The availability of ISC Class 12 Maths OP Malhotra Solutions Chapter 23 Three Dimensional Geometry Ex 23(f) encourages students to tackle difficult exercises.
S Chand Class 12 ICSE Maths Solutions Chapter 23 Three Dimensional Geometry Ex 23(f)
Find the length of the shortest distance between the lines.
Question 1.
\(\frac{x-3}{1}\) = \(\frac{y-5}{-2}\) = \(\frac{z-7}{1}\) and \(\frac{x+1}{7}\) = \(\frac{y+1}{-6}\) = \(\frac{z+1}{1}\)
Answer:
In Cartesian form, these eqns. can be written as
\(\frac{x-3}{1}\) = \(\frac{y-5}{-2}\) = \(\frac{z-7}{1}\)
and
\(\frac{x+1}{7}\) = \(\frac{y+1}{-6}\) = \(\frac{z+1}{1}\)
Any point on line (1) is P(t + 3, -2 t + 5, t + 7)
and any point on line (2) is Q(7 s – 1, -6 s – 1, s – 1)
∴ D’ratios of line PQ are < 7 s – t – 4, -6 s + 2 t – 6, s – t – 8 >
Now line PQ is ⊥ to line (1).
∴ (7 s – t – 4) 1 + (-6 s + 2 t – 6)(-2) + (s – t + 8) 1 = 0
⇒ 20 s – 6 t = 0
also line PQ is ⊥ to line (2)
∴ (7 s – t – 4) + 7 + (-6 s + 2 t – 6)(-6) + (s – t – 8) 1 = 0
⇒ 86 s – 20 t = 0
On solving (3) and (4); s = t = 0
∴ points on line (1) is P(3, 5, 7) and on line (2) be, Q}
(-1, -1, -1)
Thus |P Q| = \(\sqrt{(3+1)^2+(5+1)^2+(7+1)^2}\)
= \(\sqrt{16+36+64}\) = \(\sqrt{116}\) = 2 \(\sqrt{29}\)
Question 2.
\(\frac{x+3}{-4}\) = \(\frac{y-6}{3}\) = \(\frac{z}{2}\) and \(\frac{x+2}{-4}\) = \(\frac{y}{1}\) = \(\frac{z-7}{1}\)
Answe:
Given eqns of lines are
\(\frac{x+3}{-4}\) = \(\frac{y-6}{3}\) = \(\frac{z}{2}\)
and
\(\frac{x+2}{-4}\) = \(\frac{y}{1}\) = \(\frac{z-7}{1}\)
Let < l, m, n> be the direction cosines of the line of shortest distance. Now line of shortest distance P Q is ⊥ to both given lines.
-4 l + 3 m + 2 n = 0
-4 l + m + n = 0
Using cross-multiplication method, we have
\(\frac{l}{3-2}\) = \(\frac{m}{-8+4}\) = \(\frac{n}{-4+12}\)
∴ \(\frac{l}{1}\) = \(\frac{m}{-4}\) = \(\frac{n}{8}\)
= \(\frac{1}{9}\)
i.e., \(\frac{l}{1}\) = \(\frac{m}{-4}\)
= \(\frac{n}{8}\) = \(\frac{\sqrt{l^2+m^2+n^2}}{\sqrt{1+16+81}}\)
∴ length of S.D. = projection of line joining A(-3, 6, 0) and B(-2, 0, 7) on line PQ
= \(|l(x_2-x_1)+m(y_2-y_1)+n(z_2-z_1)|\)
= \(|\frac{1}{9}(-2+3)-\frac{4}{9}(0-6)+\frac{8}{9}(7-0)|\)
= \(|\frac{1}{9}+\frac{24}{9}+\frac{56}{9}|\)
= 9 units
Aliter: We know that shortest distance between lines
\(\frac{x-x_1}{l_1}\)
= \(\frac{y-y_1}{m_1}\)
= \(\frac{z-z_1}{n_1}\) and \(\frac{x-x_2}{l_2}\)
= \(\frac{y-y_2}{m_2}\)
= \(\frac{z-z_2}{n_2}\)
is given by \(\frac{|\begin{array}{ccc}x_2-x_1 y_2-y_1 z_2-z_1 l_1 m_1 n_1 l_2 m_2 n_2\end{array}|}{\sqrt{(l_1 m_2-l_2 m_1)^2+(m_1 n_2-m_2 n_1)^2+(n_1 l_2-l_1 n_2)^2}}\)
Here,
x1 = -3 ; y1 = 6 ; z1 = 0 ;
l1 = -4 ; m1 = 3 ; n1 = 2
x2 = -2 ; y2 = 0 ; z2 = 7 ;
l2 = -4 ; m2 = 1 ; n2 = 1
∴ Required S.D. =\(\frac{1}{\sqrt{(-4+12)^2+(3-2)^2+(-8+4)^2}}|\begin{array}{ccc}-2+3 0-6 7-0 -4 3 2 -4 1 1\end{array}|\)
= \(\frac{1}{\sqrt{64+1+16}}\)
= \(|\begin{array}{rrr}1 -6 7 -4 3 2 -4 1 1\end{array}|\);
Expanding along R1
= \(\frac{1}{9}\)[1(3 – 2) + 6(-4 + 8) + 7(-4 + 12)]
= \(\frac{1}{9}\)[1 + 24 + 56]
= 9 units
Aliter : any point on line (1) be
\(\frac{x+3}{-4}\)
= \(\frac{y-6}{3}\)
= \(\frac{z}{2}\) = t (say)
i.e., P(-4 t – 3, 3 t + 6, 2 t)
and any point on line (2) be
\(\frac{x+2}{-4}\)
= \(\frac{y}{1}\)
= \(\frac{z-7}{1}\) = r (say)
i.e., Q(-4 r – 2, r, r + 7)
Then line PQ be the line of shortest distance
∴ direction ratios of line PQ are
< -4 r – 2 + 4 t + 3, r – 3 t – 6, r + 7 – 2 t >
i.e., < -4 r + 4 t + 1, r – 3 t – 6, r – 2 t + 7 >
Now, line PQ be the line of S.D. and it is ⊥ to line (1) and (2).
∴ (-4 r + 4 t + 1)(-4) + (r – 3 t – 6) 3 + (r – 2 t + 7) 2 = 0
i.e., 16 r – 16 t – 4 + 3 r – 9 t – 18 + 2 r – 4 t + 14 = 0
⇒ 21 r – 29 t – 8 = 0
and (-4 r + 4 t + 1)(-4) + (r – 3 t – 6)1 + (r – 2 t + 7) 1 = 0
i.e., 16 r – 16 t – 4 + r – 3 t – 6 + r – 2 t + 7 = 0
⇒ 18 r – 21 t – 3 = 0
⇒ 6 r – 7 t – 1 = 0
On solving eqn. (1) and eqn. (2); we have
\(\frac{r}{29-56}\)
= \(\frac{t}{-48+21}\) = \(\frac{1}{-147+174}\)
i.e., \(\frac{r}{-27}\) = \(\frac{t}{-27}\)
= \(\frac{1}{+27}\)
i .e., r = -1 ; t = -1
Thus, coordinates of P and Q becomes (1, 3, -2) and (2, -1, 6)
∴ |PQ| = \(\sqrt{(2-1)^2+(-1-3)^2+(6+2)^2}\)
= \(\sqrt{1+16+64}\) =9 units
Find the magnitude and the equations of the line of shortest distance between the two lines:
Question 3.
\(\frac{x-3}{-1}\) = \(\frac{y-4}{2}\) = \(\frac{z+2}{1}\) and \(\frac{x-1}{1}\) = \(\frac{y+7}{2}\) = \(\frac{z+2}{1}\)
Answer:
The equations of given lines are
and
\(\frac{x-3}{-1}\) = \(\frac{y-4}{2}\)
= \(\frac{z+2}{1}\) = t (say)
\(\frac{x-1}{1}\) = \(\frac{y+7}{3}\) = \(\frac{z+2}{2}\) = r (say)
So, any points on lines (1) and (2) are P(-t + 3, 2 t + 4, t – 2) and Q(r + 1, 3 r – 7, 2 r – 2)
Then PQ be the line of S.D. if it is ⊥ to both line (1) and (2).
∴ D ratios of line PQ are < r + 1 + t – 3, 3 r – 7 – 2 t – 4, 2 r – 2 – t + 2)
i.e., < r + t – 2, 3 r – 2 t – 11, 2 r – t >
Also, line PQ is ⊥ to line (1) and line (2).
∴ -1(r + t – 2) + 2(3 r – 2 t – 11) + 1(2 r – t) = 0
⇒ 7 r – 6 t – 20 = 0
and 1(r + t – 2) + 3(3 r – 2 t – 11) + 2(2 r – t) = 0
⇒ 14 r – 7 t – 35 = 0
i.e., 2 r – t – 5 = 0
On solving eqn. (3) and eqn. (4); we have
∴ r = 2, t = -1
\(\frac{r}{30-20}\) = \(\frac{t}{-40+35}\)
= \(\frac{1}{-7+12}\)
i.e., \(\frac{r}{10}\)
= \(\frac{t}{-5}\)
= \(\frac{1}{5}\)
Hence, the coordinates of P and Q are
(1 + 3, -2 + 4, -1 – 2)
i.e., (4, 2, -3) and (2 + 1, 6 – 7, 4 – 2)
i.e., (3, -1, 2)
∴ required S.D. between lines = |P Q| = \(\sqrt{(3-4)^2+(-1-2)^2+(2+3)^2}\)
= \(\sqrt{1+9+25}\)
= \(\sqrt{35}\) units
∴ D ratios of line PQ be < 3 – 4, -1 – 2, 2 + 3 >
i.e., < -1 , -3, 5 >
i.e., < 1, 3, -5 >
Hence the required eqnn. of line of S.D. which pass through (4, 2, -3) and having direction ratios < 1, 3, -5 > is given by
\(\frac{x-4}{1}\)
= \(\frac{y-2}{3}\)
= \(\frac{z+3}{-5}\)
Question 4.
Obtain the coordinates of the points where the shortest distance between the following lines meets them:
\(\frac{x-23}{-6}\) = \(\frac{y-19}{-4}\) = \(\frac{z-25}{3}\);
\(\frac{x-12}{-9}\) = \(\frac{y-1}{4}\) = \(\frac{z-5}{2}\)
Answer:
Given eqns. of lines are
\(\frac{x-23}{-6}\) = \(\frac{y-19}{-4}\) = \(\frac{z-25}{3}\) = t (say)
and
\(\frac{x-12}{-9}\) = \(\frac{y-1}{4}\) = \(\frac{z-5}{2}\) = r(say)
So, any point on lines (1) and (2) are
P(-6 t + 23, -4 t + 19, 3 t + 25) and Q(-9 + 12, 4 r + 1, 2 r + 5)
Then PQ be the line of shortest distance if PQ be perpendicular to both lines (1) and (2).
∴ direction ratios of line PQ are
i.e., < -9 r + 12 + 6 t – 23, 4 r + 1 + 4 t – 19, 2 r + 5 – 3 t – 25 >
< -9 r + 6 t – 11, 4 r + 4 t – 18, 2 r – 3 t – 20 >
Since, PQ is ⊥ to both lines (1) and (2).
∴ (-9 r + 6 t – 11)(-6) + (4 r + 4 t – 18)(-4) + (2 r – 3 t – 20) 3 = 0
⇒ 54 r – 36 t + 66 – 16 r – 16 t + 72 + 6 r – 9 t – 60 = 0
⇒ 44 r – 61 t + 78 = 0
and (-9 r + 6 t – 11)(-9) + (4 r + 4t – 18) 4 + (2 r – 3 t – 20) 2 = 0
101 r – 44 t – 13 = 0
On solving eqn. (3) and (4); we have
r = 1, t = 2
Thus, required coordinates of points be
(-12 + 23, -8 + 19, 6 + 35)
i.e., (11, 11, 3) and (-9 + 12, 4 + 1, 2 + 5)
i.e., (3, 5, 7).
Find the shortest distance between the pairs of lines whose equations are
Question 5.
\(\vec{r}\) = 6 \(\hat{i}\) + 2 \(\hat{j}\) + 2 \(\hat{k}\) + λ(\(\hat{i}\) – 2 \(\hat{j}\) + 2 \(\hat{k}\)) and \(\vec{r}\) = -4 \(\hat{i}\) – \(\hat{k}\) + µ(3 \(\hat{j}\) – 2 \(\hat{k}\) – 2 \(\hat{k}\))
Answer:
Comparing given eqns. with \(\overrightarrow{r_1}\) = \(\overrightarrow{a_1}\) + λ\(\overrightarrow{b_1}\) and \(\overrightarrow{r_2}\) = \(\overrightarrow{a_2}\) + λ\(\overrightarrow{b_2}\); we have
\(\overrightarrow{a_1}\) = 6 \(\hat{i}\) + 2 \(\hat{j}\) + 2 \(\hat{k}\);
\(\overrightarrow{a_2}\) = -4 \(\hat{i}\) – \(\hat{k}\)
\(\overrightarrow{b_1}\) = \(\hat{i}\) + 2 \(\hat{j}\) + 2 \(\hat{k}\);
\(\overrightarrow{b_2}\) = 3 \(\hat{i}\) – 2 \(\hat{j}\) – 2 \(\hat{k}\)
Question 6.
\(\vec{r}\) = 2 \(\hat{i}\) – 5 \(\hat{j}\) + \(\hat{k}\) + µ(3 \(\hat{j}\) + 2 \(\hat{k}\) + 6 \(\hat{k}\)) and \(\vec{r}\) = 7 \(\hat{i}\) – 6 \(\hat{k}\) + µ(\(\hat{i}\) + 2 \(\hat{j}\) + 2\(\hat{k}\))
Answer:
Given equations of lines are
\(\vec{r}\) = 2 \(\hat{i}\) – 5 \(\hat{j}\) + \(\hat{k}\) + λ(3\(\hat{i}\) + 2 \(\hat{j}\) + 6 \(\hat{k}\)) and \(\vec{r}\)
= 7 \(\hat{i}\) – 6 \(\hat{k}\) + µ(\(\hat{i}\) + 2 \(\hat{j}\) + 2 \(\hat{k}\))
On comparing with \(\vec{r}\) = \(\overrightarrow{a_1}\) + λ\(\overrightarrow{b_1}\) and \(\vec{r}\)
= \(\overrightarrow{a_2}\) + λ\(\overrightarrow{b_2}\)
Here,
Here, \(\overrightarrow{a_1}\) = 2 \(\hat{i}\) – 5 \(\hat{j}\) + \(\hat{k}\);
\(\overrightarrow{b_1}\) = 3 \(\hat{i}\) + 2 \(\hat{j}\) + 6 \(\hat{k}\)
\(\overrightarrow{a_2}\) = 7 \(\hat{i}\) – 6 \(\hat{k}\);
\(\overrightarrow{b_2}\) = \(\hat{i}\) + 2 \(\hat{j}\) + 2 \(\hat{k}\)
∴ \(\overrightarrow{a_2}\) – \(\overrightarrow{a_1}\)
= (7 \(\hat{i}\) – 6 \(\hat{k}\)) – (2 \(\hat{i}\) – 5 \(\hat{j}\) + \(\hat{k}\))
= 5 \(\hat{i}\) + 5 \(\hat{j}\) – 7 \(\hat{k}\)
\(\overrightarrow{b_1}\) × \(\overrightarrow{b_2}\) = \(|\begin{array}{ccc}\hat{i} \hat{j} \hat{k} 3 2 6 1 2 2\end{array}|\);
Expanding along R1 = \(\hat{i}\)(4 – 12) – \(\hat{j}\)(6 – 6) + \(\hat{k}\)(6 – 2)
= -8 \(\hat{i}\) + 0 \(\hat{j}\) + 4 \(\hat{k}\)
∴ \(|\overrightarrow{b_1} \times \overrightarrow{b_2}|\) = \(\sqrt{(-8)^2+0+4^2}\)
= \(\sqrt{64+16}\) = \(\sqrt{80}\) = 4 \(\sqrt{5}\)
Now,
(\(\overrightarrow{a_2}\) – \(\overrightarrow{a_1}\)) (\(\overrightarrow{b_1}\) × \(\overrightarrow{b_2}\))
= (5 \(\hat{i}\) + 5 \(\hat{j}\) – 7 \(\hat{k}\)) (-8 \(\hat{i}\) + 0 \(\hat{j}\) + 4 \(\hat{k}\))
= 5(-8) + 5(0) – 7(4) = -68
Thus required S.D. between them = \(\frac{|(\overrightarrow{a_2}-\overrightarrow{a_1})(\overrightarrow{b_1} \times \overrightarrow{b_2})|}{|\overrightarrow{b_1} \times \overrightarrow{b_2}|}\)
= \(\frac{|-68|}{4 \sqrt{5}}\)
= \(\frac{17}{\sqrt{5}}\) units
Question 7.
\(\vec{r}\) = (8 + 3λ) \(\hat{i}\) – (9 + 16λ) \(\hat{j}\) + (10 + 7λ) \(\hat{k}\) and \(\vec{r}\) = 15 \(\hat{i}\) + 29 \(\hat{j}\) + 5 \(\hat{k}\) + µ(3 \(\hat{i}\) + 8 \(\hat{j}\) – 5 \(\hat{k}\))
Answer:
Given eqns. can be written as
\(\vec{r}\) = (8 \(\hat{i}\) – 9 \(\hat{j}\) + 10 \(\hat{k}\)) + λ(3 \(\hat{i}\) – 16 \(\hat{j}\) + 7 \(\hat{k}\)) and \(\vec{r}\) = (15 \(\hat{i}\) + 29 \(\hat{j}\) + 5 \(\hat{k}\)) + µ(3 \(\hat{i}\) + 8 \(\hat{j}\) – 5 \(\hat{k}\))
On comparing given eqns. with
\(\vec{r}\) = \(\overrightarrow{a_1}\) + λ\(\overrightarrow{b_1}\) and \(\vec{r}\) = \(\overrightarrow{a_2}\) + λ\(\overrightarrow{b_2}\)
∴ \(\vec{a}_1\) = 8 \(\hat{i}\) – 9 \(\hat{j}\) + 10 \(\hat{k}\);
\(\overrightarrow{a_2}\) = 15 \(\hat{i}\) + 29 \(\hat{j}\) + 5 \(\hat{k}\);
\(\overrightarrow{b_1}\) = 3 \(\hat{i}\) – 16 \(\hat{j}\) + 7 \(\hat{k}\);
\(\overrightarrow{b_2}\) = 3 \(\hat{i}\) + 8 \(\hat{j}\) – 5 \(\hat{k}\)
∴ \(\overrightarrow{a_2}\) – \(\overrightarrow{a_1}\) = 7 \(\hat{i}\) + 38 \(\hat{j}\) – 5 \(\hat{k}\)
\(\overrightarrow{b_1}\) × \(\overrightarrow{b_2}\) = \(|\begin{array}{rrr}
\hat{i} \hat{j} \hat{k}
3 -16 7
3 8 -5
\end{array}|\)
= 24 \(\hat{i}\) + 36 \(\hat{j}\) + 72 \(\hat{k}\)
= 12(2 \(\hat{i}\) + 3 \(\hat{j}\) + 6 \(\hat{k}\))
∴ |\(\overrightarrow{b_1}\) × \(\overrightarrow{b_2}\)|
= 12\(\sqrt{4+9+36}\) = 12 × 7 = 84
and
= \(\frac{1}{7}\) 2(7) + 3(38) + 6(-5)
= \(\frac{1}{7}\)[14 + 114 – 30]
= \(\frac{98}{7}\) = 14
Question 8.
Show that the lines \(\vec{r}\) = (\(\hat{i}\) – \(\hat{j}\)) + λ(2 \(\hat{i}\) + \(\hat{k}\)) and \(\vec{r}\) =(2 \(\hat{i}\) – \(\hat{j}\)) + µ(\(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\)) do not intersect.
Answer:
Given lines are
\(\vec{r}\) = (\(\hat{i}\) – \(\hat{j}\)) + λ(2 \(\hat{i}\) + \(\hat{k}\))
and
\(\vec{r}\) = (2 \(\hat{i}\) – \(\hat{j}\)) + µ(\(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\))
Comparing eqn. (1) and (2) with \(\vec{r}\) = \(\overrightarrow{a_1}\) + λ\(\overrightarrow{b_1}\);
\(\vec{r}\) = \(\overrightarrow{a_2}\) + λ\(\overrightarrow{b_2}\)
where, \(\overrightarrow{a_1}\) = \(\hat{i}\) – \(\hat{j}\);
\(\overrightarrow{b_1}\) = 2 \(\hat{i}\) + \(\hat{k}\);
\(\overrightarrow{a_2}\) = 2 \(\hat{i}\) – \(\hat{j}\);
\(\overrightarrow{b_2}\) = \(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\)
Here \(\overrightarrow{a_2}\) – \(\overrightarrow{a_1}\) = 2 \(\hat{i}\) – \(\hat{j}\) – \(\hat{i}\) + \(\hat{j}\) = \(\hat{i}\) + 0 \(\hat{j}\)
and
\(\overrightarrow{b_1}\) × \(\overrightarrow{b_2}\) = \(|\begin{array}{ccc}
\hat{i} \hat{j} \hat{k}
2 0 1
1 1 -1
\end{array}|\) = \(\hat{i}\)(0 – 1) – \(\hat{j}\)(-2 – 1) + \(\hat{k}\)(2 – 1)
= \(-\hat{i}\) + 3 \(\hat{j}\) + 2 \(\hat{k}\)
Here, (\(\overrightarrow{a_2}\) – \(\overrightarrow{a_1}\)) (\(\overrightarrow{b_1}\) × \(\overrightarrow{b_2}\))
= (\(\hat{i}\) + 0 \(\hat{k}\)) (\(-\hat{i}\) + 3 \(\hat{j}\) + 2 \(\hat{k}\)) = -1 ≠ 0
Thus given lines do not intersects.
Question 9.
By computing the shortest distance determine whether the following pairs of lines intersect or not.
(i) \(\vec{r}\) = (\(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\)) + λ(3 \(\hat{i}\) – \(\hat{j}\)) and \(\vec{r}\)
= (4 \(\hat{i}\) – \(\hat{k}\)) + µ(2 \(\hat{i}\) + 3 \(\hat{j}\))
(ii) \(\frac{x-5}{4}\) = \(\frac{y-7}{-5}\) = \(\frac{z+3}{-5}\), \(\frac{x-8}{7}\) = \(\frac{y-7}{1}\) = \(\frac{z-5}{3}\)
(iii) \(\vec{r}\) = (3 – t) \(\hat{i}\) + (4 + 2 t) \(\hat{j}\) + (t – 2) \(\hat{k}\),
\(\vec{r}\) = (1 + s) \(\hat{i}\) + (3s – 7) \(\hat{j}\) + (2s – 2) \(\hat{k}\)
Answer:
(i) Given lines are, \(\vec{r}\) = (\(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\)) + λ(3 \(\hat{i}\) – \(\hat{j}\))
and
\(\vec{r}\) = (4 \(\hat{i}\) – \(\hat{k}\)) + (2 \(\hat{i}\) + 3 \(\hat{k}\))
Comparing eqn. (1) and eqn. (2) with \(\vec{r}\)
= \(\overrightarrow{a_1}\) + λ\(\overrightarrow{b_1}\);
\(\vec{r}\) = \(\overrightarrow{a_2}\) + λ\(\overrightarrow{b_2}\)
Here
\(\overrightarrow{a_1}\) = \(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\);
\(\overrightarrow{b_1}\) = 3 \(\hat{i}\) – \(\hat{j}\);
\(\overrightarrow{a_2}\) = 4 \(\hat{i}\) + 0 \(\hat{j}\) – \(\hat{k}\);
\(\overrightarrow{b_2}\) = 2 \(\hat{i}\) + 0 \(\hat{j}\) + 3 \(\hat{k}\)
\(\overrightarrow{a_2}\) – \(\overrightarrow{a_1}\)
= (4 \(\hat{i}\) + 0 \(\hat{j}\) – \(\hat{k}\)) – (\(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\))
= 3 \(\hat{i}\) – \(\hat{j}\) + 0 \(\hat{k}\)
\(\overrightarrow{b_1}\) × \(\vec{b}_2\) = \(|\begin{array}{ccc}
\hat{i} \hat{j} \hat{k}
3 -1 0
2 0 3
\end{array}|\)
= \(\hat{i}\)(-3 – 0) – \(\hat{j}\)(9 – 0) + \(\hat{k}\)(0 + 2)
= -3 \(\hat{i}\) + 9 \(\hat{j}\) + 2 \(\hat{k}\)
∴ (\(\overrightarrow{a_2}\) – \(\overrightarrow{a_1}\)) (\(\overrightarrow{b_1}\) × \(\overrightarrow{b_2}\))
= (3 \(\hat{i}\) – \(\hat{j}\) + 0 \(\hat{k}\)) (-3 \(\hat{i}\) – 9 \(\hat{j}\) + 2 \(\hat{k}\))
= 3(-3) – 1(-9) + 0(2) = -9 + 9 = 0
Thus S.D. = \(|\frac{(\overrightarrow{a_2}-\overrightarrow{a_1}) \cdot(\overrightarrow{b_1} \times \overrightarrow{b_2})}{|\overrightarrow{b_1} \times \overrightarrow{b_2}|}|\) = 0
Thus the given lines are intersecting.
(iii) Eqn’s of given lines are \(\vec{r}\) = 3 \(\hat{i}\) + 4 \(\hat{j}\) – 2 \(\hat{k}\) + t(\(-\hat{i}\) + 2 \(\hat{j}\) + \(\hat{k}\)) and \(\vec{r}\) = \(\hat{i}\) – 7 \(\hat{j}\) – 2 \(\hat{k}\) + r(\(\hat{i}\) + 3 \(\hat{j}\) + 2 \(\hat{k}\))
On comparing with
\(\overrightarrow{r_1}\) = \(\overrightarrow{a_1}\) + λ\(\overrightarrow{b_1}\) and \(\overrightarrow{r_2}\) = \(\overrightarrow{a_2}\) + µ \(\overrightarrow{b_2}\)
Here,
\(\overrightarrow{a_1}\) = 3 \(\hat{i}\) + 4 \(\hat{j}\) – 2 \(\hat{k}\);
\(\overrightarrow{b_1}\) = \(-\hat{i}\) + 2 \(\hat{j}\) + \(\hat{k}\)
\(\overrightarrow{a_2}\) = \(\hat{i}\) – 7 \(\hat{j}\) – 2 \(\hat{k}\);
\(\overrightarrow{b_2}\) = \(\hat{i}\) + 3 \(\hat{j}\) + 2 \(\hat{k}\)
∴ \(\overrightarrow{a_2}\) – \(\overrightarrow{a_1}\) = -2 \(\hat{i}\) – 11 \(\hat{j}\) + 0 \(\hat{k}\)
\(\overrightarrow{b_1}\) × \(\overrightarrow{b_2}\)
= \(|\begin{array}{rrr}
\hat{i} \hat{j} \hat{k}
-1 2 1
1 3 2
\end{array}|\)
= \(\hat{i}\)(4 – 3) – \(\hat{j}\)(-2 – 1) + \(\hat{k}\)(-3 – 2)
= \(\hat{i}\) + 3 \(\hat{j}\) – 5 \(\hat{k}\)
∴ |\(\overrightarrow{b_1}\) × \(\overrightarrow{b_2}\)| = \(\sqrt{1+9+25}\)
= \(\sqrt{35}\)
∴ (\(\overrightarrow{a_2}\) – \(\overrightarrow{a_1}\)) (\(\overrightarrow{b_1}\) × \(\overrightarrow{b_2}\))
= (-2 \(\hat{i}\) – 11 \(\hat{j}\) + 0 \(\hat{k}\)) (\(\hat{i}\) + 3 \(\hat{j}\) – 5 \(\hat{k}\))
= 2(1) – 11(3) + 0(-5) = -35
∴ S.D. between given lines = \(\frac{|(\overrightarrow{a_2}-\overrightarrow{a_1}) \cdot(\overrightarrow{b_1} \times \overrightarrow{b_2})|}{|\overrightarrow{b_1} \times \overrightarrow{b_2}|}\)
= \(\frac{|-35|}{\sqrt{35}}\) = \(\sqrt{35}\) ≠ 0
Hence, the given lines do not intersects.
Question 10.
Find the shortest distance between the following pairs of parallel lines:
(i) \(\vec{r}\) = (\(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\)) + λ(\(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\)) and \(\vec{r}\) = (2 \(\hat{i}\) – \(\hat{j}\) – \(\hat{k}\)) + µ(\(-\hat{i}\) + \(\hat{j}\) – \(\hat{k}\))
(ii) \(\vec{r}\) = (\(\hat{i}\) + \(\hat{j}\)) + λ(2 \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\)) and \(\vec{r}\) = (2 \(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\)) + µ(4 \(\hat{i}\) – 2 \(\hat{j}\) + 2 \(\hat{k}\))
Answer:
(i) Given lines are
\(\vec{r}\) = (\(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\)) + λ(\(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\))
\(\vec{r}\) = (2 \(\hat{i}\) – \(\hat{j}\) – \(\hat{k}\)) + µ(\(-\hat{i}\) + \(\hat{j}\) – \(\hat{k}\))
⇒ \(\vec{r}\) = (2 \(\hat{i}\) – \(\hat{j}\) – \(\hat{k}\)) + µ(\(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\))
where µ = -µ
Hence both given lines are || to \(\vec{b}\) = \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\)
Here \(\vec{a}_1\) = \(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\);
\(\overrightarrow{a_2}\) = 2 \(\hat{i}\) – \(\hat{j}\) – \(\hat{k}\)
Now \(\overrightarrow{a_2}\) – \(\vec{a}_1\)
= (2 \(\hat{i}\) – \(\hat{j}\) – \(\hat{k}\)) – (\(\hat{i}\) + 2 \(\hat{j}\) + 3 \(\hat{k}\))
= \(\hat{i}\) – 3 \(\hat{j}\) – 4 \(\hat{k}\)
∴ (\(\overrightarrow{a_2}\) – \(\overrightarrow{a_1}\)) × \(\vec{b}\)
= \(\hat{i}\)(-3 – 4) – \(\hat{j}\)(1 + 4) + \(\hat{k}\)(-1 + 3)
= -7 \(\hat{i}\) – 5 \(\hat{j}\) + 2 \(\hat{k}\)
|(\(\overrightarrow{a_2}\) – \(\overrightarrow{a_1}\)) × \(\vec{b}\)| = \(\sqrt{(-7)^2+(-5)^2+2^2}\)
= \(\sqrt{49+25+4}\)
= \(\sqrt{78}\)
|\(\vec{b}\)| = \(\sqrt{1^2+1^2+1^2}\)
= \(\sqrt{3}\)
∴ required S.D. = \(\frac{|(\overrightarrow{a_2}-\overrightarrow{a_1}) \times \vec{b}|}{|\vec{b}|}\)
= \(\frac{\sqrt{78}}{\sqrt{3}}\) = \(\sqrt{26}\)
(ii) The equation of given lines are;
\(\vec{r}\) = (\(\hat{i}\) + \(\hat{j}\)) + λ(2 \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\))
\(\vec{r}\) = (2 \(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\)) + µ(4 \(\hat{i}\) – 2 \(\hat{j}\) + 2 \(\hat{k}\))
⇒ \(\vec{r}\) = ( 2 \(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\)) + µ(2 \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\))
where µ = 2 µ
Clearly given lines (1) and (2) are parallel and || to vector \(\vec{b}\) = 2 \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\)
Here \(\overrightarrow{a_1}\) = \(\hat{i}\) + \(\hat{j}\);
\(\overrightarrow{a_2}\) = 2 \(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\)
∴ \(\overrightarrow{a_2}\) – \(\overrightarrow{a_1}\) = \(\hat{i}\) + 0 \(\hat{j}\) – \(\hat{k}\);
(\(\overrightarrow{a_2}\) – \(\overrightarrow{a_1}\)) × \(\vec{b}\) =
= \(\hat{i}\)(0 – 1) – \(\hat{j}\)(1 + 2) + \(\hat{k}\)(-1 – 0)
= \(-\hat{i}\) – 3 \(\hat{j}\) – \(\hat{k}\)
∴ |(\(\overrightarrow{a_2}\) – \(\overrightarrow{a_1}\)) × \(\vec{b}\)| = \(\sqrt{(-1)^2+(-3)^2+(-1)^2}\)
= \(\sqrt{11}\)
|\(\vec{b}\)| = \(\sqrt{2^2+(-1)^2+(1)^2}\) = \(\sqrt{6}\)
and
|\(\vec{b}\)| = \(\sqrt{2^2+(-1)^2+(1)^2}\) = \(\sqrt{6}\)
Thus required S.D. = \(\frac{\left|\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right) \times \vec{b}\right|}{|\vec{b}|}\)
= \(\frac{\sqrt{11}}{\sqrt{6}}\)
Question 11.
Define the line of shortest distance between two skew lines. Find the shortest distance and the vector equation of the line of shortest distance between the lines given by:
(i) \(\vec{r}\) = (8 + 3λ) \(\hat{i}\) – (9 + 16λ) \(\hat{j}\) + (10 + 7λ) \(\hat{k}\) and \(\vec{r}\) = 15 \(\hat{i}\) + 29 \(\hat{j}\) + 5 \(\hat{k}\) + µ(3 \(\hat{i}\) + 8 \(\hat{j}\) – 5 \(\hat{k}\))
(ii) \(\vec{r}\) = (3 \(\hat{i}\) + 8 \(\hat{j}\) + 3 \(\hat{k}\)) + λ(3 \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\)) and \(\vec{r}\)
= (-3µ – 3) \(\hat{i}\) + (2µ – 7) \(\hat{j}\) + (4µ + 6) \(\hat{k}\)
Answer:
Line of shortest distance : Let l and m be two skew lines then there is one and only one line p which is ⊥ to l and m. Then line p is called line of shortest distance.
Aliter
(i) Given equs. can be written as
\(\vec{r}\) = 8 \(\hat{i}\) – 9 \(\hat{j}\) + 10 \(\hat{k}\) + λ(3 \(\hat{i}\) – 16 \(\hat{j}\) + 7 \(\hat{k}\)) and \(\vec{r}\)
= (15 \(\hat{i}\) + 29 \(\hat{j}\) + 5 \(\hat{k}\)) + µ(3 \(\hat{i}\) + 8 \(\hat{j}\) -5 \(\hat{k}\))
and in Cartesian form are;
\(\frac{x-8}{3}\) = \(\frac{y+9}{-16}\)
= \(\frac{z-10}{7}\) = t (say)
\(\frac{x-15}{3}\) = \(\frac{y-29}{8}\)
= \(\frac{z-5}{-5}\) = r (say)
Any point on lines (1) and (2) are given by
P(3 t + 8, -16 t – 9,7 t + 10)
Q(3 r + 15, 8 r + 29), -5 r + 5)
and
Q(3 r + 15, 8 r + 29), -5 r + 5)
Then PQ be the line of S.D. if it is ⊥ to both lines (1) and (2)
Now direction ratios of P Q are
< 3 r + 15 – 3 t – 8, 8 r + 29 + 16 t + 9, -5 r + 5 – 7 t – 10 >
i.e. < 3 r – 3 t + 7, 8 r + 16 t + 38, -5 r – 7 t – 5 >
Since P Q is ⊥ to both lines (1) and (2).
(3 r – 3 t + 7) 3 + (8 r + 16 t + 38)(-16) + (-5 r – 7 t – 5) 7 = 0
⇒ -154 r – 314 t – 622 = 0
⇒ 77 r + 157 t + 311 = 0
Also, (3 r – 3 t + 7) 3 + (8 r + 16 t + 38) 8 + (-5 r – 7 t – 5)(-5) = 0
⇒ 98 r + 154 t + 350 = 0
⇒ 49 r + 77 t + 175 = 0
On solving eqn. (3) and (4); we have
r = -2 ; t = -1
Hence the coordinates of P and Q are:
P(-3 + 8, 16 – 9, -7 + 10)
i.e. P(5, 7, 3)
and Q(-6 + 15, -16 + 29, 10 + 5)
i.e. Q(9, 13, 15)
∴ required S.D. between lines = |P Q|
= \(\sqrt{(9-5)^2+(13-7)^2+(15-3)^2}\)
= \(\sqrt{16+36+144}\)
= \(\sqrt{196}\) = 14 units
∴ direction ratios of line PQ are
< 9 – 5, 13 – 7, 15 – 3 >
i.e. < 4, 6, 12 >
i.e. < 2, 3, 6 >
Hence, required vector eqn. of line of S.D. which pass through (5, 7, 3) and having direction ratio < 2, 3, 6 > is given by \(\vec{r}\) = (5 \(\hat{i}\) + 7 \(\hat{j}\) + 3 \(\hat{k}\)) + λ(2 \(\hat{i}\) + 3 \(\hat{j}\) + 6 \(\hat{k}\))
(ii) Given equations in cartesian form are;
\(\frac{x-3}{3}\) = \(\frac{y-8}{-1}\) = \(\frac{z-3}{1}\) = t (say)
and \(\frac{x+3}{-3}\) = \(\frac{y+7}{2}\) = \(\frac{z-6}{4}\) = r (say)
So any points on line (1) and (2) are
P(3 t + 3, -t + 8, t + 3)
and Q(-3 r – 3, 2 r – 7, 4 r + 6)
Then line PQ be the line of S.D. if it is ⊥ to both lines (1) and (2);
Now direction ratios of line P Q are
< -3 r – 3 – 3 t – 3, 2 r – 7 + t – 8, 4 r + 6 – t – 3 >
i.e. < -3 r – 3 t – 6, 2 r + t – 15, 4 r – t + 3 >
Since line PQ is ⊥ to line (1)
(-3 r – 3 t – 6) 3 + (2 r + t – 15)(-1) + (4 r – t + 3) 1 = 0
⇒ -7 r – 11 t = 0
Also line PQ is ⊥ to line (2)
(-3 r – 3 t – 6)(-3) + (2 r + t – 15) 2 + (4 r – t + 3) 4 = 0
⇒ +29 r + 7 t = 0
On solving eqn. (3) and eqn. (4); we have
r = t = 0
Hence coordinates of P and Q are P(3, 8, 3) and Q(-3, -7, 6)
∴ required S.D. between lines = |P Q| = \(\sqrt{(-3-3)^2+(-7-8)^2+(6-3)^2}\)
= \(\sqrt{36+225+9}\)
= \(\sqrt{270}\)
= \(\sqrt{30 \times 9}\)
= 3 \(\sqrt{30}\) units
and vector eqn. of S.D. which pass through the point whose P.V. be 3 \(\hat{i}\) + 8 \(\hat{j}\) + 3 \(\hat{k}\) and || to vector \(\vec{b}\) = (-3 – 3) \(\hat{i}\) + (-7 – 8) \(\hat{j}\) + (6 – 3) \(\hat{k}\) is given by
= -6 \(\hat{i}\) – 15 \(\hat{j}\) + 3 \(\hat{k}\)
\(\vec{r}\) = (3 \(\hat{i}\) + 8 \(\hat{j}\) + 3 \(\hat{k}\)) + λ(-6 \(\hat{i}\) – 15 \(\hat{j}\) + 3 \(\hat{k}\))
⇒ \(\vec{r}\) = (3 \(\hat{i}\) + 8 \(\hat{j}\) + 3 \(\hat{k}\)) + µ(2\(\hat{i}\) + 5 \(\hat{j}\) – \(\hat{k}\))
where µ = -2λ