Class 12 Ch 11: Three Dimensional Geometry - Exercise 11.2

Q1

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Show that the three lines with direction cosines $\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}$; $\frac{4}{13}, \frac{12}{13}, \frac{3}{13}$; $\frac{3}{13}, \frac{-4}{13}, \frac{12}{13}$ are mutually perpendicular.

Two lines with direction cosines $l_1, m_1, n_1$ and $l_2, m_2, n_2$ are perpendicular if $l_1l_2 + m_1m_2 + n_1n_2 = 0$.

Lines 1 & 2: $\frac{12}{13}(\frac{4}{13}) + \frac{-3}{13}(\frac{12}{13}) + \frac{-4}{13}(\frac{3}{13}) = \frac{48 - 36 - 12}{169} = 0$.

Lines 2 & 3: $\frac{4}{13}(\frac{3}{13}) + \frac{12}{13}(\frac{-4}{13}) + \frac{3}{13}(\frac{12}{13}) = \frac{12 - 48 + 36}{169} = 0$.

Lines 3 & 1: $\frac{3}{13}(\frac{12}{13}) + \frac{-4}{13}(\frac{-3}{13}) + \frac{12}{13}(\frac{-4}{13}) = \frac{36 + 12 - 48}{169} = 0$.

$\boxed{\text{All pairs are perpendicular}}$

Q2

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Show that the line through the points $(1, -1, 2), (3, 4, -2)$ is perpendicular to the line through the points $(0, 3, 2)$ and $(3, 5, 6)$.

DRs of Line 1: $(3-1, 4-(-1), -2-2) = (2, 5, -4)$.

DRs of Line 2: $(3-0, 5-3, 6-2) = (3, 2, 4)$.

Product sum: $a_1a_2 + b_1b_2 + c_1c_2 = 2(3) + 5(2) + (-4)(4) = 6 + 10 - 16 = 0$.

$\boxed{\text{Lines are perpendicular}}$

Q3

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Show that the line through the points $(4, 7, 8), (2, 3, 4)$ is parallel to the line through the points $(-1, -2, 1), (1, 2, 5)$.

DRs of Line 1: $(2-4, 3-7, 4-8) = (-2, -4, -4)$.

DRs of Line 2: $(1-(-1), 2-(-2), 5-1) = (2, 4, 4)$.

Ratios: $\frac{-2}{2} = -1, \frac{-4}{4} = -1, \frac{-4}{4} = -1$.

Since direction ratios are proportional, the lines are parallel.

$\boxed{\text{Lines are parallel}}$

Q4

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Find the equation of the line which passes through the point $(1, 2, 3)$ and is parallel to the vector $3i + 2j - 2k$.

Position vector of point $\vec{a} = i + 2j + 3k$.

Parallel vector $\vec{b} = 3i + 2j - 2k$.

Equation of line is $\vec{r} = \vec{a} + \lambda\vec{b}$.

$\boxed{\vec{r} = (i + 2j + 3k) + \lambda(3i + 2j - 2k)}$

Q5

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Find the equation of the line in vector and in cartesian form that passes through the point with position vector $2i - j + 4k$ and is in the direction $i + 2j - k$.

Vector Form: $\vec{r} = \vec{a} + \lambda\vec{b} \Rightarrow \vec{r} = (2i - j + 4k) + \lambda(i + 2j - k)$.

Cartesian Form: Point $(2, -1, 4)$, DRs $(1, 2, -1)$.

$\boxed{\frac{x-2}{1} = \frac{y+1}{2} = \frac{z-4}{-1}}$

Q6

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Find the cartesian equation of the line which passes through the point $(-2, 4, -5)$ and parallel to the line given by $\frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6}$.

The given line has direction ratios proportional to $3, 5, 6$.

The required line passes through $(-2, 4, -5)$ and has same DRs.

$\boxed{\frac{x+2}{3} = \frac{y-4}{5} = \frac{z+5}{6}}$

Q7

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The cartesian equation of a line is $\frac{x-5}{3} = \frac{y+4}{7} = \frac{z-6}{2}$. Write its vector form.

Comparing with $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$, we get point $(5, -4, 6)$ and DRs $(3, 7, 2)$.

$\boxed{\vec{r} = (5i - 4j + 6k) + \lambda(3i + 7j + 2k)}$

Q8

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Find the vector and the cartesian equations of the lines that passes through the origin and $(5, -2, 3)$.

Line passes through $O(0,0,0)$ and $P(5, -2, 3)$.

Direction ratios: $(5-0, -2-0, 3-0) = (5, -2, 3)$.

Vector Form: $\vec{r} = \vec{0} + \lambda(5i - 2j + 3k)$.

Cartesian Form: $\frac{x-0}{5} = \frac{y-0}{-2} = \frac{z-0}{3}$.

$\boxed{\vec{r} = \lambda(5i - 2j + 3k); \quad \frac{x}{5} = \frac{y}{-2} = \frac{z}{3}}$

Q9

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Find the vector and the cartesian equations of the line that passes through the points $(3, -2, -5), (3, -2, 6)$.

DRs: $(3-3, -2-(-2), 6-(-5)) = (0, 0, 11)$. We can use $(0, 0, 1)$.

Vector Form: $\vec{r} = (3i - 2j - 5k) + \lambda k$.

Cartesian Form: $\frac{x-3}{0} = \frac{y+2}{0} = \frac{z+5}{1}$.

$\boxed{\vec{r} = (3i - 2j - 5k) + \lambda k; \quad \frac{x-3}{0} = \frac{y+2}{0} = \frac{z+5}{1}}$

Q10

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Find the angle between the pair of lines:
$\vec{r} = 2i - 5j + k + \lambda(3i + 2j + 6k)$ and $\vec{r} = 7i - 6k + \mu(i + 2j + 2k)$.

Direction vectors: $\vec{b_1} = 3i + 2j + 6k$ and $\vec{b_2} = i + 2j + 2k$.

$\cos \theta = \left| \frac{\vec{b_1} \cdot \vec{b_2}}{|\vec{b_1}||\vec{b_2}|} \right| = \left| \frac{3(1) + 2(2) + 6(2)}{\sqrt{9+4+36}\sqrt{1+4+4}} \right|$.

$\cos \theta = \frac{3+4+12}{\sqrt{49}\sqrt{9}} = \frac{19}{7 \times 3} = \frac{19}{21}$.

$\boxed{\theta = \cos^{-1}\left(\frac{19}{21}\right)}$

Q11

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Find the angle between the pair of lines:
$\frac{x-2}{2} = \frac{y-1}{5} = \frac{z+3}{-3}$ and $\frac{x+2}{-1} = \frac{y-4}{8} = \frac{z-5}{4}$.

DRs: $\vec{b_1} = (2, 5, -3)$ and $\vec{b_2} = (-1, 8, 4)$.

$\cos \theta = \left| \frac{2(-1) + 5(8) + (-3)(4)}{\sqrt{4+25+9}\sqrt{1+64+16}} \right| = \left| \frac{-2+40-12}{\sqrt{38}\sqrt{81}} \right|$.

$\cos \theta = \frac{26}{9\sqrt{38}}$.

$\boxed{\theta = \cos^{-1}\left(\frac{26}{9\sqrt{38}}\right)}$

Q12

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Find the values of $p$ so that the lines $\frac{1-x}{3} = \frac{7y-14}{2p} = \frac{z-3}{2}$ and $\frac{7-7x}{3p} = \frac{y-5}{1} = \frac{6-z}{5}$ are at right angles.

Rewrite in standard form:
$L_1: \frac{x-1}{-3} = \frac{y-2}{2p/7} = \frac{z-3}{2} \Rightarrow \text{DRs } (-3, \frac{2p}{7}, 2)$.
$L_2: \frac{x-1}{-3p/7} = \frac{y-5}{1} = \frac{z-6}{-5} \Rightarrow \text{DRs } (-\frac{3p}{7}, 1, -5)$.

Perpendicular condition: $a_1a_2 + b_1b_2 + c_1c_2 = 0$.

$(-3)(-\frac{3p}{7}) + (\frac{2p}{7})(1) + (2)(-5) = 0 \Rightarrow \frac{9p}{7} + \frac{2p}{7} - 10 = 0$.

$\frac{11p}{7} = 10 \Rightarrow 11p = 70$.

$\boxed{p = \frac{70}{11}}$

Q13

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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.

DRs are $(7, -5, 1)$ and $(1, 2, 3)$.

Product sum: $7(1) + (-5)(2) + 1(3) = 7 - 10 + 3 = 0$.

$\boxed{\text{Lines are perpendicular}}$

Q14

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Find the shortest distance between the lines $\vec{r} = (i + 2j + k) + \lambda(i - j + k)$ and $\vec{r} = (2i - j - k) + \mu(2i + j + 2k)$.

$\vec{a_1} = (1, 2, 1), \vec{b_1} = (1, -1, 1)$. $\vec{a_2} = (2, -1, -1), \vec{b_2} = (2, 1, 2)$.

$\vec{a_2} - \vec{a_1} = (1, -3, -2)$.

$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} i & j & k \\ 1 & -1 & 1 \\ 2 & 1 & 2 \end{vmatrix} = -3i + 3k$. Magnitude $= 3\sqrt{2}$.

Distance $d = \left| \frac{(\vec{b_1} \times \vec{b_2}) \cdot (\vec{a_2} - \vec{a_1})}{|\vec{b_1} \times \vec{b_2}|} \right| = \left| \frac{-3(1) + 0(-3) + 3(-2)}{3\sqrt{2}} \right| = \frac{9}{3\sqrt{2}} = \frac{3}{\sqrt{2}}$.

$\boxed{\frac{3\sqrt{2}}{2}}$

Q15

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Find the shortest distance between the lines $\frac{x+1}{7} = \frac{y+1}{-6} = \frac{z+1}{1}$ and $\frac{x-3}{1} = \frac{y-5}{-2} = \frac{z-7}{1}$.

$\vec{a_1} = (-1, -1, -1), \vec{b_1} = (7, -6, 1)$. $\vec{a_2} = (3, 5, 7), \vec{b_2} = (1, -2, 1)$.

$\vec{a_2} - \vec{a_1} = (4, 6, 8)$.

$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} i & j & k \\ 7 & -6 & 1 \\ 1 & -2 & 1 \end{vmatrix} = -4i - 6j - 8k$. Magnitude $= \sqrt{16+36+64} = \sqrt{116} = 2\sqrt{29}$.

Dot product: $-4(4) - 6(6) - 8(8) = -16 - 36 - 64 = -116$.

Distance $d = \frac{|-116|}{2\sqrt{29}} = \frac{58}{\sqrt{29}} = 2\sqrt{29}$.

$\boxed{2\sqrt{29}}$

Q16

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Find the shortest distance between the lines $\vec{r} = (i + 2j + 3k) + \lambda(i - 3j + 2k)$ and $\vec{r} = (4i + 5j + 6k) + \mu(2i + 3j + k)$.

$\vec{a_1} = (1, 2, 3), \vec{b_1} = (1, -3, 2)$. $\vec{a_2} = (4, 5, 6), \vec{b_2} = (2, 3, 1)$.

$\vec{a_2} - \vec{a_1} = (3, 3, 3)$.

$\vec{b_1} \times \vec{b_2} = -9i + 3j + 9k$. Magnitude $= \sqrt{81+9+81} = \sqrt{171} = 3\sqrt{19}$.

Dot product: $-9(3) + 3(3) + 9(3) = -27 + 9 + 27 = 9$.

Distance $d = \frac{9}{3\sqrt{19}} = \frac{3}{\sqrt{19}}$.

$\boxed{\frac{3}{\sqrt{19}}}$

Q17

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Find the shortest distance between the lines $\vec{r} = (1-t)i + (t-2)j + (3-2t)k$ and $\vec{r} = (s+1)i + (2s-1)j - (2s+1)k$.

Rearrange: $L_1: \vec{r} = (i - 2j + 3k) + t(-i + j - 2k)$.

$L_2: \vec{r} = (i - j - k) + s(i + 2j - 2k)$.

$\vec{a_2} - \vec{a_1} = (0, 1, -4)$. $\vec{b_1} = (-1, 1, -2), \vec{b_2} = (1, 2, -2)$.

$\vec{b_1} \times \vec{b_2} = 2i - 4j - 3k$. Magnitude $= \sqrt{4+16+9} = \sqrt{29}$.

Dot product: $2(0) - 4(1) - 3(-4) = -4 + 12 = 8$.

$\boxed{\frac{8}{\sqrt{29}}}$

Exercise 11.2 Practice