OP Malhotra Class 12 Maths Solutions Chapter 23 Three Dimensional Geometry Ex 23(e)

Regular engagement with ISC Class 12 Maths OP Malhotra Solutions Chapter 23 Three Dimensional Geometry Ex 23(e) can boost students’ confidence in the subject.

S Chand Class 12 ICSE Maths Solutions Chapter 23 Three Dimensional Geometry Ex 23(e)

Question 1.
Show that the lines \(\frac{x-1}{2}\) = \(\frac{y-2}{3}\) = \(\frac{z-3}{4}\) and \(\frac{x-4}{5}\) = \(\frac{y-1}{2}\) = z intersect each other. Find their point of intersection.
Answer:
Eqn. of given lines are ;
\(\frac{x-1}{2}\) = \(\frac{y-2}{3}\) = \(\frac{z-3}{4}\) = t(say )
\(\frac{x-4}{5}\) = \(\frac{y-1}{2}\) = \(\frac{z}{1}\) = r(say )
Clearly \(\frac{2}{5}\) ≠ \(\frac{3}{2}\) ≠ \(\frac{4}{1}\)
∴ both lines are not parallel.
Any point on line (1) be P(2 t + 1, 3 t + 2, 4 t + 3)
and any point on line (2) be M(5 r + 4, 2 r + 1, r)
Now both lines (1) and (2) intersects if P and M coincides.
i.e., if 2 t + 1 = 5 r + 4 ; 3 t + 2 = 2 r + 1 and 4 t + 3 = r
⇒ 2 t – 5 r = 3 ; 3 t – 2 r = -1 and 4 t – r + 3 = 0
on solving first two eqns, t = -1 ; r = -1 and these values of t and r also satisfies the 3 nd eqn.
Hence both lines intersects and required point of interesection be (-2 + 1, -3 + 2, -4 + 3), i.e., (-1, -1, -1).

Question 2.
Prove that the lines \(\frac{x-4}{1}\) = \(\frac{y+3}{-4}\) = \(\frac{z+1}{7}\); \(\frac{x-1}{2}\) = \(\frac{y+1}{-3}\) = \(\frac{z+10}{8}\) intersect and find the coordinates of their point of intersection.
Answer:
Given eqns. of line are ;
\(\frac{x-4}{1}\) = \(\frac{y+3}{-4}\) = \(\frac{z+1}{7}\) = t (say)
and
\(\frac{x-1}{2}\) = \(\frac{y+1}{-3}\) = \(\frac{z+10}{8}\) = r (say)
Clearly \(\frac{1}{2}\) ≠ \(\frac{-4}{-3}\) ≠ \(\frac{7}{8}\)
∴ both lines (1) and (2) are not parallel.
Any point on line (1) are (t + 4, -4 t – 3, 7 t – 1)
and any point on line (2) are (2 r + 1, -3 r – 1, 8 r – 10)
Now both lines intersects if both points coincides.
i.e., if t + 4 = 2 r + 1 ; -4 t – 3 = -3 r – 1 ; 7 t – 1 = 8 r – 10
if t – 2 r = -3
-4 t + 3 r = 2
and 7 t – 8 r = -9
On solving eqn. (3) and eqn. (4); we have
r = 2 ; t = 1
These values of r and t also satisfies eqn. (5).
Hence, both lines intersects and required point of intersection be (1 + 4, -4 – 3, 7 – 1) i.e., (5, -7, 6).

Question 3.
Show that the lines \(\frac{x+3}{2}\) = \(\frac{y+5}{3}\) = \(\frac{z-7}{-3}\); and \(\frac{x+1}{4}\)=\frac{y+1}{5}[/latex] = \(\frac{z+1}{-1}\) are coplanar.
Answer:
Given lines are,
\(\frac{x+3}{2}\) = \(\frac{y+5}{3}\) = \(\frac{z-7}{-3}\)
and
\(\frac{x+1}{4}\) = \(\frac{y+1}{5}\) = \(\frac{z+1}{-1}\)
We know that, the lines
\(\frac{x-x_1}{a_1}\) = \(\frac{y-y_1}{b_1}\)
= \(\frac{z-z_1}{c_1}\) and \(\frac{x+x_2}{a_2}\) = \(\frac{y-y_2}{b_2}\)
= \(\frac{z-z_2}{c_2}\)
are coplanar if \( |\begin{array}{ccc}x_2-x_1 & y_2-y_1 & z_2-z_1 a_1 & b_1 & c_1 a_2 & b_2 & c_2\end{array}|\) = 0
Here,
a₁ = +2 ; b₁ = 3 ; c₁ = -3
a₂ = 4 ; b₂ = 5 ; c₂ = -1
Here,

= 2(-3 + 15) – 4(-2 + 12) – 8(10 – 12)
= 24 – 40 + 16 = 0
Thus, given lines are coplanar.

Question 4.
Show that the lines \(\frac{x-1}{2}\) = \(\frac{y-3}{4}\) = \(\frac{z}{-1}\) and \(\frac{x+1}{5}\) = \(\frac{y-2}{1}\), z = 2 do not intersect each other.
Answer:

Question 5.
Show that the lines \(\frac{x-1}{2}\) = \(\frac{y-3}{4}\) = \(\frac{z}{-1}\) and \(\frac{x-4}{3}\) = \(\frac{y-1}{-2}\) = \(\frac{z-1}{1}\) are coplanar.
Answer:

Question 6.
Find the equations of the line which intersects the lines \(\frac{x-1}{2}\) = \(\frac{y-2}{-2}\) = \(\frac{z-3}{4}\) and \(\frac{x+2}{1}\) = \(\frac{y-3}{2}\) = \(\frac{z+1}{4}\) and passes through (1, 1, 1).
Answer:
Eqn. of line passing through the point (1, 1, 1) and having direction ratios < a, b, c > is given by
\(\frac{x-1}{a}\) = \(\frac{y-1}{b}\) = \(\frac{z-1}{c}\)
Now line (1) intersects with the line \(\frac{x-1}{2}\) = \(\frac{y-2}{3}\) = \(\frac{z-3}{4}\)
∴ \(\frac{a}{2}\) ≠ \(\frac{b}{3}\) ≠ \(\frac{c}{4}\) and
\(|\begin{array}{ccc}
1-1 & 2-1 & 3-1
a & b & c
2 & 3 & 4
\end{array}|\) = 0;
Expanding along R₁
0(4 b – 3 c) – 1(4 a – 2 c) + 2(3 a – 2 b) = 0
⇒ 2 a – 4 b + 2 c = 0
⇒ a – 2 b + c = 0
Now, line (1) is also intersects with the given line
\(\frac{x+2}{1}\) = \(\frac{y-3}{2}\)
= \(\frac{z+1}{4}\)
∴ \(\frac{a}{1}\) ≠ \(\frac{b}{2}\) ≠ \(\frac{c}{4}\)
and \(|\begin{array}{ccc}
-2-1 & 3-1 & -1-1
a & b & c
1 & 2 & 4
\end{array}|\) = 0
i.e., \(|\begin{array}{rrr}
-3 & 2 & -2
a & b & c
1 & 2 & 4
\end{array}|\) = 0;
Expanding along R₁
⇒ -3(4 b – 2 c) – 2(4 a – c) – 2(2 a – b) = 0
⇒ -12 b + 6 c – 8 a + 2 c – 4 a + 2 b = 0
-12 a – 10 b + 8 c = 0
⇒ 6 a + 5 b – 4 c = 0
On solving eqn. (2) and (3) by cross-multiplication method, we have
\(\frac{a}{8-5}\) = \(\frac{b}{6+4}\)
= \(\frac{c}{5+12}\)
i.e., \(\frac{a}{3}\) = \(\frac{b}{10}\) = \(\frac{c}{17}\)
Hence, the required eqn. of line using eqn. (1) be given by
\(\frac{x-1}{3}\) = \(\frac{y-1}{10}\) = \(\frac{z-1}{17}\)