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NCERT Solutions

Three Dimensional Geometry

Madhya Pradesh Board · Class 12 · Mathematics

NCERT Solutions for Three Dimensional Geometry — Madhya Pradesh Board Class 12 Mathematics.

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A 3D Cartesian coordinate system showing x, y, and z axes with a directed line L passing through the origin. The angles alpha, beta, and gamma are shown between line L and the positive x, y, and z axe
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25 Questions Solved · 3 Sections

Exercise 11.1

1If a line makes angles 90∘,135∘,45∘90^\circ, 135^\circ, 45^\circ with the x,yx, y and zz-axes respectively, find its direction cosines.Show solution
Given: A line makes angles α=90∘\alpha = 90^\circ, β=135∘\beta = 135^\circ, γ=45∘\gamma = 45^\circ with the xx-, yy- and zz-axes respectively.

Formula: Direction cosines are l=cos⁡α, m=cos⁡β, n=cos⁡γl = \cos\alpha,\ m = \cos\beta,\ n = \cos\gamma.

Working:
l=cos⁡90∘=0l = \cos 90^\circ = 0
m=cos⁡135∘=−12m = \cos 135^\circ = -\frac{1}{\sqrt{2}}
n=cos⁡45∘=12n = \cos 45^\circ = \frac{1}{\sqrt{2}}

Verification: l2+m2+n2=0+12+12=1l^2 + m^2 + n^2 = 0 + \dfrac{1}{2} + \dfrac{1}{2} = 1 ✓

Answer: The direction cosines are 0, −12, 120,\ -\dfrac{1}{\sqrt{2}},\ \dfrac{1}{\sqrt{2}}.
2Find the direction cosines of a line which makes equal angles with the coordinate axes.Show solution
Given: The line makes equal angles with all three coordinate axes, i.e., α=β=γ\alpha = \beta = \gamma.

Concept: If l,m,nl, m, n are direction cosines, then l2+m2+n2=1l^2 + m^2 + n^2 = 1.

Working:
Since α=β=γ\alpha = \beta = \gamma, we have l=m=n=cos⁡αl = m = n = \cos\alpha.

Substituting in the identity:
l2+l2+l2=1  ⟹  3l2=1  ⟹  l=±13l^2 + l^2 + l^2 = 1 \implies 3l^2 = 1 \implies l = \pm\frac{1}{\sqrt{3}}

Answer: The direction cosines are 13, 13, 13\dfrac{1}{\sqrt{3}},\ \dfrac{1}{\sqrt{3}},\ \dfrac{1}{\sqrt{3}} or −13, −13, −13-\dfrac{1}{\sqrt{3}},\ -\dfrac{1}{\sqrt{3}},\ -\dfrac{1}{\sqrt{3}}.
3If a line has the direction ratios −18,12,−4-18, 12, -4, then what are its direction cosines?Show solution
Given: Direction ratios are a=−18, b=12, c=−4a = -18,\ b = 12,\ c = -4.

Formula:
l=aa2+b2+c2,m=ba2+b2+c2,n=ca2+b2+c2l = \frac{a}{\sqrt{a^2+b^2+c^2}},\quad m = \frac{b}{\sqrt{a^2+b^2+c^2}},\quad n = \frac{c}{\sqrt{a^2+b^2+c^2}}

Working:
a2+b2+c2=(−18)2+(12)2+(−4)2=324+144+16=484=22\sqrt{a^2+b^2+c^2} = \sqrt{(-18)^2+(12)^2+(-4)^2} = \sqrt{324+144+16} = \sqrt{484} = 22

l=−1822=−911,m=1222=611,n=−422=−211l = \frac{-18}{22} = -\frac{9}{11},\quad m = \frac{12}{22} = \frac{6}{11},\quad n = \frac{-4}{22} = -\frac{2}{11}

Answer: The direction cosines are −911, 611, −211-\dfrac{9}{11},\ \dfrac{6}{11},\ -\dfrac{2}{11}.
4Show that the points (2,3,4), (−1,−2,1), (5,8,7)(2, 3, 4),\ (-1, -2, 1),\ (5, 8, 7) are collinear.Show solution
Given: Points A(2,3,4)A(2,3,4), B(−1,−2,1)B(-1,-2,1), C(5,8,7)C(5,8,7).

Concept: Three points are collinear if the direction ratios of ABAB and BCBC are proportional.

Direction ratios of AB:
(−1−2, −2−3, 1−4)=(−3, −5, −3)(-1-2,\ -2-3,\ 1-4) = (-3,\ -5,\ -3)

Direction ratios of BC:
(5−(−1), 8−(−2), 7−1)=(6, 10, 6)(5-(-1),\ 8-(-2),\ 7-1) = (6,\ 10,\ 6)

Check proportionality:
−36=−510=−36=−12\frac{-3}{6} = \frac{-5}{10} = \frac{-3}{6} = -\frac{1}{2}

The direction ratios of ABAB and BCBC are proportional, so AB∥BCAB \parallel BC. Since point BB is common to both, the points AA, BB, CC are collinear. \hfill■\hfill\blacksquare
5Find the direction cosines of the sides of the triangle whose vertices are (3,5,−4), (−1,1,2)(3, 5, -4),\ (-1, 1, 2) and (−5,−5,−2)(-5, -5, -2).Show solution
Given: Vertices A(3,5,−4)A(3,5,-4), B(−1,1,2)B(-1,1,2), C(−5,−5,−2)C(-5,-5,-2).

Side AB:
Direction ratios: (−1−3, 1−5, 2−(−4))=(−4, −4, 6)(-1-3,\ 1-5,\ 2-(-4)) = (-4,\ -4,\ 6)
∣AB∣=16+16+36=68=217|AB| = \sqrt{16+16+36} = \sqrt{68} = 2\sqrt{17}
Direction cosines of ABAB:
−4217, −4217, 6217=−217, −217, 317\frac{-4}{2\sqrt{17}},\ \frac{-4}{2\sqrt{17}},\ \frac{6}{2\sqrt{17}} = -\frac{2}{\sqrt{17}},\ -\frac{2}{\sqrt{17}},\ \frac{3}{\sqrt{17}}

Side BC:
Direction ratios: (−5−(−1), −5−1, −2−2)=(−4, −6, −4)(-5-(-1),\ -5-1,\ -2-2) = (-4,\ -6,\ -4)
∣BC∣=16+36+16=68=217|BC| = \sqrt{16+36+16} = \sqrt{68} = 2\sqrt{17}
Direction cosines of BCBC:
−4217, −6217, −4217=−217, −317, −217\frac{-4}{2\sqrt{17}},\ \frac{-6}{2\sqrt{17}},\ \frac{-4}{2\sqrt{17}} = -\frac{2}{\sqrt{17}},\ -\frac{3}{\sqrt{17}},\ -\frac{2}{\sqrt{17}}

Side CA:
Direction ratios: (3−(−5), 5−(−5), −4−(−2))=(8, 10, −2)(3-(-5),\ 5-(-5),\ -4-(-2)) = (8,\ 10,\ -2)
∣CA∣=64+100+4=168=242|CA| = \sqrt{64+100+4} = \sqrt{168} = 2\sqrt{42}
Direction cosines of CACA:
8242, 10242, −2242=442, 542, −142\frac{8}{2\sqrt{42}},\ \frac{10}{2\sqrt{42}},\ \frac{-2}{2\sqrt{42}} = \frac{4}{\sqrt{42}},\ \frac{5}{\sqrt{42}},\ -\frac{1}{\sqrt{42}}

Answer:
- Direction cosines of ABAB: −217, −217, 317-\dfrac{2}{\sqrt{17}},\ -\dfrac{2}{\sqrt{17}},\ \dfrac{3}{\sqrt{17}}
- Direction cosines of BCBC: −217, −317, −217-\dfrac{2}{\sqrt{17}},\ -\dfrac{3}{\sqrt{17}},\ -\dfrac{2}{\sqrt{17}}
- Direction cosines of CACA: 442, 542, −142\dfrac{4}{\sqrt{42}},\ \dfrac{5}{\sqrt{42}},\ -\dfrac{1}{\sqrt{42}}

Exercise 11.2

1Show that the three lines with direction cosines 1213,−313,−413\dfrac{12}{13}, \dfrac{-3}{13}, \dfrac{-4}{13}; 413,1213,313\dfrac{4}{13}, \dfrac{12}{13}, \dfrac{3}{13}; 313,−413,1213\dfrac{3}{13}, \dfrac{-4}{13}, \dfrac{12}{13} are mutually perpendicular.Show solution
Concept: Two lines are perpendicular if l1l2+m1m2+n1n2=0l_1 l_2 + m_1 m_2 + n_1 n_2 = 0.

Let the three lines be L1L_1, L2L_2, L3L_3 with the given direction cosines.

Check L1⊥L2L_1 \perp L_2:
1213⋅413+−313⋅1213+−413⋅313=48−36−12169=0169=0✓\frac{12}{13}\cdot\frac{4}{13} + \frac{-3}{13}\cdot\frac{12}{13} + \frac{-4}{13}\cdot\frac{3}{13} = \frac{48 - 36 - 12}{169} = \frac{0}{169} = 0 \checkmark

Check L2⊥L3L_2 \perp L_3:
413⋅313+1213⋅−413+313⋅1213=12−48+36169=0169=0✓\frac{4}{13}\cdot\frac{3}{13} + \frac{12}{13}\cdot\frac{-4}{13} + \frac{3}{13}\cdot\frac{12}{13} = \frac{12 - 48 + 36}{169} = \frac{0}{169} = 0 \checkmark

Check L1⊥L3L_1 \perp L_3:
1213⋅313+−313⋅−413+−413⋅1213=36+12−48169=0169=0✓\frac{12}{13}\cdot\frac{3}{13} + \frac{-3}{13}\cdot\frac{-4}{13} + \frac{-4}{13}\cdot\frac{12}{13} = \frac{36 + 12 - 48}{169} = \frac{0}{169} = 0 \checkmark

Since each pair of lines satisfies the perpendicularity condition, the three lines are mutually perpendicular. \hfill■\hfill\blacksquare
2Show that the line through the points (1,−1,2)(1, -1, 2), (3,4,−2)(3, 4, -2) is perpendicular to the line through the points (0,3,2)(0, 3, 2) and (3,5,6)(3, 5, 6).Show solution
Direction ratios of line L1L_1 through (1,−1,2)(1,-1,2) and (3,4,−2)(3,4,-2):
(3−1, 4−(−1), −2−2)=(2, 5, −4)(3-1,\ 4-(-1),\ -2-2) = (2,\ 5,\ -4)

Direction ratios of line L2L_2 through (0,3,2)(0,3,2) and (3,5,6)(3,5,6):
(3−0, 5−3, 6−2)=(3, 2, 4)(3-0,\ 5-3,\ 6-2) = (3,\ 2,\ 4)

Check perpendicularity: a1a2+b1b2+c1c2=0a_1 a_2 + b_1 b_2 + c_1 c_2 = 0
2(3)+5(2)+(−4)(4)=6+10−16=0✓2(3) + 5(2) + (-4)(4) = 6 + 10 - 16 = 0 \checkmark

Hence, the two lines are perpendicular to each other. \hfill■\hfill\blacksquare
3Show that the line through the points (4,7,8)(4, 7, 8), (2,3,4)(2, 3, 4) is parallel to the line through the points (−1,−2,1)(-1, -2, 1), (1,2,5)(1, 2, 5).Show solution
Direction ratios of line L1L_1 through (4,7,8)(4,7,8) and (2,3,4)(2,3,4):
(2−4, 3−7, 4−8)=(−2, −4, −4)(2-4,\ 3-7,\ 4-8) = (-2,\ -4,\ -4)

Direction ratios of line L2L_2 through (−1,−2,1)(-1,-2,1) and (1,2,5)(1,2,5):
(1−(−1), 2−(−2), 5−1)=(2, 4, 4)(1-(-1),\ 2-(-2),\ 5-1) = (2,\ 4,\ 4)

Check proportionality:
−22=−44=−44=−1\frac{-2}{2} = \frac{-4}{4} = \frac{-4}{4} = -1

The direction ratios are proportional, hence the two lines are parallel. \hfill■\hfill\blacksquare
4Find the equation of the line which passes through the point (1,2,3)(1, 2, 3) and is parallel to the vector 3i^+2j^−2k^3\hat{i} + 2\hat{j} - 2\hat{k}.Show solution
Given: Point A(1,2,3)A(1,2,3), so position vector a⃗=i^+2j^+3k^\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}; direction vector b⃗=3i^+2j^−2k^\vec{b} = 3\hat{i} + 2\hat{j} - 2\hat{k}.

Vector equation of the line:
r⃗=a⃗+λb⃗\vec{r} = \vec{a} + \lambda\vec{b}
r⃗=(i^+2j^+3k^)+λ(3i^+2j^−2k^)\boxed{\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(3\hat{i} + 2\hat{j} - 2\hat{k})}

Cartesian form:
x−13=y−22=z−3−2\frac{x-1}{3} = \frac{y-2}{2} = \frac{z-3}{-2}
5Find the equation of the line in vector and in cartesian form that passes through the point with position vector 2i^−j^+4k^2\hat{i} - \hat{j} + 4\hat{k} and is in the direction i^+2j^−k^\hat{i} + 2\hat{j} - \hat{k}.Show solution
Given: a⃗=2i^−j^+4k^\vec{a} = 2\hat{i} - \hat{j} + 4\hat{k}, b⃗=i^+2j^−k^\vec{b} = \hat{i} + 2\hat{j} - \hat{k}.

Vector equation:
r⃗=a⃗+λb⃗\vec{r} = \vec{a} + \lambda\vec{b}
r⃗=(2i^−j^+4k^)+λ(i^+2j^−k^)\boxed{\vec{r} = (2\hat{i} - \hat{j} + 4\hat{k}) + \lambda(\hat{i} + 2\hat{j} - \hat{k})}

Cartesian equation: The point is (2,−1,4)(2, -1, 4) and direction ratios are (1,2,−1)(1, 2, -1).
x−21=y+12=z−4−1\boxed{\frac{x-2}{1} = \frac{y+1}{2} = \frac{z-4}{-1}}
6Find the cartesian equation of the line which passes through the point (−2,4,−5)(-2, 4, -5) and parallel to the line given by x+33=y−45=z+86\dfrac{x+3}{3} = \dfrac{y-4}{5} = \dfrac{z+8}{6}.Show solution
Given: Point (−2,4,−5)(-2, 4, -5); the given line has direction ratios (3,5,6)(3, 5, 6).

Since the required line is parallel to the given line, it has the same direction ratios (3,5,6)(3, 5, 6).

Cartesian equation:
x+23=y−45=z+56\boxed{\frac{x+2}{3} = \frac{y-4}{5} = \frac{z+5}{6}}
7The cartesian equation of a line is x−53=y+47=z−62\dfrac{x-5}{3} = \dfrac{y+4}{7} = \dfrac{z-6}{2}. Write its vector form.Show solution
Given: Cartesian equation x−53=y+47=z−62\dfrac{x-5}{3} = \dfrac{y+4}{7} = \dfrac{z-6}{2}.

The line passes through the point (5,−4,6)(5, -4, 6) and has direction ratios (3,7,2)(3, 7, 2).

So a⃗=5i^−4j^+6k^\vec{a} = 5\hat{i} - 4\hat{j} + 6\hat{k} and b⃗=3i^+7j^+2k^\vec{b} = 3\hat{i} + 7\hat{j} + 2\hat{k}.

Vector equation:
r⃗=(5i^−4j^+6k^)+λ(3i^+7j^+2k^)\boxed{\vec{r} = (5\hat{i} - 4\hat{j} + 6\hat{k}) + \lambda(3\hat{i} + 7\hat{j} + 2\hat{k})}
8Find the angle between the following pairs of lines:
(i) r⃗=2i^−5j^+k^+λ(3i^+2j^+6k^)\vec{r} = 2\hat{i} - 5\hat{j} + \hat{k} + \lambda(3\hat{i} + 2\hat{j} + 6\hat{k}) and r⃗=7i^−6k^+μ(i^+2j^+2k^)\vec{r} = 7\hat{i} - 6\hat{k} + \mu(\hat{i} + 2\hat{j} + 2\hat{k})
(ii) r⃗=3i^+j^−2k^+λ(i^−j^−2k^)\vec{r} = 3\hat{i} + \hat{j} - 2\hat{k} + \lambda(\hat{i} - \hat{j} - 2\hat{k}) and r⃗=2i^−j^−56k^+μ(3i^−5j^−4k^)\vec{r} = 2\hat{i} - \hat{j} - 56\hat{k} + \mu(3\hat{i} - 5\hat{j} - 4\hat{k})Show solution
Formula: cos⁡θ=∣b⃗1⋅b⃗2∣b⃗1∣∣b⃗2∣∣\cos\theta = \left|\dfrac{\vec{b}_1 \cdot \vec{b}_2}{|\vec{b}_1||\vec{b}_2|}\right|

(i) b⃗1=3i^+2j^+6k^\vec{b}_1 = 3\hat{i}+2\hat{j}+6\hat{k}, b⃗2=i^+2j^+2k^\vec{b}_2 = \hat{i}+2\hat{j}+2\hat{k}

b⃗1⋅b⃗2=3(1)+2(2)+6(2)=3+4+12=19\vec{b}_1 \cdot \vec{b}_2 = 3(1)+2(2)+6(2) = 3+4+12 = 19
∣b⃗1∣=9+4+36=49=7,∣b⃗2∣=1+4+4=9=3|\vec{b}_1| = \sqrt{9+4+36} = \sqrt{49} = 7,\quad |\vec{b}_2| = \sqrt{1+4+4} = \sqrt{9} = 3
cos⁡θ=∣197×3∣=1921\cos\theta = \left|\frac{19}{7 \times 3}\right| = \frac{19}{21}
θ=cos⁡−1(1921)\boxed{\theta = \cos^{-1}\left(\frac{19}{21}\right)}

(ii) b⃗1=i^−j^−2k^\vec{b}_1 = \hat{i}-\hat{j}-2\hat{k}, b⃗2=3i^−5j^−4k^\vec{b}_2 = 3\hat{i}-5\hat{j}-4\hat{k}

b⃗1⋅b⃗2=1(3)+(−1)(−5)+(−2)(−4)=3+5+8=16\vec{b}_1 \cdot \vec{b}_2 = 1(3)+(-1)(-5)+(-2)(-4) = 3+5+8 = 16
∣b⃗1∣=1+1+4=6,∣b⃗2∣=9+25+16=50=52|\vec{b}_1| = \sqrt{1+1+4} = \sqrt{6},\quad |\vec{b}_2| = \sqrt{9+25+16} = \sqrt{50} = 5\sqrt{2}
cos⁡θ=∣166⋅52∣=16512=16103=853\cos\theta = \left|\frac{16}{\sqrt{6}\cdot 5\sqrt{2}}\right| = \frac{16}{5\sqrt{12}} = \frac{16}{10\sqrt{3}} = \frac{8}{5\sqrt{3}}
θ=cos⁡−1(853)\boxed{\theta = \cos^{-1}\left(\frac{8}{5\sqrt{3}}\right)}
9Find the angle between the following pair of lines:
(i) x−22=y−15=z+3−3\dfrac{x-2}{2} = \dfrac{y-1}{5} = \dfrac{z+3}{-3} and x+2−1=y−48=z−54\dfrac{x+2}{-1} = \dfrac{y-4}{8} = \dfrac{z-5}{4}
(ii) x2=y2=z1\dfrac{x}{2} = \dfrac{y}{2} = \dfrac{z}{1} and x−54=y−21=z−38\dfrac{x-5}{4} = \dfrac{y-2}{1} = \dfrac{z-3}{8}Show solution
Formula: cos⁡θ=∣a1a2+b1b2+c1c2a12+b12+c12 a22+b22+c22∣\cos\theta = \left|\dfrac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\,\sqrt{a_2^2+b_2^2+c_2^2}}\right|

(i) Direction ratios: (a1,b1,c1)=(2,5,−3)(a_1,b_1,c_1)=(2,5,-3) and (a2,b2,c2)=(−1,8,4)(a_2,b_2,c_2)=(-1,8,4)

a1a2+b1b2+c1c2=2(−1)+5(8)+(−3)(4)=−2+40−12=26a_1a_2+b_1b_2+c_1c_2 = 2(-1)+5(8)+(-3)(4) = -2+40-12 = 26
4+25+9=38,1+64+16=81=9\sqrt{4+25+9}=\sqrt{38},\quad \sqrt{1+64+16}=\sqrt{81}=9
cos⁡θ=26938\cos\theta = \frac{26}{9\sqrt{38}}
θ=cos⁡−1(26938)\boxed{\theta = \cos^{-1}\left(\frac{26}{9\sqrt{38}}\right)}

(ii) Direction ratios: (a1,b1,c1)=(2,2,1)(a_1,b_1,c_1)=(2,2,1) and (a2,b2,c2)=(4,1,8)(a_2,b_2,c_2)=(4,1,8)

a1a2+b1b2+c1c2=8+2+8=18a_1a_2+b_1b_2+c_1c_2 = 8+2+8 = 18
4+4+1=9=3,16+1+64=81=9\sqrt{4+4+1}=\sqrt{9}=3,\quad \sqrt{16+1+64}=\sqrt{81}=9
cos⁡θ=183×9=1827=23\cos\theta = \frac{18}{3\times 9} = \frac{18}{27} = \frac{2}{3}
θ=cos⁡−1(23)\boxed{\theta = \cos^{-1}\left(\frac{2}{3}\right)}
10Find the values of pp so that the lines 1−x3=7y−142p=z−32\dfrac{1-x}{3} = \dfrac{7y-14}{2p} = \dfrac{z-3}{2} and 7−7x3p=y−51=6−z5\dfrac{7-7x}{3p} = \dfrac{y-5}{1} = \dfrac{6-z}{5} are at right angles.Show solution
Rewrite in standard form:

Line 1: x−1−3=y−22p/7=z−32\dfrac{x-1}{-3} = \dfrac{y-2}{2p/7} = \dfrac{z-3}{2}

So direction ratios: a1=−3, b1=2p7, c1=2a_1 = -3,\ b_1 = \dfrac{2p}{7},\ c_1 = 2.

Line 2: x−1−3p/7=y−51=z−6−5\dfrac{x-1}{-3p/7} = \dfrac{y-5}{1} = \dfrac{z-6}{-5}

So direction ratios: a2=−3p7, b2=1, c2=−5a_2 = -\dfrac{3p}{7},\ b_2 = 1,\ c_2 = -5.

Condition for perpendicularity: a1a2+b1b2+c1c2=0a_1a_2 + b_1b_2 + c_1c_2 = 0

(−3)(−3p7)+2p7(1)+(2)(−5)=0(-3)\left(-\frac{3p}{7}\right) + \frac{2p}{7}(1) + (2)(-5) = 0
9p7+2p7−10=0\frac{9p}{7} + \frac{2p}{7} - 10 = 0
11p7=10\frac{11p}{7} = 10
p=7011p = \frac{70}{11}

p=7011\boxed{p = \frac{70}{11}}
11Show that the lines x−57=y+2−5=z1\dfrac{x-5}{7} = \dfrac{y+2}{-5} = \dfrac{z}{1} and x1=y2=z3\dfrac{x}{1} = \dfrac{y}{2} = \dfrac{z}{3} are perpendicular to each other.Show solution
Direction ratios of L1L_1: (7,−5,1)(7, -5, 1)

Direction ratios of L2L_2: (1,2,3)(1, 2, 3)

Check: a1a2+b1b2+c1c2=7(1)+(−5)(2)+1(3)=7−10+3=0a_1a_2 + b_1b_2 + c_1c_2 = 7(1) + (-5)(2) + 1(3) = 7 - 10 + 3 = 0

Since the dot product of direction ratios is zero, the two lines are perpendicular to each other. \hfill■\hfill\blacksquare
12Find the shortest distance between the lines
r⃗=(i^+2j^+k^)+λ(i^−j^+k^)\vec{r} = (\hat{i}+2\hat{j}+\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})
r⃗=2i^−j^−k^+μ(2i^+j^+2k^)\vec{r} = 2\hat{i}-\hat{j}-\hat{k}+\mu(2\hat{i}+\hat{j}+2\hat{k})Show solution
Given:
a⃗1=i^+2j^+k^\vec{a}_1 = \hat{i}+2\hat{j}+\hat{k}, b⃗1=i^−j^+k^\vec{b}_1 = \hat{i}-\hat{j}+\hat{k}
a⃗2=2i^−j^−k^\vec{a}_2 = 2\hat{i}-\hat{j}-\hat{k}, b⃗2=2i^+j^+2k^\vec{b}_2 = 2\hat{i}+\hat{j}+2\hat{k}

Formula: d=∣(b⃗1×b⃗2)⋅(a⃗2−a⃗1)∣b⃗1×b⃗2∣∣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|

Step 1: a⃗2−a⃗1=(2−1)i^+(−1−2)j^+(−1−1)k^=i^−3j^−2k^\vec{a}_2 - \vec{a}_1 = (2-1)\hat{i}+(-1-2)\hat{j}+(-1-1)\hat{k} = \hat{i}-3\hat{j}-2\hat{k}

Step 2: b⃗1×b⃗2=∣i^amp;j^amp;k^1amp;−1amp;12amp;1amp;2∣\vec{b}_1\times\vec{b}_2 = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&-1&1\\2&1&2\end{vmatrix}
=i^[(−1)(2)−(1)(1)]−j^[(1)(2)−(1)(2)]+k^[(1)(1)−(−1)(2)]= \hat{i}[(-1)(2)-(1)(1)] - \hat{j}[(1)(2)-(1)(2)] + \hat{k}[(1)(1)-(-1)(2)]
=i^(−2−1)−j^(2−2)+k^(1+2)= \hat{i}(-2-1) - \hat{j}(2-2) + \hat{k}(1+2)
=−3i^+0j^+3k^= -3\hat{i} + 0\hat{j} + 3\hat{k}

Step 3: ∣b⃗1×b⃗2∣=9+0+9=32|\vec{b}_1\times\vec{b}_2| = \sqrt{9+0+9} = 3\sqrt{2}

Step 4: (b⃗1×b⃗2)⋅(a⃗2−a⃗1)=(−3)(1)+(0)(−3)+(3)(−2)=−3+0−6=−9(\vec{b}_1\times\vec{b}_2)\cdot(\vec{a}_2-\vec{a}_1) = (-3)(1)+(0)(-3)+(3)(-2) = -3+0-6 = -9

Step 5: d=∣−932∣=932=32=322d = \left|\dfrac{-9}{3\sqrt{2}}\right| = \dfrac{9}{3\sqrt{2}} = \dfrac{3}{\sqrt{2}} = \dfrac{3\sqrt{2}}{2}

d=322\boxed{d = \frac{3\sqrt{2}}{2}}
13Find the shortest distance between the lines
x+17=y+1−6=z+11andx−31=y−5−2=z−71\frac{x+1}{7} = \frac{y+1}{-6} = \frac{z+1}{1} \quad \text{and} \quad \frac{x-3}{1} = \frac{y-5}{-2} = \frac{z-7}{1}Show solution
Given:
(x1,y1,z1)=(−1,−1,−1)(x_1,y_1,z_1)=(-1,-1,-1), (a1,b1,c1)=(7,−6,1)(a_1,b_1,c_1)=(7,-6,1)
(x2,y2,z2)=(3,5,7)(x_2,y_2,z_2)=(3,5,7), (a2,b2,c2)=(1,−2,1)(a_2,b_2,c_2)=(1,-2,1)

Formula (Cartesian):
d=∣x2−x1amp;y2−y1amp;z2−z1a1amp;b1amp;c1a2amp;b2amp;c2∣(b1c2−b2c1)2+(c1a2−c2a1)2+(a1b2−a2b1)2d = \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}}

Numerator determinant:
∣4amp;6amp;87amp;−6amp;11amp;−2amp;1∣\begin{vmatrix} 4 & 6 & 8 \\ 7 & -6 & 1 \\ 1 & -2 & 1 \end{vmatrix}
=4[(−6)(1)−(−2)(1)]−6[(7)(1)−(1)(1)]+8[(7)(−2)−(−6)(1)]= 4[(-6)(1)-(-2)(1)] - 6[(7)(1)-(1)(1)] + 8[(7)(-2)-(-6)(1)]
=4(−6+2)−6(7−1)+8(−14+6)= 4(-6+2) - 6(7-1) + 8(-14+6)
=4(−4)−6(6)+8(−8)= 4(-4) - 6(6) + 8(-8)
=−16−36−64=−116= -16 - 36 - 64 = -116

Denominator:
b1c2−b2c1=(−6)(1)−(−2)(1)=−6+2=−4b_1c_2 - b_2c_1 = (-6)(1)-(-2)(1) = -6+2 = -4
c1a2−c2a1=(1)(1)−(1)(7)=1−7=−6c_1a_2 - c_2a_1 = (1)(1)-(1)(7) = 1-7 = -6
a1b2−a2b1=(7)(−2)−(1)(−6)=−14+6=−8a_1b_2 - a_2b_1 = (7)(-2)-(1)(-6) = -14+6 = -8

(−4)2+(−6)2+(−8)2=16+36+64=116=229\sqrt{(-4)^2+(-6)^2+(-8)^2} = \sqrt{16+36+64} = \sqrt{116} = 2\sqrt{29}

d=∣−116∣229=116229=5829=5829×2929=582929=229d = \frac{|-116|}{2\sqrt{29}} = \frac{116}{2\sqrt{29}} = \frac{58}{\sqrt{29}} = \frac{58}{\sqrt{29}}\times\frac{\sqrt{29}}{\sqrt{29}} = \frac{58\sqrt{29}}{29} = 2\sqrt{29}

d=229\boxed{d = 2\sqrt{29}}
14Find the shortest distance between the lines whose vector equations are
r⃗=(i^+2j^+3k^)+λ(i^−3j^+2k^)\vec{r} = (\hat{i}+2\hat{j}+3\hat{k})+\lambda(\hat{i}-3\hat{j}+2\hat{k})
r⃗=4i^+5j^+6k^+μ(2i^+3j^+k^)\vec{r} = 4\hat{i}+5\hat{j}+6\hat{k}+\mu(2\hat{i}+3\hat{j}+\hat{k})Show solution
Given:
a⃗1=i^+2j^+3k^\vec{a}_1 = \hat{i}+2\hat{j}+3\hat{k}, b⃗1=i^−3j^+2k^\vec{b}_1 = \hat{i}-3\hat{j}+2\hat{k}
a⃗2=4i^+5j^+6k^\vec{a}_2 = 4\hat{i}+5\hat{j}+6\hat{k}, b⃗2=2i^+3j^+k^\vec{b}_2 = 2\hat{i}+3\hat{j}+\hat{k}

Step 1: a⃗2−a⃗1=3i^+3j^+3k^\vec{a}_2-\vec{a}_1 = 3\hat{i}+3\hat{j}+3\hat{k}

Step 2: b⃗1×b⃗2=∣i^amp;j^amp;k^1amp;−3amp;22amp;3amp;1∣\vec{b}_1\times\vec{b}_2 = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&-3&2\\2&3&1\end{vmatrix}
=i^[(−3)(1)−(2)(3)]−j^[(1)(1)−(2)(2)]+k^[(1)(3)−(−3)(2)]= \hat{i}[(-3)(1)-(2)(3)] - \hat{j}[(1)(1)-(2)(2)] + \hat{k}[(1)(3)-(-3)(2)]
=i^(−3−6)−j^(1−4)+k^(3+6)= \hat{i}(-3-6) - \hat{j}(1-4) + \hat{k}(3+6)
=−9i^+3j^+9k^= -9\hat{i}+3\hat{j}+9\hat{k}

Step 3: ∣b⃗1×b⃗2∣=81+9+81=171=319|\vec{b}_1\times\vec{b}_2| = \sqrt{81+9+81} = \sqrt{171} = 3\sqrt{19}

Step 4: (b⃗1×b⃗2)⋅(a⃗2−a⃗1)=(−9)(3)+(3)(3)+(9)(3)=−27+9+27=9(\vec{b}_1\times\vec{b}_2)\cdot(\vec{a}_2-\vec{a}_1) = (-9)(3)+(3)(3)+(9)(3) = -27+9+27 = 9

Step 5: d=∣9319∣=319=31919d = \left|\dfrac{9}{3\sqrt{19}}\right| = \dfrac{3}{\sqrt{19}} = \dfrac{3\sqrt{19}}{19}

d=319=31919\boxed{d = \frac{3}{\sqrt{19}} = \frac{3\sqrt{19}}{19}}
15Find the shortest distance between the lines whose vector equations are
r⃗=(1−t)i^+(t−2)j^+(3−2t)k^\vec{r} = (1-t)\hat{i}+(t-2)\hat{j}+(3-2t)\hat{k}
r⃗=(s+1)i^+(2s−1)j^−(2s+1)k^\vec{r} = (s+1)\hat{i}+(2s-1)\hat{j}-(2s+1)\hat{k}Show solution
Rewrite in standard form:

Line 1: r⃗=(i^−2j^+3k^)+t(−i^+j^−2k^)\vec{r} = (\hat{i}-2\hat{j}+3\hat{k}) + t(-\hat{i}+\hat{j}-2\hat{k})
So a⃗1=i^−2j^+3k^\vec{a}_1 = \hat{i}-2\hat{j}+3\hat{k}, b⃗1=−i^+j^−2k^\vec{b}_1 = -\hat{i}+\hat{j}-2\hat{k}

Line 2: r⃗=(i^−j^−k^)+s(i^+2j^−2k^)\vec{r} = (\hat{i}-\hat{j}-\hat{k}) + s(\hat{i}+2\hat{j}-2\hat{k})
So a⃗2=i^−j^−k^\vec{a}_2 = \hat{i}-\hat{j}-\hat{k}, b⃗2=i^+2j^−2k^\vec{b}_2 = \hat{i}+2\hat{j}-2\hat{k}

Step 1: a⃗2−a⃗1=(1−1)i^+(−1+2)j^+(−1−3)k^=0i^+j^−4k^\vec{a}_2-\vec{a}_1 = (1-1)\hat{i}+(-1+2)\hat{j}+(-1-3)\hat{k} = 0\hat{i}+\hat{j}-4\hat{k}

Step 2: b⃗1×b⃗2=∣i^amp;j^amp;k^−1amp;1amp;−21amp;2amp;−2∣\vec{b}_1\times\vec{b}_2 = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\-1&1&-2\\1&2&-2\end{vmatrix}
=i^[(1)(−2)−(−2)(2)]−j^[(−1)(−2)−(−2)(1)]+k^[(−1)(2)−(1)(1)]= \hat{i}[(1)(-2)-(-2)(2)] - \hat{j}[(-1)(-2)-(-2)(1)] + \hat{k}[(-1)(2)-(1)(1)]
=i^(−2+4)−j^(2+2)+k^(−2−1)= \hat{i}(-2+4) - \hat{j}(2+2) + \hat{k}(-2-1)
=2i^−4j^−3k^= 2\hat{i}-4\hat{j}-3\hat{k}

Step 3: ∣b⃗1×b⃗2∣=4+16+9=29|\vec{b}_1\times\vec{b}_2| = \sqrt{4+16+9} = \sqrt{29}

Step 4: (b⃗1×b⃗2)⋅(a⃗2−a⃗1)=(2)(0)+(−4)(1)+(−3)(−4)=0−4+12=8(\vec{b}_1\times\vec{b}_2)\cdot(\vec{a}_2-\vec{a}_1) = (2)(0)+(-4)(1)+(-3)(-4) = 0-4+12 = 8

Step 5: d=∣829∣=829=82929d = \left|\dfrac{8}{\sqrt{29}}\right| = \dfrac{8}{\sqrt{29}} = \dfrac{8\sqrt{29}}{29}

d=829=82929\boxed{d = \frac{8}{\sqrt{29}} = \frac{8\sqrt{29}}{29}}

Miscellaneous Exercise on Chapter 11

1Find the angle between the lines whose direction ratios are a,b,ca, b, c and b−c,c−a,a−bb-c, c-a, a-b.Show solution
Given: Direction ratios of L1L_1: (a,b,c)(a, b, c) and L2L_2: (b−c,c−a,a−b)(b-c, c-a, a-b).

Formula: cos⁡θ=∣a1a2+b1b2+c1c2a12+b12+c12 a22+b22+c22∣\cos\theta = \left|\dfrac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\,\sqrt{a_2^2+b_2^2+c_2^2}}\right|

Numerator:
a(b−c)+b(c−a)+c(a−b)=ab−ac+bc−ab+ca−cb=0a(b-c)+b(c-a)+c(a-b) = ab-ac+bc-ab+ca-cb = 0

Since the numerator is 00, we have cos⁡θ=0\cos\theta = 0, which gives θ=90∘\theta = 90^\circ.

θ=90∘\boxed{\theta = 90^\circ}

The two lines are perpendicular to each other.
2Find the equation of a line parallel to xx-axis and passing through the origin.Show solution
Given: The line passes through the origin O(0,0,0)O(0,0,0) and is parallel to the xx-axis.

The xx-axis has direction ratios (1,0,0)(1, 0, 0), so the required line also has direction ratios (1,0,0)(1, 0, 0).

Vector equation:
r⃗=0⃗+λi^  ⟹  r⃗=λi^\vec{r} = \vec{0} + \lambda\hat{i} \implies \boxed{\vec{r} = \lambda\hat{i}}

Cartesian equation:
x−01=y−00=z−00  ⟹  x1=y0=z0\frac{x-0}{1} = \frac{y-0}{0} = \frac{z-0}{0} \implies \boxed{\frac{x}{1} = \frac{y}{0} = \frac{z}{0}}

This represents the xx-axis itself (any line through the origin parallel to the xx-axis is the xx-axis).
3If the lines x−1−3=y−22k=z−32\dfrac{x-1}{-3} = \dfrac{y-2}{2k} = \dfrac{z-3}{2} and x−13k=y−11=z−6−5\dfrac{x-1}{3k} = \dfrac{y-1}{1} = \dfrac{z-6}{-5} are perpendicular, find the value of kk.Show solution
Direction ratios of L1L_1: (−3,2k,2)(-3, 2k, 2)

Direction ratios of L2L_2: (3k,1,−5)(3k, 1, -5)

Condition for perpendicularity: a1a2+b1b2+c1c2=0a_1a_2+b_1b_2+c_1c_2 = 0

(−3)(3k)+(2k)(1)+(2)(−5)=0(-3)(3k)+(2k)(1)+(2)(-5) = 0
−9k+2k−10=0-9k+2k-10 = 0
−7k=10-7k = 10
k=−107k = -\frac{10}{7}

k=−107\boxed{k = -\frac{10}{7}}
4Find the shortest distance between lines r⃗=6i^+2j^+2k^+λ(i^−2j^+2k^)\vec{r} = 6\hat{i}+2\hat{j}+2\hat{k}+\lambda(\hat{i}-2\hat{j}+2\hat{k}) and r⃗=−4i^−k^+μ(3i^−2j^−2k^)\vec{r} = -4\hat{i}-\hat{k}+\mu(3\hat{i}-2\hat{j}-2\hat{k}).Show solution
Given:
a⃗1=6i^+2j^+2k^\vec{a}_1 = 6\hat{i}+2\hat{j}+2\hat{k}, b⃗1=i^−2j^+2k^\vec{b}_1 = \hat{i}-2\hat{j}+2\hat{k}
a⃗2=−4i^+0j^−k^\vec{a}_2 = -4\hat{i}+0\hat{j}-\hat{k}, b⃗2=3i^−2j^−2k^\vec{b}_2 = 3\hat{i}-2\hat{j}-2\hat{k}

Step 1: a⃗2−a⃗1=(−4−6)i^+(0−2)j^+(−1−2)k^=−10i^−2j^−3k^\vec{a}_2-\vec{a}_1 = (-4-6)\hat{i}+(0-2)\hat{j}+(-1-2)\hat{k} = -10\hat{i}-2\hat{j}-3\hat{k}

Step 2: b⃗1×b⃗2=∣i^amp;j^amp;k^1amp;−2amp;23amp;−2amp;−2∣\vec{b}_1\times\vec{b}_2 = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&-2&2\\3&-2&-2\end{vmatrix}
=i^[(−2)(−2)−(2)(−2)]−j^[(1)(−2)−(2)(3)]+k^[(1)(−2)−(−2)(3)]= \hat{i}[(-2)(-2)-(2)(-2)] - \hat{j}[(1)(-2)-(2)(3)] + \hat{k}[(1)(-2)-(-2)(3)]
=i^(4+4)−j^(−2−6)+k^(−2+6)= \hat{i}(4+4) - \hat{j}(-2-6) + \hat{k}(-2+6)
=8i^+8j^+4k^= 8\hat{i}+8\hat{j}+4\hat{k}

Step 3: ∣b⃗1×b⃗2∣=64+64+16=144=12|\vec{b}_1\times\vec{b}_2| = \sqrt{64+64+16} = \sqrt{144} = 12

Step 4: (b⃗1×b⃗2)⋅(a⃗2−a⃗1)=(8)(−10)+(8)(−2)+(4)(−3)=−80−16−12=−108(\vec{b}_1\times\vec{b}_2)\cdot(\vec{a}_2-\vec{a}_1) = (8)(-10)+(8)(-2)+(4)(-3) = -80-16-12 = -108

Step 5: d=∣−10812∣=9d = \left|\dfrac{-108}{12}\right| = 9

d=9\boxed{d = 9}
5Find the vector equation of the line passing through the point (1,2,−4)(1, 2, -4) and perpendicular to the two lines:
x−83=y+19−16=z−107andx−153=y−298=z−5−5\frac{x-8}{3} = \frac{y+19}{-16} = \frac{z-10}{7} \quad \text{and} \quad \frac{x-15}{3} = \frac{y-29}{8} = \frac{z-5}{-5}Show solution
Given: The required line passes through A(1,2,−4)A(1, 2, -4) and is perpendicular to both given lines.

Let direction vector of required line be b⃗=xi^+yj^+zk^\vec{b} = x\hat{i}+y\hat{j}+z\hat{k}.

Direction vectors of given lines: b⃗1=3i^−16j^+7k^\vec{b}_1 = 3\hat{i}-16\hat{j}+7\hat{k} and b⃗2=3i^+8j^−5k^\vec{b}_2 = 3\hat{i}+8\hat{j}-5\hat{k}.

Since b⃗\vec{b} is perpendicular to both, b⃗=b⃗1×b⃗2\vec{b} = \vec{b}_1 \times \vec{b}_2:

b⃗=∣i^amp;j^amp;k^3amp;−16amp;73amp;8amp;−5∣\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\3&-16&7\\3&8&-5\end{vmatrix}
=i^[(−16)(−5)−(7)(8)]−j^[(3)(−5)−(7)(3)]+k^[(3)(8)−(−16)(3)]= \hat{i}[(-16)(-5)-(7)(8)] - \hat{j}[(3)(-5)-(7)(3)] + \hat{k}[(3)(8)-(-16)(3)]
=i^(80−56)−j^(−15−21)+k^(24+48)= \hat{i}(80-56) - \hat{j}(-15-21) + \hat{k}(24+48)
=24i^+36j^+72k^= 24\hat{i}+36\hat{j}+72\hat{k}
=12(2i^+3j^+6k^)= 12(2\hat{i}+3\hat{j}+6\hat{k})

So direction vector is b⃗=2i^+3j^+6k^\vec{b} = 2\hat{i}+3\hat{j}+6\hat{k}.

Vector equation of the required line:
r⃗=(i^+2j^−4k^)+λ(2i^+3j^+6k^)\vec{r} = (\hat{i}+2\hat{j}-4\hat{k}) + \lambda(2\hat{i}+3\hat{j}+6\hat{k})

r⃗=(i^+2j^−4k^)+λ(2i^+3j^+6k^)\boxed{\vec{r} = (\hat{i}+2\hat{j}-4\hat{k}) + \lambda(2\hat{i}+3\hat{j}+6\hat{k})}

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