frac{2p}{7}} = \frac{z-3}{2} \)
So, the direction ratios are \( (a_1, b_1, c_1) = (-3, \frac{2p}{7}, 2) \).
For the second line:
\( \frac{7(1-x)}{3p} = \frac{y-5}{1} = \frac{-(z-6)}{5} \)
\( \implies \frac{-(x-1)}{3p/7} = \frac{y-5}{1} = \frac{z-6}{-5} \)
\( \implies \frac{x-1}{-3p/7} = \frac{y-5}{1} = \frac{z-6}{-5} \)
So, the direction ratios are \( (a_2, b_2, c_2) = (-\frac{3p}{7}, 1, -5) \).
For the lines to be at right angles (perpendicular), the condition is \( a_1a_2 + b_1b_2 + c_1c_2 = 0 \).
\( (-3)\left(-\frac{3p}{7}\right) + \left(\frac{2p}{7}\right)(1) + (2)(-5) = 0 \)
\( \frac{9p}{7} + \frac{2p}{7} - 10 = 0 \)
Multiply by 7 to clear the denominators:
\( 9p + 2p - 70 = 0 \)
\( 11p - 70 = 0 \)
\( 11p = 70 \)
\( \implies p = \frac{70}{11} \).
In simple words: First, we change the given line equations into the standard Cartesian form, which means having (x - x1) in the numerator, not (7x - 14). Once we get the direction ratios for both lines, we use the rule for perpendicular lines: multiply the matching ratios and add them up, setting the total to zero. Then, we just solve this equation to find 'p'.

Exam Tip: Always convert line equations to the standard form \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \) before identifying direction ratios. A common mistake is to directly take coefficients without factoring out constants or ensuring the \( x, y, z \) terms are positive.

Question 13. Show that the lines \( \frac{x-5}{7} = \frac{y+2}{-5} = \frac{z}{1} \) and \( \frac{x}{1} = \frac{y}{2} = \frac{z}{3} \) are perpendicular to each other.
Answer: For the first line, \( \frac{x-5}{7} = \frac{y+2}{-5} = \frac{z}{1} \), the direction ratios are \( (a_1, b_1, c_1) = (7, -5, 1) \).
So, the direction vector \( \vec{b}_1 = 7\hat {i} – 5\hat {j} + \hat {k} \).
For the second line, \( \frac{x}{1} = \frac{y}{2} = \frac{z}{3} \), the direction ratios are \( (a_2, b_2, c_2) = (1, 2, 3) \).
So, the direction vector \( \vec{b}_2 = \hat {i} + 2\hat {j} + 3\hat {k} \).
Two lines are perpendicular if \( a_1a_2 + b_1b_2 + c_1c_2 = 0 \).
Let's check this condition:
\( (7)(1) + (-5)(2) + (1)(3) \)
\( = 7 - 10 + 3 \)
\( = 0 \)
Since the sum is 0, the lines are perpendicular to each other.
Alternatively, using the dot product of direction vectors:
\( \vec{b}_1 \cdot \vec{b}_2 = (7\hat {i} – 5\hat {j} + \hat {k}) \cdot (\hat {i} + 2\hat {j} + 3\hat {k}) \)
\( = 7(1) + (-5)(2) + 1(3) \)
\( = 7 - 10 + 3 = 0 \)
Since the dot product is 0, the angle \( \theta \) between them is given by \( \cos \theta = \frac{|\vec{b}_1 \cdot \vec{b}_2|}{|\vec{b}_1| |\vec{b}_2|} = \frac{0}{|\vec{b}_1| |\vec{b}_2|} = 0 \).
Thus, \( \theta = \cos^{-1}(0) = \frac{\pi}{2} \). The lines are perpendicular.
In simple words: We get the direction ratios for each line from their equations. Then, we multiply the corresponding ratios for the two lines and add them all up. If the total is zero, it means the lines are at a right angle to each other, so they are perpendicular.

Exam Tip: For lines in Cartesian form, the denominators represent the direction ratios. The perpendicularity condition \( a_1a_2 + b_1b_2 + c_1c_2 = 0 \) is a quick way to check, or you can use the vector dot product.

Question 14. Find the shortest distance between the lines \( \vec{r} = (\hat {i} + 2\hat {j} + \hat {k} ) + \lambda(\hat {i} – \hat {j} + \hat {k}) \) and \( \vec{r} = (2\hat {i} – \hat {j} – \hat {k} ) + \mu(2\hat {i} + \hat {j} + 2\hat {k}) \).
Answer: The shortest distance between two skew lines \( \vec{r} = \vec{a}_1 + \lambda\vec{b}_1 \) and \( \vec{r} = \vec{a}_2 + \mu\vec{b}_2 \) is given by the formula:
\( d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|} \).
Comparing the given equations with the standard form, we have:
For the first line:
\( \vec{a}_1 = \hat {i} + 2\hat {j} + \hat {k} \)
\( \vec{b}_1 = \hat {i} – \hat {j} + \hat {k} \)
For the second line:
\( \vec{a}_2 = 2\hat {i} – \hat {j} – \hat {k} \)
\( \vec{b}_2 = 2\hat {i} + \hat {j} + 2\hat {k} \)
Now, let's calculate the required terms:
1. \( \vec{a}_2 - \vec{a}_1 \):
\( \vec{a}_2 - \vec{a}_1 = (2\hat {i} – \hat {j} – \hat {k}) - (\hat {i} + 2\hat {j} + \hat {k}) \)
\( = (2-1)\hat {i} + (-1-2)\hat {j} + (-1-1)\hat {k} \)
\( = \hat {i} – 3\hat {j} – 2\hat {k} \)
2. \( \vec{b}_1 \times \vec{b}_2 \):
\( \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 2 & 1 & 2 \end{vmatrix} \)
\( = \hat{i}((-1)(2) - (1)(1)) - \hat{j}((1)(2) - (1)(2)) + \hat{k}((1)(1) - (-1)(2)) \)
\( = \hat{i}(-2-1) - \hat{j}(2-2) + \hat{k}(1+2) \)
\( = -3\hat {i} + 0\hat {j} + 3\hat {k} = -3\hat {i} + 3\hat {k} \)
3. \( |\vec{b}_1 \times \vec{b}_2| \):
\( |\vec{b}_1 \times \vec{b}_2| = \sqrt{(-3)^2 + 0^2 + 3^2} = \sqrt{9 + 0 + 9} = \sqrt{18} = 3\sqrt{2} \)
4. \( (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) \):
\( (\hat {i} – 3\hat {j} – 2\hat {k} ) \cdot (-3\hat {i} + 3\hat {k} ) \)
\( = (1)(-3) + (-3)(0) + (-2)(3) \)
\( = -3 + 0 - 6 = -9 \)
Now, substitute these values into the shortest distance formula:
\( d = \frac{|-9|}{3\sqrt{2}} = \frac{9}{3\sqrt{2}} = \frac{3}{\sqrt{2}} \)
To rationalize the denominator, multiply by \( \frac{\sqrt{2}}{\sqrt{2}} \):
\( d = \frac{3\sqrt{2}}{2} \).
In simple words: To find the shortest distance between two lines that do not cross each other, we first identify the starting points and direction vectors for each line. Then, we calculate the vector connecting the starting points, the cross product of the direction vectors, and the magnitude of this cross product. Finally, we use a formula involving these values to get the shortest distance.

Exam Tip: Be meticulous with vector operations, especially the cross product and dot product, as a small error can propagate through the entire calculation. Rationalize the denominator in your final answer if it contains a square root.

Question 15. Find the shortest distance between the lines \( \frac{x+1}{7} = \frac{y+1}{-6} = \frac{z+1}{1} \) and \( \frac{x-3}{1} = \frac{y-5}{-2} = \frac{z-7}{1} \).
Answer: The shortest distance between two lines given in Cartesian form \( \frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1} \) and \( \frac{x-x_2}{a_2} = \frac{y-y_2}{b_2} = \frac{z-z_2}{c_2} \) is given by:
\( d = \frac{\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix}}{\sqrt{(b_1c_2-b_2c_1)^2 + (c_1a_2-c_2a_1)^2 + (a_1b_2-a_2b_1)^2}} \).
From the first line, \( \frac{x+1}{7} = \frac{y+1}{-6} = \frac{z+1}{1} \):
Point \( (x_1, y_1, z_1) = (-1, -1, -1) \)
Direction ratios \( (a_1, b_1, c_1) = (7, -6, 1) \)
From the second line, \( \frac{x-3}{1} = \frac{y-5}{-2} = \frac{z-7}{1} \):
Point \( (x_2, y_2, z_2) = (3, 5, 7) \)
Direction ratios \( (a_2, b_2, c_2) = (1, -2, 1) \)
Calculate \( (x_2-x_1, y_2-y_1, z_2-z_1) \):
\( x_2-x_1 = 3 - (-1) = 4 \)
\( y_2-y_1 = 5 - (-1) = 6 \)
\( z_2-z_1 = 7 - (-1) = 8 \)
Numerator \( N = \begin{vmatrix} 4 & 6 & 8 \\ 7 & -6 & 1 \\ 1 & -2 & 1 \end{vmatrix} \)
\( N = 4((-6)(1) - (1)(-2)) - 6((7)(1) - (1)(1)) + 8((7)(-2) - (-6)(1)) \)
\( N = 4(-6 + 2) - 6(7 - 1) + 8(-14 + 6) \)
\( N = 4(-4) - 6(6) + 8(-8) \)
\( N = -16 - 36 - 64 = -116 \)
Denominator \( D = \sqrt{(b_1c_2-b_2c_1)^2 + (c_1a_2-c_2a_1)^2 + (a_1b_2-a_2b_1)^2} \)
This is the magnitude of \( \vec{b}_1 \times \vec{b}_2 \). Let's calculate \( \vec{b}_1 \times \vec{b}_2 \) first:
\( \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & -6 & 1 \\ 1 & -2 & 1 \end{vmatrix} \)
\( = \hat{i}((-6)(1) - (1)(-2)) - \hat{j}((7)(1) - (1)(1)) + \hat{k}((7)(-2) - (-6)(1)) \)
\( = \hat{i}(-6 + 2) - \hat{j}(7 - 1) + \hat{k}(-14 + 6) \)
\( = -4\hat{i} - 6\hat{j} - 8\hat{k} \)
\( D = |\vec{b}_1 \times \vec{b}_2| = \sqrt{(-4)^2 + (-6)^2 + (-8)^2} \)
\( = \sqrt{16 + 36 + 64} = \sqrt{116} \)
Now, the shortest distance \( d = \frac{|N|}{D} \):
\( d = \frac{|-116|}{\sqrt{116}} = \frac{116}{\sqrt{116}} = \sqrt{116} \)
We can simplify \( \sqrt{116} = \sqrt{4 \times 29} = 2\sqrt{29} \).
So, \( d = 2\sqrt{29} \).
In simple words: When lines are given in Cartesian form, we find the starting points and direction ratios for each. Then we use a special determinant formula for the numerator and the magnitude of the cross product of direction vectors for the denominator. Dividing these gives the shortest distance between the lines.

Exam Tip: This formula is specific for shortest distance between two skew lines in Cartesian form. Ensure that \( x_2-x_1, y_2-y_1, z_2-z_1 \) form the first row of the determinant and that you correctly compute the determinant and the magnitude of the cross product.

Question 16. Find the shortest distance the lines whose vector equations are \( \vec{r} = (\hat {i} + 2\hat {j} + 3\hat {k} ) + \lambda(\hat {i} – 3\hat {j} + 2\hat {k}) \) and \( \vec{r} = (4\hat {i} + 5\hat {j} + 6\hat {k} ) + \mu(2\hat {i} + 3\hat {j} + \hat {k}) \).
Answer: We know that the shortest distance between two lines \( \vec{r} = \vec{a}_1 + \lambda\vec{b}_1 \) and \( \vec{r} = \vec{a}_2 + \mu\vec{b}_2 \) is given by the formula:
\( d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|} \).
From the given equations:
For the first line:
\( \vec{a}_1 = \hat {i} + 2\hat {j} + 3\hat {k} \)
\( \vec{b}_1 = \hat {i} – 3\hat {j} + 2\hat {k} \)
For the second line:
\( \vec{a}_2 = 4\hat {i} + 5\hat {j} + 6\hat {k} \)
\( \vec{b}_2 = 2\hat {i} + 3\hat {j} + \hat {k} \)
Now, let's calculate the necessary components:
1. \( \vec{a}_2 - \vec{a}_1 \):
\( \vec{a}_2 - \vec{a}_1 = (4\hat {i} + 5\hat {j} + 6\hat {k} ) - (\hat {i} + 2\hat {j} + 3\hat {k} ) \)
\( = (4-1)\hat {i} + (5-2)\hat {j} + (6-3)\hat {k} \)
\( = 3\hat {i} + 3\hat {j} + 3\hat {k} \)
2. \( \vec{b}_1 \times \vec{b}_2 \):
\( \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ 2 & 3 & 1 \end{vmatrix} \)
\( = \hat{i}((-3)(1) - (2)(3)) - \hat{j}((1)(1) - (2)(2)) + \hat{k}((1)(3) - (-3)(2)) \)
\( = \hat{i}(-3 - 6) - \hat{j}(1 - 4) + \hat{k}(3 + 6) \)
\( = -9\hat {i} + 3\hat {j} + 9\hat {k} \)
3. \( |\vec{b}_1 \times \vec{b}_2| \):
\( |\vec{b}_1 \times \vec{b}_2| = \sqrt{(-9)^2 + 3^2 + 9^2} = \sqrt{81 + 9 + 81} = \sqrt{171} = \sqrt{9 \times 19} = 3\sqrt{19} \)
4. \( (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) \):
\( (3\hat {i} + 3\hat {j} + 3\hat {k} ) \cdot (-9\hat {i} + 3\hat {j} + 9\hat {k} ) \)
\( = (3)(-9) + (3)(3) + (3)(9) \)
\( = -27 + 9 + 27 = 9 \)
Finally, substitute these values into the shortest distance formula:
\( d = \frac{|9|}{3\sqrt{19}} = \frac{9}{3\sqrt{19}} = \frac{3}{\sqrt{19}} \)
Rationalizing the denominator:
\( d = \frac{3\sqrt{19}}{19} \).
In simple words: We find the starting points and direction vectors for each line from their vector equations. Then we compute the vector joining the starting points, the cross product of the direction vectors, and its magnitude. Using these, we calculate the shortest distance with the specific formula.

Exam Tip: When simplifying square roots, always look for perfect square factors. Also, remember to rationalize the denominator by multiplying both the numerator and denominator by the radical term.

Question 17. Find the shortest distance between the lines whose vector equations are \( \vec{r} = (1 - t)\hat {i} + (t – 2)\hat {j} + (3 – 2)\hat {k} \) and \( \vec{r} = (\hat {i} – \hat {j} – \hat {k} ) + s(\hat {i} + 2\hat {j} – 2\hat {k}) \).
Answer: The shortest distance between two lines \( \vec{r} = \vec{a}_1 + \lambda\vec{b}_1 \) and \( \vec{r} = \vec{a}_2 + \mu\vec{b}_2 \) is given by the formula:
\( d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|} \).
First, rewrite the given equations in the standard form \( \vec{r} = \vec{a} + \text{parameter}\cdot\vec{b} \).
For the first line:
\( \vec{r} = (1 - t)\hat {i} + (t – 2)\hat {j} + (3 – 2t)\hat {k} \)
\( \vec{r} = (\hat {i} - 2\hat {j} + 3\hat {k}) + t(-\hat {i} + \hat {j} - 2\hat {k}) \)
So, \( \vec{a}_1 = \hat {i} - 2\hat {j} + 3\hat {k} \)
And \( \vec{b}_1 = -\hat {i} + \hat {j} - 2\hat {k} \)
For the second line:
\( \vec{r} = (\hat {i} – \hat {j} – \hat {k} ) + s(\hat {i} + 2\hat {j} – 2\hat {k}) \)
So, \( \vec{a}_2 = \hat {i} – \hat {j} – \hat {k} \)
And \( \vec{b}_2 = \hat {i} + 2\hat {j} – 2\hat {k} \)
Now, let's calculate the required components:
1. \( \vec{a}_2 - \vec{a}_1 \):
\( \vec{a}_2 - \vec{a}_1 = (\hat {i} – \hat {j} – \hat {k}) - (\hat {i} - 2\hat {j} + 3\hat {k}) \)
\( = (1-1)\hat {i} + (-1-(-2))\hat {j} + (-1-3)\hat {k} \)
\( = 0\hat {i} + \hat {j} – 4\hat {k} = \hat {j} – 4\hat {k} \)
2. \( \vec{b}_1 \times \vec{b}_2 \):
\( \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & -2 \\ 1 & 2 & -2 \end{vmatrix} \)
\( = \hat{i}((1)(-2) - (-2)(2)) - \hat{j}((-1)(-2) - (-2)(1)) + \hat{k}((-1)(2) - (1)(1)) \)
\( = \hat{i}(-2 + 4) - \hat{j}(2 + 2) + \hat{k}(-2 - 1) \)
\( = 2\hat {i} – 4\hat {j} – 3\hat {k} \)
3. \( |\vec{b}_1 \times \vec{b}_2| \):
\( |\vec{b}_1 \times \vec{b}_2| = \sqrt{2^2 + (-4)^2 + (-3)^2} = \sqrt{4 + 16 + 9} = \sqrt{29} \)
4. \( (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) \):
\( (\hat {j} – 4\hat {k} ) \cdot (2\hat {i} – 4\hat {j} – 3\hat {k} ) \)
\( = (0)(2) + (1)(-4) + (-4)(-3) \)
\( = 0 - 4 + 12 = 8 \)
Finally, substitute these values into the shortest distance formula:
\( d = \frac{|8|}{\sqrt{29}} = \frac{8}{\sqrt{29}} \)
Rationalizing the denominator:
\( d = \frac{8\sqrt{29}}{29} \).
In simple words: We first rewrite the given lines in the standard vector form to easily pick out the starting points and direction vectors. Then, we calculate the vector between the starting points and the cross product of the direction vectors. After finding the magnitude of the cross product, we use the formula involving these values to determine the shortest distance between the lines.

Exam Tip: Be careful when rewriting parameterized line equations. Collect terms without the parameter to form the position vector \( \vec{a} \) and terms with the parameter to form the direction vector \( \vec{b} \).