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

Question 1. Find the angle between the following pairs of lines:
(i) \( \frac{x+4}{3} = \frac{y-1}{5} = \frac{z+3}{4} \); \( \frac{x+1}{1} = \frac{y-4}{1} = \frac{z-5}{2} \)
(ii) \( \frac{x-2}{3} = \frac{y+3}{-2}, z = 5 \); \( \frac{x+1}{1} = \frac{2 y-3}{3} = \frac{z-5}{2} \)
Answer:
(i) For the lines \( \frac{x+4}{3} = \frac{y-1}{5} = \frac{z+3}{4} \) and \( \frac{x+1}{1} = \frac{y-4}{1} = \frac{z-5}{2} \):
Let \( \vec{b}_1 \) and \( \overrightarrow{b_2} \) be the vectors parallel to line (1) and (2) respectively.
So, \( \vec{b}_1 = 3 \hat{i} + 5 \hat{j} + 4 \hat{k} \)
And \( \overrightarrow{b_2} = \hat{i} + \hat{j} + 2 \hat{k} \).
Let \( \theta \) be the angle between the given lines.
The cosine of the angle \( \theta \) between two vectors \( \vec{b}_1 \) and \( \overrightarrow{b_2} \) is given by the formula:
\( \cos\theta = \frac{\overrightarrow{b_1} \cdot \overrightarrow{b_2}}{|\overrightarrow{b_1}| |\overrightarrow{b_2}|} \)
\( \implies \cos\theta = \frac{(3 \hat{i}+5 \hat{j}+4 \hat{k}) \cdot (\hat{i}+\hat{j}+2 \hat{k})}{\sqrt{3^2+5^2+4^2} \sqrt{1^2+1^2+2^2}} \)
\( \implies \cos\theta = \frac{3(1)+5(1)+4(2)}{\sqrt{9+25+16} \sqrt{1+1+4}} \)
\( \implies \cos\theta = \frac{3+5+8}{\sqrt{50} \sqrt{6}} \)
\( \implies \cos\theta = \frac{16}{\sqrt{300}} \)
\( \implies \cos\theta = \frac{16}{10\sqrt{3}} \)
\( \implies \cos\theta = \frac{8}{5\sqrt{3}} \)
So, \( \theta = \cos^{-1} \left( \frac{8}{5\sqrt{3}} \right) \). This angle can be used to describe the orientation between the two lines.
(ii) The given lines are \( \frac{x-2}{3} = \frac{y+3}{-2} \), \( z = 5 \) and \( \frac{x+1}{1} = \frac{2 y-3}{3} = \frac{z-5}{2} \).
We can write the first line as \( \frac{x-2}{3} = \frac{y+3}{-2} = \frac{z-5}{0} \) because \( z=5 \) means the change in \( z \) is zero.
The second line can be rewritten in standard form: \( \frac{x+1}{1} = \frac{y - \frac{3}{2}}{\frac{3}{2}} = \frac{z-5}{2} \). To avoid fractions, we can multiply the direction ratios by 2, so it becomes \( \frac{x+1}{2} = \frac{y - \frac{3}{2}}{3} = \frac{z-5}{4} \).
Let \( \vec{b}_1 \) and \( \overrightarrow{b_2} \) be the vectors parallel to the given lines.
Then \( \vec{b}_1 = 3 \hat{i} - 2 \hat{j} + 0 \hat{k} \)
And \( \overrightarrow{b_2} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k} \).
Let \( \theta \) be the angle between the given lines. The angle \( \theta \) is also the angle between \( \overrightarrow{b_1} \) and \( \overrightarrow{b_2} \).
\( \cos\theta = \frac{\overrightarrow{b_1} \cdot \overrightarrow{b_2}}{|\overrightarrow{b_1}| |\overrightarrow{b_2}|} \)
\( \implies \cos\theta = \frac{(3 \hat{i}-2 \hat{j}+0 \hat{k}) \cdot (2 \hat{i}+3 \hat{j}+4 \hat{k})}{\sqrt{3^2+(-2)^2+0^2} \sqrt{2^2+3^2+4^2}} \)
\( \implies \cos\theta = \frac{3(2)+(-2)(3)+0(4)}{\sqrt{9+4+0} \sqrt{4+9+16}} \)
\( \implies \cos\theta = \frac{6-6+0}{\sqrt{13} \sqrt{29}} \)
\( \implies \cos\theta = \frac{0}{\sqrt{13} \sqrt{29}} \)
\( \implies \cos\theta = 0 \)
Since \( \cos\theta = 0 \), the angle \( \theta \) is \( \frac{\pi}{2} \) radians, or 90 degrees. This means the lines are perpendicular.
In simple words: For part (i), we find the direction vectors for each line and use a formula to calculate the cosine of the angle between them. Then, we find the angle itself using the inverse cosine. For part (ii), after writing the lines in a clear format, we find their direction vectors. When we calculate the cosine of the angle, we get zero, which means the lines cross at a 90-degree angle, making them perpendicular.

🎯 Exam Tip: When lines are given in different forms (Cartesian or vector), first convert them to a standard form to easily identify their direction ratios or vectors. Remember the dot product formula for finding the angle between them.

Question 2. Find the angle between the following pairs of lines:
(i) \( \vec{r} = 4 \hat{i} – \hat{j} + \lambda(\hat{i} + 2 \hat{j} – 2 \hat{k}) \); \( \vec{r} = \hat{i} – \hat{j} + 2 \hat{k} – \mu(2 \hat{i} + 4 \hat{j} – 4 \hat{k}) \)
(ii) \( \vec{r} = 3 \hat{i} + 2 \hat{j} – 4 \hat{k} + \lambda(\hat{i} + 2 \hat{j} + 2 \hat{k}) \); \( \vec{r} = 5 \hat{i} – 2 \hat{j} + \mu(3 \hat{i} + 2 \hat{j} + 6 \hat{k}) \)
(iii) \( \vec{r} = \lambda(\hat{i} + \hat{j} + 2 \hat{k}) \); \( \vec{r} = 2 \hat{j} + \mu[(\sqrt{3} – 1) \hat{i} – (\sqrt{3} + 1) \hat{j} + 4 \hat{k}] \)
Answer:
(i) Given equations of lines are:
\( \vec{r} = (4 \hat{i} – \hat{j}) + \lambda(\hat{i} + 2 \hat{j} – 2 \hat{k}) \)
\( \vec{r} = (\hat{i} – \hat{j} + 2 \hat{k}) – \mu(2 \hat{i} + 4 \hat{j} – 4 \hat{k}) \)
Let \( \overrightarrow{m_1} \) and \( \overrightarrow{m_2} \) be vectors parallel to line (1) and (2).
So, \( \overrightarrow{m_1} = \hat{i} + 2 \hat{j} – 2 \hat{k} \)
And \( \overrightarrow{m_2} = 2 \hat{i} + 4 \hat{j} – 4 \hat{k} \).
Let \( \theta \) be the angle between the given lines, which is the same as the angle between \( \overrightarrow{m_1} \) and \( \overrightarrow{m_2} \).
\( \cos\theta = \frac{\overrightarrow{m_1} \cdot \overrightarrow{m_2}}{|\overrightarrow{m_1}| |\overrightarrow{m_2}|} \)
\( \implies \cos\theta = \frac{(\hat{i}+2 \hat{j}-2 \hat{k}) \cdot (2 \hat{i}+4 \hat{j}-4 \hat{k})}{\sqrt{1^2+2^2+(-2)^2} \sqrt{2^2+4^2+(-4)^2}} \)
\( \implies \cos\theta = \frac{1(2)+2(4)-2(-4)}{\sqrt{1+4+4} \sqrt{4+16+16}} \)
\( \implies \cos\theta = \frac{2+8+8}{\sqrt{9} \sqrt{36}} \)
\( \implies \cos\theta = \frac{18}{3 \times 6} \)
\( \implies \cos\theta = \frac{18}{18} \)
\( \implies \cos\theta = 1 \)
When \( \cos\theta = 1 \), the angle \( \theta \) is 0°. This means the given pair of lines are parallel to each other.
(ii) The given pair of lines are:
\( \vec{r} = 3 \hat{i} + 2 \hat{j} – 4 \hat{k} + \lambda(\hat{i} + 2 \hat{j} + 2 \hat{k}) \)
\( \vec{r} = 5 \hat{i} – 2 \hat{j} + \mu(3 \hat{i} + 2 \hat{j} + 6 \hat{k}) \)
Let \( \vec{b} \) be the direction vector for the first line and \( \vec{d} \) for the second line.
Here, \( \vec{b} = \hat{i} + 2 \hat{j} + 2 \hat{k} \).
The magnitude of \( \vec{b} \) is \( |\vec{b}| = \sqrt{1^2+2^2+2^2} = \sqrt{1+4+4} = \sqrt{9} = 3 \).
And \( \vec{d} = 3 \hat{i} + 2 \hat{j} + 6 \hat{k} \).
The magnitude of \( \vec{d} \) is \( |\vec{d}| = \sqrt{3^2+2^2+6^2} = \sqrt{9+4+36} = \sqrt{49} = 7 \).
Now, calculate the dot product \( \vec{b} \cdot \vec{d} \):
\( \vec{b} \cdot \vec{d} = (1)(3) + (2)(2) + (2)(6) = 3 + 4 + 12 = 19 \).
If \( \theta \) is the required angle, then:
\( \cos\theta = \frac{\vec{b} \cdot \vec{d}}{|\vec{b}| |\vec{d}|} \)
\( \implies \cos\theta = \frac{19}{3 \times 7} = \frac{19}{21} \).
So, \( \theta = \cos^{-1} \left( \frac{19}{21} \right) \). This gives the specific angle between these two lines.
(iii) Given lines are:
\( \vec{r} = \lambda(\hat{i} + \hat{j} + 2 \hat{k}) \)
\( \vec{r} = 2 \hat{j} + \mu[(\sqrt{3} – 1) \hat{i} – (\sqrt{3} + 1) \hat{j} + 4 \hat{k}] \)
Let \( \vec{b}_1 \) and \( \overrightarrow{b_2} \) be the vectors parallel to line (1) and line (2).
So, \( \vec{b}_1 = \hat{i} + \hat{j} + 2 \hat{k} \)
And \( \overrightarrow{b_2} = (\sqrt{3}-1) \hat{i} – (\sqrt{3}+1) \hat{j} + 4 \hat{k} \).
Let \( \theta \) be the angle between the given lines.
\( \cos\theta = \frac{\overrightarrow{b_1} \cdot \overrightarrow{b_2}}{|\overrightarrow{b_1}| |\overrightarrow{b_2}|} \)
\( \implies \cos\theta = \frac{(\hat{i}+\hat{j}+2 \hat{k}) \cdot [(\sqrt{3}-1) \hat{i}-(\sqrt{3}+1) \hat{j}+4 \hat{k}]}{\sqrt{1^2+1^2+2^2} \sqrt{(\sqrt{3}-1)^2+(\sqrt{3}+1)^2+4^2}} \)
\( \implies \cos\theta = \frac{1(\sqrt{3}-1) + 1(-(\sqrt{3}+1)) + 2(4)}{\sqrt{1+1+4} \sqrt{(3+1-2\sqrt{3})+(3+1+2\sqrt{3})+16}} \)
\( \implies \cos\theta = \frac{\sqrt{3}-1-\sqrt{3}-1+8}{\sqrt{6} \sqrt{4-2\sqrt{3}+4+2\sqrt{3}+16}} \)
\( \implies \cos\theta = \frac{6}{\sqrt{6} \sqrt{24}} \)
\( \implies \cos\theta = \frac{6}{\sqrt{144}} \)
\( \implies \cos\theta = \frac{6}{12} \)
\( \implies \cos\theta = \frac{1}{2} \)
When \( \cos\theta = \frac{1}{2} \), the angle \( \theta \) is \( \frac{\pi}{3} \) radians, or 60 degrees. This angle tells us how much the lines are tilted with respect to each other.
In simple words: For all three parts, we find the special vectors that show the direction of each line. Then, we use a formula involving these vectors to calculate the cosine of the angle between the lines. Finally, we use the inverse cosine to find the actual angle. If the cosine is 1, lines are parallel; if it's 0.5, they meet at 60 degrees.

🎯 Exam Tip: Remember to simplify the vector equations to their basic direction vectors before applying the angle formula. Pay close attention to signs and magnitudes in calculations to avoid errors.

Question 3. Find the angle between the pairs of lines with direction ratios:
(i) 2, 2, 1 and 4, 1, 8
(ii) 1, 2, -2 and -2, 2, 1.
Answer:
(i) Given direction ratios are \( (a_1, b_1, c_1) = (2, 2, 1) \) and \( (a_2, b_2, c_2) = (4, 1, 8) \).
Let \( \theta \) be the angle between the given pair of lines. The formula for the cosine of the angle between two lines with direction ratios is:
\( \cos\theta = \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}} \)
\( \implies \cos\theta = \frac{2(4)+2(1)+1(8)}{\sqrt{2^2+2^2+1^2} \sqrt{4^2+1^2+8^2}} \)
\( \implies \cos\theta = \frac{8+2+8}{\sqrt{4+4+1} \sqrt{16+1+64}} \)
\( \implies \cos\theta = \frac{18}{\sqrt{9} \sqrt{81}} \)
\( \implies \cos\theta = \frac{18}{3 \times 9} \)
\( \implies \cos\theta = \frac{18}{27} \)
\( \implies \cos\theta = \frac{2}{3} \)
So, \( \theta = \cos^{-1} \left( \frac{2}{3} \right) \). This angle specifies the degree of separation between the two lines.
(ii) Given direction ratios are \( (a_1, b_1, c_1) = (1, 2, -2) \) and \( (a_2, b_2, c_2) = (-2, 2, 1) \).
Let \( \theta \) be the angle between the given pair of lines.
Using the same formula:
\( \cos\theta = \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}} \)
\( \implies \cos\theta = \frac{1(-2)+2(2)+(-2)(1)}{\sqrt{1^2+2^2+(-2)^2} \sqrt{(-2)^2+2^2+1^2}} \)
\( \implies \cos\theta = \frac{-2+4-2}{\sqrt{1+4+4} \sqrt{4+4+1}} \)
\( \implies \cos\theta = \frac{0}{\sqrt{9} \sqrt{9}} \)
\( \implies \cos\theta = 0 \)
Since \( \cos\theta = 0 \), the angle \( \theta \) is \( \frac{\pi}{2} \) radians, or 90 degrees. This indicates that the lines are perpendicular to each other.
In simple words: To find the angle between lines using direction ratios, we use a formula that takes the dot product of the ratios and divides it by the product of their magnitudes. If the result is 2/3, we find the angle whose cosine is 2/3. If the result is 0, it means the lines are at a perfect right angle (90 degrees).

🎯 Exam Tip: When dealing with direction ratios, be careful with signs. If the dot product of direction ratios is zero, the lines are always perpendicular.

Question 4. Prove that the two lines \( \frac{x}{1} = \frac{y}{2} = \frac{z}{1} \) and \( \frac{x}{1} = \frac{y}{-1} = \frac{z}{1} \) are at right angles.
Answer:
Given lines are:
(1) \( \frac{x}{1} = \frac{y}{2} = \frac{z}{1} \)
(2) \( \frac{x}{1} = \frac{y}{-1} = \frac{z}{1} \)
The direction ratios of line (1) are \( (a_1, b_1, c_1) = (1, 2, 1) \).
The direction ratios of line (2) are \( (a_2, b_2, c_2) = (1, -1, 1) \).
To check if two lines are at right angles (perpendicular), we verify if the dot product of their direction ratios is zero:
\( a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 \)
Substitute the values:
\( (1)(1) + (2)(-1) + (1)(1) = 0 \)
\( 1 - 2 + 1 = 0 \)
\( 0 = 0 \)
Since the sum is 0, both lines (1) and (2) are at right angles to each other. This is a property of perpendicular lines in 3D space.
In simple words: To show that two lines meet at a right angle, we look at their direction numbers. If we multiply the direction numbers for each line and add them up, and the answer is zero, then the lines are definitely perpendicular.

🎯 Exam Tip: Remember that for perpendicular lines, the sum of the products of their corresponding direction ratios must be zero. This is a quick test for perpendicularity.

Question 5. P, Q, R, S are the points (-2, 3, 4), (-4, 4, 6), (4, 3, 5) and (0, 1, 2). Show that P Q is perpendicular to R S.
Answer:
Given points are P(-2, 3, 4), Q(-4, 4, 6), R(4, 3, 5) and S(0, 1, 2).
First, find the direction ratios of the line segment PQ.
Direction ratios of PQ are \( \langle x_2-x_1, y_2-y_1, z_2-z_1 \rangle \):
\( \langle -4 - (-2), 4 - 3, 6 - 4 \rangle \)
\( \langle -4 + 2, 1, 2 \rangle \)
So, the direction ratios of PQ are \( \langle -2, 1, 2 \rangle \).
Next, find the direction ratios of the line segment RS.
Direction ratios of RS are \( \langle x_2-x_1, y_2-y_1, z_2-z_1 \rangle \):
\( \langle 0 - 4, 1 - 3, 2 - 5 \rangle \)
So, the direction ratios of RS are \( \langle -4, -2, -3 \rangle \).
To show that PQ is perpendicular to RS, we check if the dot product of their direction ratios is zero:
\( a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 \)
Substitute the direction ratios:
\( (-2)(-4) + (1)(-2) + (2)(-3) = 0 \)
\( 8 - 2 - 6 = 0 \)
\( 0 = 0 \)
Since the sum of the products is zero, line PQ is perpendicular to line RS. This proves that the two line segments form a right angle where they would intersect.
In simple words: We find the direction numbers for the line PQ and the line RS by subtracting the coordinates of their end points. Then, we multiply the matching direction numbers together and add them up. If the total is zero, it means the lines are perpendicular, or form a right angle.

🎯 Exam Tip: Always calculate direction ratios carefully, ensuring the subtraction order is consistent. The perpendicularity test is a straightforward dot product check.

Question 6.
(i) Determine the value of k so that the line joining the points A(k, -1, -1), B(2, 0, 2 k) is perpendicular to the line joining the points C(4, 2 k, 1) and D(2, 3, 2).
(ii) For what value of p will the line through (4, 1, 2) and (5, p, 0) be perpendicular to the line through (2, 1, 1) and (3, 3, -1) ?
Answer:
(i) The direction ratios of the line joining points A(k, -1, -1) and B(2, 0, 2k) are \( \langle x_2-x_1, y_2-y_1, z_2-z_1 \rangle \):
\( \langle 2 - k, 0 - (-1), 2k - (-1) \rangle \)
So, the direction ratios of AB are \( \langle 2 - k, 1, 2k + 1 \rangle \). Let these be \( (a_1, b_1, c_1) \).
The direction ratios of the line joining points C(4, 2k, 1) and D(2, 3, 2) are:
\( \langle 2 - 4, 3 - 2k, 2 - 1 \rangle \)
So, the direction ratios of CD are \( \langle -2, 3 - 2k, 1 \rangle \). Let these be \( (a_2, b_2, c_2) \).
Since line AB is perpendicular to line CD, the dot product of their direction ratios must be zero:
\( a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 \)
\( \implies (2 - k)(-2) + (1)(3 - 2k) + (2k + 1)(1) = 0 \)
\( \implies -4 + 2k + 3 - 2k + 2k + 1 = 0 \)
\( \implies 2k + 0 = 0 \)
\( \implies 2k = 0 \)
\( \implies k = 0 \). The value of k must be 0 for the lines to be perpendicular.
(ii) The direction ratios of the line through (4, 1, 2) and (5, p, 0) are:
\( \langle 5 - 4, p - 1, 0 - 2 \rangle \)
So, the direction ratios are \( \langle 1, p - 1, -2 \rangle \). Let these be \( (a_1, b_1, c_1) \).
The direction ratios of the line through (2, 1, 1) and (3, 3, -1) are:
\( \langle 3 - 2, 3 - 1, -1 - 1 \rangle \)
So, the direction ratios are \( \langle 1, 2, -2 \rangle \). Let these be \( (a_2, b_2, c_2) \).
Since line AB is perpendicular to line CD, the dot product of their direction ratios must be zero:
\( a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 \)
\( \implies (1)(1) + (p - 1)(2) + (-2)(-2) = 0 \)
\( \implies 1 + 2p - 2 + 4 = 0 \)
\( \implies 2p + 3 = 0 \)
\( \implies 2p = -3 \)
\( \implies p = -\frac{3}{2} \). So, for the lines to be perpendicular, the value of p must be \( -\frac{3}{2} \).
In simple words: For both parts, we first find the direction numbers for each line segment by subtracting the coordinates. Because the lines are perpendicular, we know that if we multiply the matching direction numbers and add them up, the total must be zero. We then solve this equation to find the unknown value (k or p).

🎯 Exam Tip: When finding direction ratios between two points \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \), they are \( (x_2-x_1, y_2-y_1, z_2-z_1) \). Apply the perpendicularity condition \( a_1a_2 + b_1b_2 + c_1c_2 = 0 \) carefully, especially with negative numbers and variables.

Question 7. Find the value of \( \lambda \), so that the following lines are perpendicular to each other :
(i) \( \frac{x-5}{5 \lambda+2} = \frac{2-y}{5} = \frac{1-z}{-1} \); \( \frac{x}{1} = \frac{2 y+1}{4 \lambda} = \frac{1-z}{-3} \)
(ii) \( \frac{1-x}{3} = \frac{7 y-14}{2 \lambda} = \frac{5 z-10}{11} \); \( \frac{7-7 x}{3 \lambda} = \frac{y-5}{1} = \frac{6-z}{5} \)
Answer:
(i) First, rewrite the given lines in standard symmetric form \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \).
Line 1: \( \frac{x-5}{5 \lambda+2} = \frac{-(y-2)}{5} = \frac{-(z-1)}{-1} \)
\( \implies \frac{x-5}{5 \lambda+2} = \frac{y-2}{-5} = \frac{z-1}{1} \)
The direction ratios of line (1) are proportional to \( \langle 5 \lambda + 2, -5, 1 \rangle \). So, \( a_1 = 5 \lambda + 2, b_1 = -5, c_1 = 1 \).
Line 2: \( \frac{x}{1} = \frac{2(y+\frac{1}{2})}{4 \lambda} = \frac{-(z-1)}{-3} \)
\( \implies \frac{x}{1} = \frac{y+\frac{1}{2}}{2 \lambda} = \frac{z-1}{3} \)
The direction ratios of line (2) are proportional to \( \langle 1, 2 \lambda, 3 \rangle \). So, \( a_2 = 1, b_2 = 2 \lambda, c_2 = 3 \).
Since lines (1) and (2) are perpendicular to each other, the dot product of their direction ratios must be zero:
\( a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 \)
\( \implies (5 \lambda + 2)(1) + (-5)(2 \lambda) + (1)(3) = 0 \)
\( \implies 5 \lambda + 2 - 10 \lambda + 3 = 0 \)
\( \implies -5 \lambda + 5 = 0 \)
\( \implies 5 \lambda = 5 \)
\( \implies \lambda = 1 \). Thus, \( \lambda \) must be 1 for the lines to be perpendicular.
(ii) Rewrite the given lines in standard symmetric form.
Line 1: \( \frac{-(x-1)}{3} = \frac{7(y-2)}{2 \lambda} = \frac{5(z-2)}{11} \)
\( \implies \frac{x-1}{-3} = \frac{y-2}{\frac{2 \lambda}{7}} = \frac{z-2}{\frac{11}{5}} \)
The direction ratios of line (1) are proportional to \( \langle -3, \frac{2 \lambda}{7}, \frac{11}{5} \rangle \). So, \( a_1 = -3, b_1 = \frac{2 \lambda}{7}, c_1 = \frac{11}{5} \).
Line 2: \( \frac{-7(x-1)}{3 \lambda} = \frac{y-5}{1} = \frac{-(z-6)}{5} \)
\( \implies \frac{x-1}{-\frac{3 \lambda}{7}} = \frac{y-5}{1} = \frac{z-6}{-5} \)
The direction ratios of line (2) are proportional to \( \langle -\frac{3 \lambda}{7}, 1, -5 \rangle \). So, \( a_2 = -\frac{3 \lambda}{7}, b_2 = 1, c_2 = -5 \).
Since lines (1) and (2) are perpendicular to each other:
\( a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 \)
\( \implies (-3)(-\frac{3 \lambda}{7}) + (\frac{2 \lambda}{7})(1) + (\frac{11}{5})(-5) = 0 \)
\( \implies \frac{9 \lambda}{7} + \frac{2 \lambda}{7} - 11 = 0 \)
\( \implies \frac{11 \lambda}{7} - 11 = 0 \)
\( \implies \frac{11 \lambda}{7} = 11 \)
\( \implies 11 \lambda = 77 \)
\( \implies \lambda = 7 \). This value of \( \lambda \) ensures that the lines are perpendicular.
In simple words: For both parts, the first step is to rewrite the line equations into a standard form to clearly see their direction numbers. Since the lines are perpendicular, we use the rule that the sum of the products of their matching direction numbers must be zero. We then solve this equation to find the value of \( \lambda \).

🎯 Exam Tip: Always make sure the line equations are in the standard symmetric form (e.g., \( (x-x_1)/a \)) before extracting direction ratios. Watch out for negative signs and fractions when rewriting the equations.

Question 8. Find the values of p and q so that the line joining the points (7, p, 2) and (q, -2, 5) may be parallel to the line joining the points (2, -3, 5) and (-6, -15, 11).
Answer:
Let A(7, p, 2) and B(q, -2, 5) be the points for the first line.
The direction ratios of line AB are \( \langle q-7, -2-p, 5-2 \rangle \), which simplifies to \( \langle q-7, -2-p, 3 \rangle \).
Let C(2, -3, 5) and D(-6, -15, 11) be the points for the second line.
The direction ratios of line CD are \( \langle -6-2, -15-(-3), 11-5 \rangle \), which simplifies to \( \langle -8, -12, 6 \rangle \).
These direction ratios can be simplified by dividing by 2: \( \langle -4, -6, 3 \rangle \).
Since line AB is parallel to line CD, their direction ratios must be proportional.
This means \( \frac{q-7}{-4} = \frac{-2-p}{-6} = \frac{3}{3} \).
From the last part, \( \frac{3}{3} = 1 \).
So we have two equations:
1) \( \frac{q-7}{-4} = 1 \)
\( \implies q-7 = -4 \)
\( \implies q = 7 - 4 \)
\( \implies q = 3 \)
2) \( \frac{-2-p}{-6} = 1 \)
\( \implies -2-p = -6 \)
\( \implies p = -2 + 6 \)
\( \implies p = 4 \)
So, the values are \( p = 4 \) and \( q = 3 \). These values ensure that the two lines run in the same direction without intersecting.
In simple words: First, we find the direction numbers for both lines. Since the lines are parallel, their direction numbers must be in the same proportion. We set up equations by comparing the ratios and then solve them to find the unknown values of 'p' and 'q'.

🎯 Exam Tip: For parallel lines, the direction ratios are proportional. Remember to set up two separate equations if there are two unknown variables, and solve them carefully.

Question 9. Find the equations of the line passing through the point (2, 1, 3) and perpendicular to the lines \( \frac{x-1}{1} = \frac{y-2}{2} = \frac{z-3}{3} \) and \( \frac{x}{-3} = \frac{y}{2} = \frac{z}{5} \).
Answer:
Let the direction ratios of the required line be \( \langle a, b, c \rangle \).
The first given line has direction ratios \( \langle 1, 2, 3 \rangle \).
The second given line has direction ratios \( \langle -3, 2, 5 \rangle \).
Since the required line is perpendicular to both given lines, its direction vector \( \langle a, b, c \rangle \) must be perpendicular to the direction vectors of the other two lines. This means their dot products are zero:
1) \( a(1) + b(2) + c(3) = 0 \implies a + 2b + 3c = 0 \)
2) \( a(-3) + b(2) + c(5) = 0 \implies -3a + 2b + 5c = 0 \)
We can solve these two equations for \( a, b, c \) using the cross-multiplication method:
\( \frac{a}{2 \times 5 - 3 \times 2} = \frac{b}{3 \times (-3) - 5 \times 1} = \frac{c}{1 \times 2 - 2 \times (-3)} \)
\( \frac{a}{10 - 6} = \frac{b}{-9 - 5} = \frac{c}{2 - (-6)} \)
\( \frac{a}{4} = \frac{b}{-14} = \frac{c}{8} \)
We can simplify these ratios by dividing by 2:
\( \frac{a}{2} = \frac{b}{-7} = \frac{c}{4} \)
So, the direction ratios of the required line are proportional to \( \langle 2, -7, 4 \rangle \).
The required line passes through the point (2, 1, 3) and has direction ratios \( \langle 2, -7, 4 \rangle \).
The equation of the line in symmetric form is:
\( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \)
\( \implies \frac{x-2}{2} = \frac{y-1}{-7} = \frac{z-3}{4} \). This is the equation of the line that is perpendicular to both given lines.
In simple words: We want to find a line that is "crosswise" to two other lines. First, we get the direction numbers for the two given lines. Then, we use a special method called cross-multiplication to find the direction numbers of our new line, which must be perpendicular to both. Finally, we use these new direction numbers and the given point (2,1,3) to write the equation for our line.

🎯 Exam Tip: When a line is perpendicular to two other lines, its direction vector is proportional to the cross product of their direction vectors. The cross-multiplication method is a systematic way to find these proportional direction ratios.

Question 10. Find the equations of the line passing through the point A(-1, 3, 7) and perpendicular to the lines \( \vec{r} = 2 \hat{i} – 3 \hat{j} + \lambda(2\hat{i} + 3 \hat{j} + \hat{k}) \) and \( \vec{r} = \hat{i} – \hat{j} + \hat{k} + \mu(7 \hat{j} – 5 \hat{k}) \).
Answer:
Let the required equation of the line pass through point A(-1, 3, 7) and have direction ratios \( \langle a, b, c \rangle \).
The general symmetric form of this line would be \( \frac{x-(-1)}{a} = \frac{y-3}{b} = \frac{z-7}{c} \), or \( \frac{x+1}{a} = \frac{y-3}{b} = \frac{z-7}{c} \).
Now, let's find the direction ratios of the given lines from their vector forms:
Line 1: \( \vec{r} = 2 \hat{i} – 3 \hat{j} + \lambda(2\hat{i} + 3 \hat{j} + \hat{k}) \). Its direction vector is \( \overrightarrow{b_1} = 2\hat{i} + 3 \hat{j} + \hat{k} \). So, its direction ratios are \( \langle 2, 3, 1 \rangle \).
Line 2: \( \vec{r} = \hat{i} – \hat{j} + \hat{k} + \mu(7 \hat{j} – 5 \hat{k}) \). Its direction vector is \( \overrightarrow{b_2} = 0\hat{i} + 7 \hat{j} – 5 \hat{k} \). So, its direction ratios are \( \langle 0, 7, -5 \rangle \).
Since the required line (with direction ratios \( \langle a, b, c \rangle \)) is perpendicular to both given lines, we have:
1) \( 2a + 3b + c = 0 \)
2) \( 0a + 7b - 5c = 0 \)
We can solve these two equations for \( a, b, c \) using the cross-multiplication method:
\( \frac{a}{(3)(-5) - (1)(7)} = \frac{b}{(1)(0) - (-5)(2)} = \frac{c}{(2)(7) - (3)(0)} \)
\( \frac{a}{-15 - 7} = \frac{b}{0 - (-10)} = \frac{c}{14 - 0} \)
\( \frac{a}{-22} = \frac{b}{10} = \frac{c}{14} \)
We can simplify these ratios by dividing by 2:
\( \frac{a}{-11} = \frac{b}{5} = \frac{c}{7} \)
So, the direction ratios of the required line are proportional to \( \langle -11, 5, 7 \rangle \).
The equation of the line passing through A(-1, 3, 7) with these direction ratios is:
\( \frac{x-(-1)}{-11} = \frac{y-3}{5} = \frac{z-7}{7} \)
\( \implies \frac{x+1}{-11} = \frac{y-3}{5} = \frac{z-7}{7} \). This line is oriented perpendicular to the two original lines.
In simple words: First, we find the direction numbers for the two lines given in vector form. Since our new line needs to be at right angles to both of these, its direction numbers can be found using a special cross-multiplication method from the equations of perpendicularity. Finally, we use these new direction numbers and the point A(-1,3,7) to write the full equation of our line.

🎯 Exam Tip: When given lines in vector form, always correctly identify their direction vectors. Pay close attention to components that are zero, like \( 0\hat{i} \), to avoid errors in the cross-multiplication method.

Question 11. Find the equations of the line passing through the point (1, -1, 1) and perpendicular to the lines joining the points (4, 3, 2),(1, -1, 0) and (1, 2, -1),(2, 1, 1).
Answer:
Let the direction ratios of the required line be proportional to \( \langle a, b, c \rangle \).
The first given line connects points (4, 3, 2) and (1, -1, 0). Its direction ratios are:
\( \langle 1-4, -1-3, 0-2 \rangle = \langle -3, -4, -2 \rangle \).
The second given line connects points (1, 2, -1) and (2, 1, 1). Its direction ratios are:
\( \langle 2-1, 1-2, 1-(-1) \rangle = \langle 1, -1, 2 \rangle \).
Since the required line is perpendicular to both given lines, its direction ratios \( \langle a, b, c \rangle \) must satisfy:
1) \( -3a - 4b - 2c = 0 \)
2) \( 1a - 1b + 2c = 0 \)
We can also write the first equation as \( 3a + 4b + 2c = 0 \).
Using the cross-multiplication method to solve for \( a, b, c \):
\( \frac{a}{(4)(2) - (2)(-1)} = \frac{b}{(2)(1) - (2)(3)} = \frac{c}{(3)(-1) - (4)(1)} \)
\( \frac{a}{8 - (-2)} = \frac{b}{2 - 6} = \frac{c}{-3 - 4} \)
\( \frac{a}{10} = \frac{b}{-4} = \frac{c}{-7} \)
So, the direction ratios of the required line are proportional to \( \langle 10, -4, -7 \rangle \).
The required line passes through the point (1, -1, 1) and has these direction ratios.
The equation of the line in symmetric form is:
\( \frac{x-1}{10} = \frac{y-(-1)}{-4} = \frac{z-1}{-7} \)
\( \implies \frac{x-1}{10} = \frac{y+1}{-4} = \frac{z-1}{-7} \). This line represents the path that is at right angles to both the specified line segments.
In simple words: We first find the direction numbers for the two lines given by pairs of points. Our new line needs to be perpendicular to both of these. We use a mathematical technique (cross-multiplication) to find the direction numbers for this perpendicular line. Finally, we use these direction numbers and the point (1, -1, 1) to write down the equation of the line.

🎯 Exam Tip: Always double-check your subtraction when calculating direction ratios from two points. The cross-multiplication method is reliable but requires careful execution of determinant-like calculations.

Question 12. Find the coordinates of the foot of the perpendicular from (1, 1, 1) on the line joining (5, 4, 4) and (1, 4, 6).
Answer:
Let P be the point (1, 1, 1). Let A be (5, 4, 4) and B be (1, 4, 6).
First, find the equation of the line AB. Its direction ratios are \( \langle 1-5, 4-4, 6-4 \rangle = \langle -4, 0, 2 \rangle \).
The equation of line AB passing through (5, 4, 4) with direction ratios \( \langle -4, 0, 2 \rangle \) is:
\( \frac{x-5}{-4} = \frac{y-4}{0} = \frac{z-4}{2} = t \) (say)
Any point M on line AB can be represented as \( (-4t + 5, 0t + 4, 2t + 4) \), which is \( (-4t + 5, 4, 2t + 4) \).
Let M be the foot of the perpendicular drawn from P(1, 1, 1) to the line AB.
The direction ratios of PM are \( \langle (-4t + 5) - 1, 4 - 1, (2t + 4) - 1 \rangle \), which simplifies to \( \langle -4t + 4, 3, 2t + 3 \rangle \).
Since PM is perpendicular to line AB, the dot product of their direction ratios must be zero:
\( (-4t + 4)(-4) + (3)(0) + (2t + 3)(2) = 0 \)
\( \implies 16t - 16 + 0 + 4t + 6 = 0 \)
\( \implies 20t - 10 = 0 \)
\( \implies 20t = 10 \)
\( \implies t = \frac{10}{20} = \frac{1}{2} \).
Now substitute \( t = \frac{1}{2} \) into the coordinates of point M:
\( M_x = -4(\frac{1}{2}) + 5 = -2 + 5 = 3 \)
\( M_y = 4 \)
\( M_z = 2(\frac{1}{2}) + 4 = 1 + 4 = 5 \)
Thus, the coordinates of the foot of the perpendicular M are (3, 4, 5). This point is the closest point on the line to P.

In simple words: First, we write the equation of the line AB. Any point on this line can be written using a variable 't'. We then form a line segment PM from the given point P to any point M on line AB. Since PM is perpendicular to AB, we set the dot product of their direction numbers to zero. Solving this equation gives us the value of 't', which we then use to find the exact coordinates of point M, which is the foot of the perpendicular.

🎯 Exam Tip: When finding the foot of the perpendicular, remember to express a general point on the line in terms of a parameter (t or \( \lambda \)). The key is that the line segment from the external point to the foot of the perpendicular is orthogonal to the given line.

Question 13. Find the foot of the perpendicular from (0, 2, 7) on the line \( \frac{x+2}{-1} = \frac{y-1}{3} = \frac{z-3}{-2} \).
Answer:
Let P be the point (0, 2, 7). The given line is \( \frac{x+2}{-1} = \frac{y-1}{3} = \frac{z-3}{-2} \).
Let L be the foot of the perpendicular drawn from P onto the given line. Any point L on the given line can be written in terms of a parameter 't':
\( \frac{x+2}{-1} = \frac{y-1}{3} = \frac{z-3}{-2} = t \)
So, \( x = -t - 2 \), \( y = 3t + 1 \), \( z = -2t + 3 \).
The coordinates of point L are \( (-t - 2, 3t + 1, -2t + 3) \).
Now, find the direction ratios of the line segment PL:
\( \langle (-t - 2) - 0, (3t + 1) - 2, (-2t + 3) - 7 \rangle \)
\( \implies \langle -t - 2, 3t - 1, -2t - 4 \rangle \).
The direction ratios of the given line are \( \langle -1, 3, -2 \rangle \).
Since line PL is perpendicular to the given line, the dot product of their direction ratios must be zero:
\( (-t - 2)(-1) + (3t - 1)(3) + (-2t - 4)(-2) = 0 \)
\( \implies t + 2 + 9t - 3 + 4t + 8 = 0 \)
\( \implies 14t + 7 = 0 \)
\( \implies 14t = -7 \)
\( \implies t = -\frac{7}{14} = -\frac{1}{2} \).
Substitute \( t = -\frac{1}{2} \) back into the coordinates of point L:
\( L_x = -(-\frac{1}{2}) - 2 = \frac{1}{2} - 2 = -\frac{3}{2} \)
\( L_y = 3(-\frac{1}{2}) + 1 = -\frac{3}{2} + 1 = -\frac{1}{2} \)
\( L_z = -2(-\frac{1}{2}) + 3 = 1 + 3 = 4 \)
Thus, the coordinates of the foot of the perpendicular L are \( (-\frac{3}{2}, -\frac{1}{2}, 4) \). This point is the projection of P onto the line.

In simple words: We treat any point on the given line using a variable 't'. We then find the direction numbers of a line connecting our point P to this general point L. Since this line PL must be perpendicular to the original line, we use the perpendicularity rule (dot product of direction numbers is zero) to find the exact value of 't'. Finally, we use this 't' to calculate the specific coordinates of L, the foot of the perpendicular.

🎯 Exam Tip: Parameterizing a point on the line is the first critical step. Ensure careful arithmetic, especially with negative numbers and fractions, throughout the calculation of 't' and the final coordinates.

Question 14. A(0, 6, -9), B(-3, -6, 3) and C(7, 4, -1) are three points. Find the equation of the line A B. If D is the foot of perpendicular drawn from C to the line A B, find coordinate of the point D.
Answer:
Given points are A(0, 6, -9), B(-3, -6, 3) and C(7, 4, -1).
First, find the direction ratios of line AB:
\( \langle -3-0, -6-6, 3-(-9) \rangle = \langle -3, -12, 12 \rangle \).
These direction ratios can be simplified by dividing by -3: \( \langle 1, 4, -4 \rangle \).
The equation of line AB passing through A(0, 6, -9) with direction ratios \( \langle 1, 4, -4 \rangle \) is:
\( \frac{x-0}{1} = \frac{y-6}{4} = \frac{z-(-9)}{-4} \)
\( \implies \frac{x}{1} = \frac{y-6}{4} = \frac{z+9}{-4} \).
Let D be the foot of the perpendicular drawn from C(7, 4, -1) to line AB.
Any point D on line AB can be written in terms of a parameter 't':
\( \frac{x}{1} = \frac{y-6}{4} = \frac{z+9}{-4} = t \)
So, \( x = t \), \( y = 4t + 6 \), \( z = -4t - 9 \).
The coordinates of point D are \( (t, 4t + 6, -4t - 9) \).
Now, find the direction ratios of the line segment CD:
\( \langle t-7, (4t+6)-4, (-4t-9)-(-1) \rangle \)
\( \implies \langle t-7, 4t+2, -4t-8 \rangle \).
The direction ratios of line AB are \( \langle 1, 4, -4 \rangle \).
Since CD is perpendicular to AB, the dot product of their direction ratios must be zero:
\( (t-7)(1) + (4t+2)(4) + (-4t-8)(-4) = 0 \)
\( \implies t - 7 + 16t + 8 + 16t + 32 = 0 \)
\( \implies 33t + 33 = 0 \)
\( \implies 33t = -33 \)
\( \implies t = -1 \).
Substitute \( t = -1 \) back into the coordinates of point D:
\( D_x = -1 \)
\( D_y = 4(-1) + 6 = -4 + 6 = 2 \)
\( D_z = -4(-1) - 9 = 4 - 9 = -5 \)
Thus, the coordinates of the foot of the perpendicular D are (-1, 2, -5). This point is the projection of C onto the line AB.

In simple words: First, we find the equation of the line AB. Then, we assume a general point D on line AB using a variable 't'. We create a line segment CD from the given point C to this general point D. Since CD is perpendicular to AB, we use the rule that their direction numbers' dot product is zero. Solving for 't' gives us its value, which we then plug back into the coordinates of D to find its exact location.

🎯 Exam Tip: Ensure that you simplify direction ratios (e.g., from \( \langle -3, -12, 12 \rangle \) to \( \langle 1, 4, -4 \rangle \)) to make calculations easier. This doesn't change the direction of the line. The method of parameterizing a point on the line is standard for this type of problem.

Question 15. Find the distance of (1, 0, 0) from the line through (1, -1, -10) whose d.c's are proportional to 2, -3, 8.
Answer:
Let P be the point (1, 0, 0). The line passes through \( A(1, -1, -10) \) and its direction ratios are \( \langle 2, -3, 8 \rangle \).
The equation of the line is \( \frac{x-1}{2} = \frac{y-(-1)}{-3} = \frac{z-(-10)}{8} \).
\( \implies \frac{x-1}{2} = \frac{y+1}{-3} = \frac{z+10}{8} = t \) (say).
Let L be the foot of the perpendicular from P to the line. Any point L on the line can be written as:
\( x = 2t + 1 \), \( y = -3t - 1 \), \( z = 8t - 10 \).
The coordinates of point L are \( (2t + 1, -3t - 1, 8t - 10) \).
The direction ratios of the line segment PL are \( \langle (2t+1)-1, (-3t-1)-0, (8t-10)-0 \rangle \).
\( \implies \langle 2t, -3t-1, 8t-10 \rangle \).
The direction ratios of the given line are \( \langle 2, -3, 8 \rangle \).
Since PL is perpendicular to the given line, their dot product is zero:
\( (2t)(2) + (-3t-1)(-3) + (8t-10)(8) = 0 \)
\( \implies 4t + 9t + 3 + 64t - 80 = 0 \)
\( \implies 77t - 77 = 0 \)
\( \implies 77t = 77 \)
\( \implies t = 1 \).
Substitute \( t = 1 \) into the coordinates of point L:
\( L_x = 2(1) + 1 = 3 \)
\( L_y = -3(1) - 1 = -4 \)
\( L_z = 8(1) - 10 = -2 \)
The coordinates of the foot of the perpendicular L are (3, -4, -2).
Now, calculate the perpendicular distance |PL| using the distance formula between P(1, 0, 0) and L(3, -4, -2):
\( |PL| = \sqrt{(3-1)^2 + (-4-0)^2 + (-2-0)^2} \)
\( |PL| = \sqrt{2^2 + (-4)^2 + (-2)^2} \)
\( |PL| = \sqrt{4 + 16 + 4} \)
\( |PL| = \sqrt{24} \)
\( |PL| = \sqrt{4 \times 6} \)
\( |PL| = 2\sqrt{6} \) units. This is the shortest distance from point P to the line.

In simple words: To find the distance from a point to a line, we first write the line's equation and represent any point on it using a variable 't'. Then we find the point L on the line that is closest to our given point P by ensuring the line segment PL is perpendicular to the original line. Once we find the coordinates of L, we use the distance formula between P and L to get the shortest distance.

🎯 Exam Tip: This problem involves two main parts: finding the foot of the perpendicular and then calculating the distance. Be careful with calculations, especially when dealing with squares and square roots for the final distance.

Question 16. Find the perpendicular distance of the point (2, 3, 4) from the line \( \frac{4-x}{2} = \frac{y}{6} = \frac{1-z}{3} \). Also find the coordinates of the foot of the perpendicular.
Answer:
Let P be the point (2, 3, 4). The given line needs to be rewritten in standard symmetric form:
\( \frac{-(x-4)}{2} = \frac{y-0}{6} = \frac{-(z-1)}{3} \)
\( \implies \frac{x-4}{-2} = \frac{y}{6} = \frac{z-1}{-3} = t \) (say).
Let M be the foot of the perpendicular from P to this line. Any point M on the line can be written as:
\( x = -2t + 4 \), \( y = 6t \), \( z = -3t + 1 \).
The coordinates of M are \( (-2t + 4, 6t, -3t + 1) \).
Now, find the direction ratios of the line segment PM:
\( \langle (-2t + 4) - 2, (6t) - 3, (-3t + 1) - 4 \rangle \)
\( \implies \langle -2t + 2, 6t - 3, -3t - 3 \rangle \).
The direction ratios of the given line are \( \langle -2, 6, -3 \rangle \).
Since PM is perpendicular to the given line, their dot product must be zero:
\( (-2t + 2)(-2) + (6t - 3)(6) + (-3t - 3)(-3) = 0 \)
\( \implies 4t - 4 + 36t - 18 + 9t + 9 = 0 \)
\( \implies 49t - 13 = 0 \)
\( \implies 49t = 13 \)
\( \implies t = \frac{13}{49} \).
Now substitute \( t = \frac{13}{49} \) into the coordinates of point M to find the foot of the perpendicular:
\( M_x = -2(\frac{13}{49}) + 4 = -\frac{26}{49} + \frac{196}{49} = \frac{170}{49} \)
\( M_y = 6(\frac{13}{49}) = \frac{78}{49} \)
\( M_z = -3(\frac{13}{49}) + 1 = -\frac{39}{49} + \frac{49}{49} = \frac{10}{49} \)
The coordinates of the foot of the perpendicular M are \( (\frac{170}{49}, \frac{78}{49}, \frac{10}{49}) \).
Next, calculate the perpendicular distance |PM| between P(2, 3, 4) and M\( (\frac{170}{49}, \frac{78}{49}, \frac{10}{49}) \):
\( |PM| = \sqrt{(\frac{170}{49}-2)^2 + (\frac{78}{49}-3)^2 + (\frac{10}{49}-4)^2} \)
\( |PM| = \sqrt{(\frac{170-98}{49})^2 + (\frac{78-147}{49})^2 + (\frac{10-196}{49})^2} \)
\( |PM| = \sqrt{(\frac{72}{49})^2 + (\frac{-69}{49})^2 + (\frac{-186}{49})^2} \)
\( |PM| = \frac{1}{49} \sqrt{72^2 + (-69)^2 + (-186)^2} \)
\( |PM| = \frac{1}{49} \sqrt{5184 + 4761 + 34596} \)
\( |PM| = \frac{1}{49} \sqrt{44541} \). (Note: \( \sqrt{44541} = \sqrt{9 \times 4949} = 3\sqrt{4949} \))
\( |PM| = \frac{3\sqrt{4949}}{49} \) units. This distance is the length of the shortest path from the point to the line.

In simple words: First, we write the line's equation in a standard way and express any point M on it using a variable 't'. We then form a line segment PM from the given point P to point M. By using the rule that PM must be perpendicular to the original line, we find the value of 't'. This 't' helps us find the exact coordinates of M (the foot of the perpendicular). Finally, we use the distance formula between P and M to calculate the shortest distance.

🎯 Exam Tip: Always rewrite the line equation into the standard form \( \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} \) correctly, especially when signs are reversed in the numerator (e.g., \( 4-x \) to \( x-4 \)). Fractions in coordinates or final distance calculations are common, so be precise with arithmetic.

Question 17. Find the image of the point (3, 5, 3) in the line \( \frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3} \).
Answer:
Let P be the given point (3, 5, 3). The given line (let's call it AB) is \( \frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3} \).
Let M be the foot of the perpendicular drawn from P to the line AB. Any point M on the line can be written using a parameter 't':
\( \frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3} = t \)
So, \( x = t \), \( y = 2t + 1 \), \( z = 3t + 2 \).
The coordinates of M are \( (t, 2t + 1, 3t + 2) \).
The direction ratios of the line segment PM are:
\( \langle t-3, (2t+1)-5, (3t+2)-3 \rangle \)
\( \implies \langle t-3, 2t-4, 3t-1 \rangle \).
The direction ratios of the line AB are \( \langle 1, 2, 3 \rangle \).
Since PM is perpendicular to AB, the dot product of their direction ratios is zero:
\( (t-3)(1) + (2t-4)(2) + (3t-1)(3) = 0 \)
\( \implies t - 3 + 4t - 8 + 9t - 3 = 0 \)
\( \implies 14t - 14 = 0 \)
\( \implies 14t = 14 \)
\( \implies t = 1 \).
Substitute \( t = 1 \) into the coordinates of M:
\( M_x = 1 \)
\( M_y = 2(1) + 1 = 3 \)
\( M_z = 3(1) + 2 = 5 \)
So, the coordinates of the foot of the perpendicular M are (1, 3, 5).
Now, let P'\( (\alpha, \beta, \gamma) \) be the image of point P(3, 5, 3) in the line AB. M is the midpoint of PP'.
Using the midpoint formula:
\( \frac{3+\alpha}{2} = 1 \implies 3+\alpha = 2 \implies \alpha = -1 \)
\( \frac{5+\beta}{2} = 3 \implies 5+\beta = 6 \implies \beta = 1 \)
\( \frac{3+\gamma}{2} = 5 \implies 3+\gamma = 10 \implies \gamma = 7 \)
Hence, the required coordinates of the image of P(3, 5, 3) in the given line are (-1, 1, 7). The image is a reflection across the line.

In simple words: First, we find the foot of the perpendicular (point M) from the given point P to the line. This point M is the closest point on the line to P. Since the image P' is a reflection of P across the line, M will be exactly in the middle of P and P'. We use the midpoint formula with P and M to find the coordinates of the image point P'.

🎯 Exam Tip: Finding the image of a point requires a two-step process: first find the foot of the perpendicular, and then use the midpoint formula. Be careful with calculations in both steps, especially when substituting values back into equations.

Question 18. Find the image of the point (2, -1, 5) in the line \( \vec{r} = (11 \hat{i} - 2 \hat{j} - 8 \hat{k}) + \lambda(10 \hat{i} - 4 \hat{j} - 11 \hat{k}) \).
Answer: Let P(2, -1, 5) be the given point. The equation of the line in Cartesian form can be written as:
\( \frac{x-11}{10} = \frac{y+2}{-4} = \frac{z+8}{-11} = t \) (say)

Any point M on the line can be written as \( M(10t + 11, -4t - 2, -11t - 8) \). The direction ratios of the line segment PM are given by:
\( < (10t + 11) - 2, (-4t - 2) - (-1), (-11t - 8) - 5 > \)
\( < 10t + 9, -4t - 1, -11t - 13 > \)
The direction ratios of the given line (AB) are \( < 10, -4, -11 > \).
Since PM is perpendicular to AB, their dot product must be zero:
\( (10t + 9)(10) + (-4t - 1)(-4) + (-11t - 13)(-11) = 0 \)
\( 100t + 90 + 16t + 4 + 121t + 143 = 0 \)
\( 237t + 237 = 0 \)
\( \implies 237t = -237 \)
\( \implies t = -1 \)
Substitute \( t = -1 \) into the coordinates of M:
\( M(10(-1) + 11, -4(-1) - 2, -11(-1) - 8) \)
\( M(-10 + 11, 4 - 2, 11 - 8) \)
\( M(1, 2, 3) \)
So, the coordinates of the foot of the perpendicular M are (1, 2, 3).
Let P'\( (\alpha, \beta, \gamma) \) be the image of P(2, -1, 5) in the line. Since M is the midpoint of PP', we can use the midpoint formula:
\( \frac{\alpha+2}{2} = 1 \)
\( \implies \alpha+2 = 2 \)
\( \implies \alpha = 0 \)
\( \frac{\beta-1}{2} = 2 \)
\( \implies \beta-1 = 4 \)
\( \implies \beta = 5 \)
\( \frac{\gamma+5}{2} = 3 \)
\( \implies \gamma+5 = 6 \)
\( \implies \gamma = 1 \)
Therefore, the coordinates of the image P' are (0, 5, 1).
In simple words: To find the image of a point in a line, first, we find the point on the line that is closest to our given point (this is called the foot of the perpendicular). We do this by setting up the line's equation and using the fact that the line connecting the point to the line is perpendicular to the given line. Once we find this foot, the image point is simply the reflection of the original point across this foot.

🎯 Exam Tip: Remember that finding the image of a point involves two main steps: first, finding the foot of the perpendicular from the point to the line (using the perpendicularity condition), and second, using the foot as the midpoint to find the reflected image point. Be careful with calculations for direction ratios and the midpoint formula.