Effective ISC Class 12 Maths OP Malhotra Solutions Chapter 23 Three Dimensional Geometry Ex 23(b) can help bridge the gap between theory and application.
S Chand Class 12 ICSE Maths Solutions Chapter 23 Three Dimensional Geometry Ex 23(b)
Question 1.
Find the equations of a line passing through the point (-1, 2, 3) and having direction ratios proportional to -4, 5, 6.
Answer:
We know that, eqn. of line through the point (x1, y1, z1) and having direction ratios < a, b, c> be given by \(\frac{x-x_1}{a}\) = \(\frac{y-y_1}{b}\) = \(\frac{z-z_1}{c}\)
∴ required eqn. of line through the point (-1, 2, 3) and having direction ratio < -4, 5, 6 > be
\(\frac{x-1}{-4}\) = \(\frac{y-2}{5}\) = \(\frac{z-3}{6}\)
Question 2.
Find the equations of a line passing through the point (2, -3, 0) and having direction cosines \(-\frac{1}{7}\), \(\frac{4}{7}\), \(-\frac{6}{7}\)
Answer:
Thus required eqn. of line passing through the point (2, -3, 0) and having direction cosines < \(-\frac{1}{7}\), \(\frac{4}{7}\), \(\frac{-6}{7}\) > by given by \(\frac{x-2}{\frac{-1}{7}}\) = \(\frac{y+3}{\frac{4}{7}}\) = \(\frac{z-0}{\frac{-6}{7}}\) or \(\frac{x-2}{-1}\) = \(\frac{y+3}{4}\) = \(\frac{z-0}{-6}\)
Question 3.
Find the equations of a line passing through the points (2, 3, 4) and (4, 6, 5).
Answr:
We know that eqn. of line passing through the points (2, 3, 4) and (4, 6, 5) is given by
\(\frac{x-2}{4-2}\) = \(\frac{y-3}{6-3}\) = \(\frac{z-4}{5-4}\)
i.e., \(\frac{x-2}{2}\) = \(\frac{y-3}{3}\) = \(\frac{z-4}{1}\)
[∵ eqn. of line through the points (x1, y1, z1) and (x2, y2, z2) be given by
\(\frac{x-x_1}{x_2-x_1}\)
= \(\frac{y-y_1}{y_2-y_1}\) = \(\frac{z-z_1}{z_2-z_1}\)
Question 4.
Find the equations of a line passing through the points (3, -2, -5) and (3, -2, 6).
Answer:
The required equation of line passing through the points (3,-2,-5) and (3,-2,6) is given by
\(\frac{x-3}{3-3}\)
= \(\frac{y+2}{-2+2}\)
= \(\frac{z+5}{6+5}\)
i.e., \(\frac{x-3}{0}\)
= \(\frac{y+2}{0}\) = \(\frac{z+5}{11}\)
Question 5.
Find the coordinates of the point, where the line through (5, 1, 6) and (3, 4, 1) crosses the
(i) y z-plane
(ii) the x y-plane
(iii) the x-plane.
Sol.
eqn. of line passing through the points (5, 1, 6) and (3, 4, 1) is given by
\(\frac{x-5}{3-5}\) = \(\frac{y-5}{4-1}\)
= \(\frac{z+6}{1-6}\)
i.e., \(\frac{x-5}{-2}\)
= \(\frac{y-1}{3}\)
= \(\frac{z+6}{-5}\)
(i) Since line (1) crosses y z plane
∴ x = 0
∴ from (1); \(\frac{0-5}{-2}\) = \(\frac{y-1}{3}\) = \(\frac{z-6}{-5}\)
⇒ 2 y – 2 = 15
⇒ y = \(\frac{17}{2}\)
and 2(6 – z) = 25
⇒ 6 – z = \(\frac{25}{2}\)
⇒ z = \(-\frac{13}{2}\)
Thus the required coordinates of point be (0, \(\frac{17}{2}\), \(\frac{-13}{2}\)).
(ii) Since line (1) crosses x y plane ∴ z = 0
i.e., putting z = 0 in eqn. (1); we have
\(\frac{x-5}{-2}\)
= \(\frac{y-1}{3}\) = \(\frac{0-6}{-5}\)
= \(\frac{6}{5}\)
⇒ 5 x – 25 = -12
⇒ x = \(\frac{13}{5}\)
and
5(y – 1) = 18
⇒ y = \(\frac{23}{5}\)
Thus the required coordinates of point be (\(\frac{13}{5}\), \(\frac{23}{5}\), 0)
(iii) Since line (1) crosses z x plane,
i.e., y = 0, putting y = 0 in eqn. (1); we have
\(\frac{x-5}{-2}\) = \(\frac{0-1}{3}\)
= \(\frac{z-6}{-5}\)
⇒ 3(x – 5) = 2
z-6 = \(\frac{5}{3}\)
⇒ z = \(\frac{23}{3}\)
and
z – 6 = \(\frac{5}{3}\) ⇒ z = \(\frac{23}{3}\)
Thus, the required coordinates of point be (\(\frac{17}{3}\), 0, \(\frac{23}{3}\))
Question 6.
The cartesian equations of a line are 6 x-2=3 y+1=2 z-2. Find the direction ratios.
Answer:
The eqn. of given line by 6 x – 2 = 3 y + 1 = 2 z – 2
⇒ 6(x – \(\frac{1}{3}\)) = 3(y + \(\frac{1}{3}\)) = 2(z – 1)
⇒ \(\frac{x-\frac{1}{3}}{1 / 6}\) = \(\frac{y+\frac{1}{3}}{1 / 3}\)
= \(\frac{z-1}{1 / 2}\)
⇒ \(\frac{x-\frac{1}{3}}{1}\)
= \(\frac{y+\frac{1}{3}}{2}\) = \(\frac{z-1}{3}\)
∴ direction ratios of line (1) are < 1, 2, 3 >
Question 7.
Find the cartesian equations of a line which
(i) passes through the point (1, 2, 3) and parallel to the line
\(\frac{-x-2}{1}\) = \(\frac{y+3}{7}\) = \(\frac{2 z-6}{3}\)
(ii) passes through the point (1,3,-2) and is parallel to the line given by
\(\frac{x+1}{3}\) = \(\frac{y+4}{5}\) = \(\frac{z+3}{-6}\)
(iii) through the point (2, -1, 1) and parallel to the line joining the points (-1, 4, 1) and (1 , 2 , 2 ).
Answer:
(i) eqn. of given line be
\(\frac{-x-2}{1}\) = \(\frac{y+3}{7}\) = \(\frac{2 z-6}{3}\)
⇒ \(\frac{x+2}{-1}\) = \(\frac{y+3}{7}\) = \(\frac{z-3}{3 / 2}\)
⇒ \(\frac{x+2}{-2}\) = \(\frac{y+3}{14}\) = \(\frac{z-3}{3}\)
∴ direction ratios of given line are < -2, 14, 3 >
Thus direction ratio of the line which is parallel to given line be < -2, 14, 3 >
Thus the required cartesian eqn. of line passes through the point (1, 2, 3) and having direction ratios < -2, 14, 3 > be \(\frac{x-1}{-2}\)
= \(\frac{y-2}{14}\)
= \(\frac{z-3}{3}\)
(ii) eqn. of given line be \(\frac{x+1}{3}\) = \(\frac{y-4}{5}\)
= \(\frac{z+3}{-6}\)
∴ D ratios of given line ( 1 ) be < 3, 5, -6 >
Thus D ratios of line parallel to line (1) be < 3, 5, -6 >.
Therefore eqn. of line passing through the point (1,3,-2) and having direction ratios < 3, 5, -6 > be given by \(\frac{x-1}{3}\) = \(\frac{y-3}{5}\)
= \(\frac{z+2}{-6}\)
(iii) Direction numbers of the line joining the points (-1, 4, 1) and (1, 2, 2) be < 1 + 1, 2 – 4, 2 – 1 >
i.e., < 2, -2, 1 >
∴ direction ratios of the line || to given line are proportional to < 2, -2, 1 >
Thus the required eqn. of line through the point (2, -1, 1) and having direction ratios < 2, -2, 1 > will be given by \(\frac{x-2}{2}\)
= \(\frac{y+1}{-2}\) = \(\frac{z-1}{1}\)
Question 8.
Prove that the points A(-1, 3, 2), B(-4, 2, -2) and C(5, 5, 10) are collinear.
Answer:
The eqn. of line through the points A(-1, 3, 2) and B(-4, 2, -2) be given by
\(\frac{x+1}{-4+1}\) = \(\frac{y-3}{2-3}\) = \(\frac{z-2}{-2-2}\)
i.e., \(\frac{x+1}{-3}\) = \(\frac{y-3}{-1}\) = \(\frac{z-2}{-4}\)
The point C(5, 5, 10) lies on line (1) if (5, 5, 10) satisfies eqn. (1)
i.e., if \(\frac{5+1}{-3}\) = \(\frac{5-3}{-1}\) = \(\frac{10-2}{-4}\)
if -2 = -2 = -2, which is true
Thus, the given points are collinear.
Question 9.
Find the value of X, for which the points A(2, 1, 3), B(5, 0, 5) and C(-4, λ,-1) are collinear.
Answer:
The eqn. of line passing through the points A(2, 1, 3) and B(5, 0, 5) is given by
\(\frac{x-2}{5-2}\) = \(\frac{y-1}{0-1}\) = \(\frac{2-3}{5-3}\)
i.e., \(\frac{x-2}{3}\) = \(\frac{y-1}{-1}\) = \(\frac{2-3}{2}\)
Since the given points A, B and C are collinear.
∴ Point C(-4, λ, -1) lies on line (1).
Thus C(-4, λ, -1) satisfies eqn. (1).
∴ \(\frac{-4-2}{3}\) = \(\frac{\lambda-1}{-1}\) = \(\frac{-1-3}{2}\)
⇒ -2 = \(\frac{\lambda-1}{-1}\) = -2
⇒ λ -1 = 2
⇒ λ = 3
Question 10.
Find the equations of a line passing through the point P(1, 2, 3) and having direction cosines \(\frac{2}{3}\), \(-\frac{2}{3}\), \(\frac{1}{3}\) Also find the coordinates of a point on the line at a distance of 6 units from P.
Answer:
The required eqn. of line passing through the point P(1, 2, 3) and having direction cosines < \(\frac{2}{3}\), \(\frac{-2}{3}\), \(\frac{1}{3}\) >
i.e., having direction ratios proportional to < 2, -2, 1 > is given by
\(\frac{x-1}{2}\) = \(\frac{y-2}{-2}\) = \(\frac{z-3}{1}\) = t (say)
So any point on line (1) be Q(2 t+1, -2 t+2, t+3)
Also it is given that |P Q| = 6 units
∴ \(\sqrt{(2 t+1-1)^2+(-2 t+2-2)^2+(t+3-3)^2}\) = 6
⇒ \(\sqrt{4 t^2+4 t^2+t^2}\) = 6;
on squaring; we have
⇒ 9 t2 = 36
⇒ t2 = 4
⇒ t = ± 2
When t = 2; coordinates of point Q be (5, -2, 5)
When t = -2; coordinates of point Q be (-3, 6, 1)
Question 11.
Find the values of p and q so that the points (p, q, 1), (-1, 4, -2) and (0, 2, -1) are collinear.
Answer:
Let the given points are A(p, q, 1), B(-1, 4, -2) and C(0, 2, -1)
∴ eqn. of line passing through the points A(p, q, 1) and B(-1, 4, -2) is given by
\(\frac{x-p}{-1-p}\) = \(\frac{y-q}{4-q}\) = \(\frac{z-1}{-2-1}\)
Since points must A, B and C are collinear
∴ point C(0, 2, -1) lies on eqn. (1).
∴ \(\frac{0-p}{-1-p}\) = \(\frac{2-q}{4-q}\)
= \(\frac{-1-1}{-3}\)
i.e., \(\frac{p}{1+p}\) = \(\frac{2-q}{4-q}\) = \(\frac{2}{3}\)
Now
\(\frac{p}{1+p}\) = \(\frac{2}{3}\)
⇒ 3p = 2 + 2 p
⇒ p = 2
\(\frac{2-q}{4-q}\) = \(\frac{2}{3}\)
⇒ 6 – 3 q = 8 – 2 q
⇒ q = -2
and
\(\frac{2-q}{4-q}\) = \(\frac{2}{3}\)
⇒ 6 – 3q = 8 – 2 q ⇒ q = -2
Question 12.
Write the following equations of a line in standard form and hence find the coordinates of a point on it and its direction cosines :
\(\frac{3-2 x}{4}\) = \(\frac{2 y-1}{2}\) = \(\frac{3+z}{2}\)
Answer:
Equation of given line be, given by
\(\frac{3-2 x}{4}\) = \(\frac{2 y-1}{2}\) = \(\frac{3+z}{2}\)
⇒ \(\frac{-2\left(x-\frac{3}{2}\right)}{4}\)
= \(\frac{2\left(y-\frac{1}{2}\right)}{2}\) = \(\frac{z+3}{2}\)
⇒ \(\frac{x-\frac{3}{2}}{-2}\) = \(\frac{y-\frac{1}{2}}{1}\)
= \(\frac{z+3}{2}\)
⇒ \(\frac{x-\frac{3}{2}}{-2}\)
= \(\frac{y-\frac{1}{2}}{1}\) = \(\frac{z-(-3)}{2}\)
which is the required line in standard form clearly line (1) passing through the point (\(\frac{3}{2}\), \(\frac{1}{2}\), – 3) and having direction ratios < -2, 1, 2 >
∴ direction cosines of line (1) are
< \(\frac{-2}{\sqrt{(-2)^2+1^2+2^2}}\), \(\frac{1}{\sqrt{(-2)^2+1^2+2^2}}\), \(\frac{2}{\sqrt{(-2)^2+1^2+2^2}}\) >
i.e., < \(\frac{-2}{3}\), \(\frac{1}{3}\), \(\frac{2}{3}\) >
Question 13.
Find the direction cosines of the line whose equations are \(\frac{x-2}{2}\) = \(\frac{2 y-5}{-3}\), z = -1.
Answer:
Given line can be written in symmetrical form as
\(\frac{x-2}{2}\) = \(\frac{2 y-5}{-3}\) = \(\frac{x+1}{0}\)
⇒ \(\frac{x-2}{2}\) = \(\frac{y-5 / 2}{-3 / 2}\) = \(\frac{z+1}{0}\)
⇒ \(\frac{x-2}{4}\) = \(\frac{y-5 / 2}{-3}\) = \(\frac{z+1}{0}\)
∴ Direction ratios of given line be < 4, -3, 0 >
∴ direction cosines of given line are; < \(\frac{4}{\sqrt{4^2+(-3)^2+0^2}}\), \(\frac{-3}{\sqrt{4^2+(-3)^2+0^2}}\), 0 >
i.e., < \(\frac{4}{5}\), \(\frac{-3}{5}\), 0 >
Question 14.
Find the equations of a line through A(1 ,-1, 5) and parallel to the line
\(\frac{x-2}{3}\) = \(\frac{y-5}{-2}\), z = -1
Answer:
Given eqn. of line can be written in symmetrical form as
\(\frac{x-2}{3}\) = \(\frac{y-5}{-2}\) = \(\frac{z+1}{0}\)
∴ direction ratios of line (1) are < 3, -2, 0 >
Thus direction ratios of the line || to line (1) are proportional to < 3, -2, 0 >
Thus, the required eqn. of line through A(1, -1, 5) and parallel to line (1) is given by
\(\frac{x-1}{3}\) = \(\frac{y+1}{-2}\) = \(\frac{z-5}{0}\)
Question 15.
The equation of a line is \(\frac{2 x-5}{4}\) = \(\frac{y+4}{3}\) = \(\frac{6-z}{6}\).
Find the direction cosines of a line parallel to the line.
Answer:
Given eqn. of line be \(\frac{2 x-5}{4}\) = \(\frac{y+4}{3}\) = \(\frac{6-z}{6}\)
eqn. (1) can be written in symmetrical form as :
\(\frac{2\left(x-\frac{5}{2}\right)}{4}\) = \(\frac{y+4}{3}\) = \(\frac{-(z-6)}{6}\)
i.e., \(\frac{x-\frac{5}{2}}{2}\) = \(\frac{y+4}{3}\) = \(\frac{z-6}{-6}\)
∴ direction ratios of line (1) are < 2, 3, -6 >
Thus the direction cosines of line (1) are ;
< \(\frac{2}{\sqrt{2^2+3^2+(-6)^2}}\), \(\frac{3}{\sqrt{2^2+3^2+(-6)^2}}\), \(\frac{-6}{\sqrt{2^2+3^2+(-6)^2}}\) >
i.e., < \(\frac{2}{7}\), \(\frac{3}{7}\), \(\frac{-6}{7}\) >