NCERT Solutions For Class 12 Maths Chapter 11 Three Dimensional Geometry Exercise 11.2 - 2025-26

```html 1. Show that the three lines with direction cosines  $\dfrac{12}{13},\dfrac{-3}{13},\dfrac{-4}{13};\dfrac{4}{13},\dfrac{12}{13},\dfrac{3}{13};\dfrac{3}{13},\dfrac{-4}{13},\dfrac{12}{13}$ are mutually perpendicular.
To prove the lines are perpendicular, check if the sum of the products of their direction cosines equals zero for each pair. - For the first pair: $ \dfrac{12}{13}\times \dfrac{4}{13} + \dfrac{-3}{13}\times \dfrac{12}{13} + \dfrac{-4}{13}\times \dfrac{3}{13} = \dfrac{48}{169} - \dfrac{36}{169} - \dfrac{12}{169} = 0 $ So, the first two lines are perpendicular. - For the second pair: $ \dfrac{4}{13}\times \dfrac{3}{13} + \dfrac{12}{13}\times \dfrac{-4}{13} + \dfrac{3}{13}\times \dfrac{12}{13} = \dfrac{12}{169} - \dfrac{48}{169} + \dfrac{36}{169} = 0 $ So, these two are also perpendicular. - For the third pair: $ \dfrac{3}{13}\times \dfrac{12}{13} + \dfrac{-4}{13}\times \dfrac{-3}{13} + \dfrac{12}{13}\times \dfrac{-4}{13} = \dfrac{36}{169} + \dfrac{12}{169} - \dfrac{48}{169} = 0 $ So, the third and first lines are perpendicular. Hence, all three lines are mutually perpendicular.

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).
Find the direction ratios for both lines and check if their dot product is zero. - For line AB (from (1, -1, 2) to (3, 4, -2)), the direction ratios are $2, 5, -4$. - For line CD (from (0, 3, 2) to (3, 5, 6)), the direction ratios are $3, 2, 4$. Their dot product: $(2 \times 3) + (5 \times 2) + (-4 \times 4) = 6 + 10 - 16 = 0$ Since the product is zero, the lines are perpendicular.

3. Show that the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line through the points (-1, -2, 1), (1, 2, 5).
Find direction ratios for both lines and check if they are proportional. - For AB: $(2-4, 3-7, 4-8) = (-2, -4, -4)$ - For CD: $(1-(-1), 2-(-2), 5-1) = (2, 4, 4)$ Now, $\dfrac{-2}{2} = \dfrac{-4}{4} = \dfrac{-4}{4} = -1$ Since all ratios are equal, the lines are parallel.

4. Find the equation of the line which passes through point (1, 2, 3) and is parallel to the vector $3\hat{i}+2\hat{j}-2\hat{k}$.
The required line passes through $(1, 2, 3)$ and has direction $3\hat{i} + 2\hat{j} - 2\hat{k}$. So, the equation is: $\vec{r} = \hat{i} + 2\hat{j} + 3\hat{k} + \lambda(3\hat{i} + 2\hat{j} - 2\hat{k})$

5. Find the equation of the line in vector and in Cartesian form that passes through the point with positive vector $2\hat{i}-\hat{j}+4\hat{k}$ and is in the direction $\hat{i}+2\hat{j}-\hat{k}$.
The line passes through $\vec{a} = 2\hat{i} - \hat{j} + 4\hat{k}$ and is parallel to $\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$. - Vector form: $\vec{r} = 2\hat{i} - \hat{j} + 4\hat{k} + \lambda(\hat{i} + 2\hat{j} - \hat{k})$ - Cartesian form: $\dfrac{x-2}{1} = \dfrac{y+1}{2} = \dfrac{z-4}{-1}$

6. Find the Cartesian equation of the line which passes through the point (-2, 4, -5) and parallel to the line given by $\dfrac{x+3}{3}=\dfrac{y-4}{5}=\dfrac{z+8}{6}$.
The new line passes through $(-2, 4, -5)$ and has direction ratios $3, 5, 6$, same as the given line. So, its equation is: $\dfrac{x+2}{3} = \dfrac{y-4}{5} = \dfrac{z+5}{6}$

7. The Cartesian equation of a line is $\dfrac{x-5}{3}=\dfrac{y+4}{7}=\dfrac{z-6}{2}$. Write its vector form.
The line passes through $(5, -4, 6)$ with direction ratios $3, 7, 2$. So, vector equation is: $\vec{r} = 5\hat{i} - 4\hat{j} + 6\hat{k} + \lambda(3\hat{i} + 7\hat{j} + 2\hat{k})$

8. Find the angle between the following pairs of lines:
(i) $\vec{r}=\left(2\hat{i}-5\hat{j}+1\hat{k}\right)+\lambda\left(3\hat{i}-2\hat{j}+6\hat{k}\right)$ and $\vec{r}=\left(7\hat{i}-6\hat{k}\right)+\mu\left(\hat{i}+2\hat{j}+2\hat{k}\right)$
(ii) $\vec{r}=\left(3\hat{i}+\hat{j}-2\hat{k}\right)+\lambda\left(\hat{i}-\hat{j}-2\hat{k}\right)$ and $\vec{r}=\left(2\hat{i}-\hat{j}-56\hat{k}\right)+\mu\left(3\hat{i}-5\hat{j}-4\hat{k}\right)$
For both, the angle between lines is calculated with the formula $\cos Q = \left| \dfrac{\vec{b}_1 \cdot \vec{b}_2}{|\vec{b}_1| \cdot |\vec{b}_2|} \right|$. - (i) Vectors: $\vec{b}_1 = 3\hat{i} - 2\hat{j} + 6\hat{k}$; $\vec{b}_2 = \hat{i} + 2\hat{j} + 2\hat{k}$ - $|\vec{b}_1| = \sqrt{49} = 7$ - $|\vec{b}_2| = \sqrt{9} = 3$ - $\vec{b}_1 \cdot \vec{b}_2 = 19$ - $\cos Q = \dfrac{19}{21}$ - $Q = \cos^{-1}\left(\dfrac{19}{21}\right)$ - (ii) Vectors: $\vec{b}_1 = \hat{i} - \hat{j} - 2\hat{k}$; $\vec{b}_2 = 3\hat{i} - 5\hat{j} - 4\hat{k}$ - $|\vec{b}_1| = \sqrt{6}$ - $|\vec{b}_2| = 5\sqrt{2}$ - $\vec{b}_1 \cdot \vec{b}_2 = 16$ - $\cos Q = \dfrac{16}{\sqrt{6} \times 5\sqrt{2}}$ - $Q = \cos^{-1}\left(\dfrac{8}{5\sqrt{3}}\right)$

9. Find the angle between following pairs of lines.
(i) $\dfrac{x-2}{2}=\dfrac{y-1}{5}=\dfrac{z+3}{-3}$ and $\dfrac{x+2}{-1}=\dfrac{y-4}{8}=\dfrac{z-5}{4}$
(ii) $\dfrac{x}{2}=\dfrac{y}{2}=\dfrac{z}{1}$ and $\dfrac{x-5}{-4}=\dfrac{y-2}{1}=\dfrac{z-3}{8}$
To find the angle between two lines, use $\cos Q = \left| \dfrac{\vec{b}_1 \cdot \vec{b}_2}{|\vec{b}_1||\vec{b}_2|} \right|$. For (i): - Vectors: $\vec{b}_1 = 2\hat{i} + 5\hat{j} - 3\hat{k}$; $\vec{b}_2 = -\hat{i} + 8\hat{j} + 4\hat{k}$ - $|\vec{b}_1| = \sqrt{38}$ - $|\vec{b}_2| = 9$ - $\vec{b}_1 \cdot \vec{b}_2 = 26$ - $\cos Q = \dfrac{26}{9\sqrt{38}}$ - $Q = \cos^{-1}\left(\dfrac{26}{9\sqrt{38}}\right)$ For (ii): - Vectors: $\vec{b}_1 = 2\hat{i} + 2\hat{j} + \hat{k}$; $\vec{b}_2 = 4\hat{i} + \hat{j} + 8\hat{k}$ - $|\vec{b}_1| = 3$ - $|\vec{b}_2| = 9$ - $\vec{b}_1 \cdot \vec{b}_2 = 18$ - $\cos Q = \dfrac{18}{3 \times 9}$ - $Q = \cos^{-1}\left(\dfrac{2}{3}\right)$

10. Find the values of p so that the lines $\dfrac{1-x}{3}=\dfrac{7y-14}{2p}=\dfrac{z-3}{2}$ and $\dfrac{7-7x}{3p}=\dfrac{y-5}{1}=\dfrac{6-z}{5}$ are at right angles.
To make the lines perpendicular, set the sum of the product of their direction ratios to zero. - First, write direction ratios for both lines: Line 1: $-3,\dfrac{2p}{7},2$ Line 2: $-\dfrac{3p}{7}, 1, -5$ - Their dot product is: $(-3)\left(-\dfrac{3p}{7}\right) + \left(\dfrac{2p}{7}\times 1\right) + (2\times -5) = 0$ - Simplified: $\dfrac{9p}{7} + \dfrac{2p}{7} - 10 = 0$ $11p - 70 = 0$ $p = \dfrac{70}{11}$ So, $p = \dfrac{70}{11}$.

11. Show that the lines $\dfrac{x-5}{7}=\dfrac{y+2}{-5}=\dfrac{z}{1}$ and $\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3}$ are perpendicular to each other.
- Direction ratios: Line 1: $7, -5, 1$ Line 2: $1, 2, 3$ - Dot product: $7\times 1 + (-5)\times 2 + 1\times 3 = 7 - 10 + 3 = 0$ So, the lines are perpendicular.

12. Find the shortest distance between the lines $\vec{r}=\left(\hat{i}+2\hat{j}+\hat{k}\right)+\lambda\left(\hat{i}-\hat{j}+\hat{k}\right)$ and $\vec{r}=\left(2\hat{i}-\hat{j}-\hat{k}\right)+\mu\left(2\hat{i}+\hat{j}+2\hat{k}\right)$ .
Summary: The shortest distance is calculated using the formula $d = \left| \dfrac{(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1)}{|\vec{b}_1 \times \vec{b}_2|} \right|$. - Find position vectors and direction vectors: - $\vec{a}_1 = \hat{i} + 2\hat{j} + \hat{k}$ - $\vec{b}_1 = \hat{i} - \hat{j} + \hat{k}$ - $\vec{a}_2 = 2\hat{i} - \hat{j} - \hat{k}$ - $\vec{b}_2 = 2\hat{i} + \hat{j} + 2\hat{k}$ - Difference: $\vec{a}_2 - \vec{a}_1 = \hat{i} - 3\hat{j} - 2\hat{k}$ - Cross product: $\vec{b}_1 \times \vec{b}_2 = -3\hat{i} + 3\hat{k}$ - Magnitude: $|\vec{b}_1 \times \vec{b}_2| = 3\sqrt{2}$ - Dot product: $(-3) \times 1 + 3 \times (-2) = -3 - 6 = -9$ - Shortest distance: $d = \left|\dfrac{-9}{3\sqrt{2}}\right| = \dfrac{3}{\sqrt{2}}$ So, the shortest distance is $\dfrac{3}{\sqrt{2}}$ units.

13. Find the shortest distance between the lines $\dfrac{x+1}{7}=\dfrac{y+1}{-6}=\dfrac{z+1}{1}$ and $\dfrac{x-3}{1}=\dfrac{y-5}{-2}=\dfrac{z-7}{1}$.
Summary: Use the determinant formula to find the shortest distance between two lines in Cartesian form. - Points: - First line: $(-1, -1, -1)$ with direction ratios $(7, -6, 1)$ - Second line: $(3, 5, 7)$ with direction ratios $(1, -2, 1)$ - Set up determinant: $\left| \begin{matrix} 4 & 6 & 8 \\ 7 & -6 & 1 \\ 1 & -2 & 1 \\ \end{matrix} \right| = -116$ - Denominator: ${\sqrt{116}} = 2\sqrt{29}$ - Distance: $d = \dfrac{| -116 |}{2\sqrt{29}} = \dfrac{58}{\sqrt{29}} = 2\sqrt{29}$ So, the shortest distance is $2\sqrt{29}$ units.

14. Find the shortest distance between the lines whose vector equations are $\vec{r}=\left(\hat{i}+2\hat{j}+3\hat{k}\right)+\lambda\left(\hat{i}-3\hat{j}+2\hat{k}\right)$ and $\vec{r}=\left(4\hat{i}+5\hat{j}+6\hat{k}\right)+\mu\left(2\hat{i}+3\hat{j}+\hat{k}\right)$ .
Summary: Apply the vector shortest distance formula to get the answer. - $\vec{a}_1 = \hat{i} + 2\hat{j} + 3\hat{k}$ - $\vec{b}_1 = \hat{i} - 3\hat{j} + 2\hat{k}$ - $\vec{a}_2 = 4\hat{i} + 5\hat{j} + 6\hat{k}$ - $\vec{b}_2 = 2\hat{i} + 3\hat{j} + \hat{k}$ - Difference: $\vec{a}_2 - \vec{a}_1 = 3\hat{i} + 3\hat{j} + 3\hat{k}$ - Cross product: $\vec{b}_1 \times \vec{b}_2 = -9\hat{i} + 3\hat{j} + 9\hat{k}$ - Magnitude: $3\sqrt{19}$ - Dot product: $(-9) \times 3 + 3 \times 3 + 9 \times 3 = -27 + 9 + 27 = 9$ - Distance: $d = \left| \dfrac{9}{3\sqrt{19}} \right| = \dfrac{3}{\sqrt{19}}$ The answer is $\dfrac{3}{\sqrt{19}}$ units.

15. Find the shortest distance between the lines whose vector equations are $ \vec{r}=(1-t)\hat{i} +(t-2)\hat{j}+(3-2t)\hat{k}$ $ \vec{r}=(s+1)\hat{i} +(2s-1)\hat{j}-(2s+1)\hat{k}$
Summary: First, write both lines in standard form, find difference and direction vectors, then use the vector formula for distance. - Rewriting, - Line 1: $\vec{r} = \hat{i} - 2\hat{j} + 3\hat{k} + t(-\hat{i} + \hat{j} - 2\hat{k})$ - Line 2: $\vec{r} = \hat{i} - \hat{j} - \hat{k} + s(\hat{i} + 2\hat{j} - 2\hat{k})$ - So, - $\vec{a}_1 = \hat{i} - 2\hat{j} + 3\hat{k}$, $\vec{b}_1 = -\hat{i} + \hat{j} - 2\hat{k}$ - $\vec{a}_2 = \hat{i} - \hat{j} - \hat{k}$, $\vec{b}_2 = \hat{i} + 2\hat{j} - 2\hat{k}$ - Difference: $\vec{a}_2 - \vec{a}_1 = \hat{j} - 4\hat{k}$ - Cross product: $\vec{b}_1 \times \vec{b}_2 = 2\hat{i} - 4\hat{j} - 3\hat{k}$ - Magnitude: $\sqrt{4 + 16 + 9} = \sqrt{29}$ - Dot product: $(2\hat{i} - 4\hat{j} - 3\hat{k}) \cdot (\hat{j} - 4\hat{k}) = -4 + 12 = 8$ - Distance: $d = \left| \dfrac{8}{\sqrt{29}} \right| = \dfrac{8}{\sqrt{29}}$

``` --- ```html

Key Points Learned from Class 12 Three Dimensional Geometry Exercise 11.2

• You can use direction ratios to check if lines are parallel or perpendicular.
• Equation of a line in 3D can be written in both vector and Cartesian forms.
• Shortest distance between skew lines is found using the cross product formula.
• Class 12 maths exercise 11.2 includes questions on angles and distances between lines.
• Practicing these NCERT questions helps with CBSE exams and entrance tests.

```