Cartesian Equation of a Line Class 12 OP Malhotra Exe-23B ISC Maths Solutions Ch-23 Three Dimensional Geometry. In this article you would learn to solve questions / problems/ example on cartesian form a line passing through a given point and having given direction cosines. Visit official Website CISCE for detail information about ISC Board Class-12 Mathematics.
Cartesian Equation of a Line Class 12 OP Malhotra Exe-23B ISC Maths Solutions Ch-23 Three Dimensional Geometry
| Board | ISC |
| Publications | S Chand |
| Subject | Maths |
| Class | 12th |
| Chapter-23 | Three Dimensional Geometry |
| Writer | OP Malhotra |
| Exe-23(b) | cartesian form a line passing through a given point and having given direction cosines |
Cartesian Equation of a Line Passing Through a Given Point and Having Given Direction Cosines
Three Dimensional Geometry Class 12 OP Malhotra Exe-23B Solutions
Que-1: Find the equations of a line passing through the point (-1, 2, 3) and having direction ratios proportional to -4, 5, 6.
Sol: We know that, eqn. of line through the point (x1, y1, z1) and having direction ratios < a, b, c> be given by x−x1/a = y−y1/b = z−z1/c
∴ required eqn. of line through the point (-1, 2, 3) and having direction ratio < -4, 5, 6 > be
x−1/−4 = y−2/5 = z−3/6
Que-2: Find the equations of a line passing through the point (2, -3, 0) and having direction cosines −1/7, 4/7, −6/7
Sol: Thus required eqn. of line passing through the point (2, -3, 0) and having direction cosines < −1/7, 4/7, −6/7 > by given by x−2/−(1/7) = y+3/(4/7) = z−0/−(6/7) or x−2/−1 = y+3/4 = z−0/−6
Que-3: Find the equations of a line passing through the points (2, 3, 4) and (4, 6, 5).
Sol: We know that eqn. of line passing through the points (2, 3, 4) and (4, 6, 5) is given by
x−2/4−2 = y−3/6−3 = z−4/5−4
i.e., x−2/2 = y−3/3 = z−4/1
[∵ eqn. of line through the points (x1, y1, z1) and (x2, y2, z2) be given by
x−x1/x2−x1
= y−y1/y2−y1 = z−z1/z2−z1
Que-4: Find the equations of a line passing through the points (3, -2, -5) and (3, -2, 6).
Sol: The required equation of line passing through the points (3,-2,-5) and (3,-2,6) is given by
x−3/3−3
= y+2/−2+2
= z+5/6+5
i.e., x−3/0
= y+2/0 = z+5/11
Que-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
x−5/3−5 = y−5/4−1
= z+6/1−6
i.e., x−5/−2
= y−1/3
= z+6/−5
(i) Since line (1) crosses y z plane
∴ x = 0
∴ from (1); 0−5/−2 = y−1/3 = z−6/−5
⇒ 2 y – 2 = 15
⇒ y = 17/2
and 2(6 – z) = 25
⇒ 6 – z = 25/2
⇒ z = −13/2
Thus the required coordinates of point be (0, 17/2, −13/2).
(ii) Since line (1) crosses x y plane ∴ z = 0
i.e., putting z = 0 in eqn. (1); we have
x−5/−2
= y−1/3 = 0−6/−5
= 6/5
⇒ 5x – 25 = -12
⇒ x = 13/5
and
5(y – 1) = 18
⇒ y = 23/5
Thus the required coordinates of point be (13/5, 23/5, 0)
(iii) Since line (1) crosses z x plane,
i.e., y = 0, putting y = 0 in eqn. (1); we have
x−5/−2 = 0−1/3
= z−6/−5
⇒ 3(x – 5) = 2
z-6 = 5/3
⇒ z = 23/3
and
z – 6 = 5/3 ⇒ z = 23/3
Thus, the required coordinates of point be (17/3, 0, 23/3)
Que-6: The cartesian equations of a line are 6 x-2=3 y+1=2 z-2. Find the direction ratios.
Sol: The eqn. of given line by 6 x – 2 = 3 y + 1 = 2 z – 2
⇒ 6(x – 1/3) = 3(y + 1/3) = 2(z – 1)
⇒ x−1/3/(1/6) = y+1/3/(1/3)
= z−1/(1/2)
⇒ (x−(1/3))/1
=( y+(1/3))/2 = z−1/3
∴ direction ratios of line (1) are < 1, 2, 3 >
Que-7: Find the cartesian equations of a line which
(i) passes through the point (1, 2, 3) and parallel to the line
−x−2/1 = y+3/7 = 2z−6/3
(ii) passes through the point (1,3,-2) and is parallel to the line given by
x+1/3 = y+4/5 = 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 ).
Sol: (i) eqn. of given line be
−x−2/1 = y+3/7 = 2z−6/3
⇒ x+2/−1 = y+3/7 = z−3/(3/2)
⇒ x+2/−2 = y+3/14 = 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 x−1/−2
= y−2/14
= z−3/3
(ii) eqn. of given line be x+1/3 = y−4/5
= 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 x−1/3 = y−3/5
= 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 x−2/2
= y+1/−2 = z−1/1
Que-8: Prove that the points A(-1, 3, 2), B(-4, 2, -2) and C(5, 5, 10) are collinear.
Sol: The eqn. of line through the points A(-1, 3, 2) and B(-4, 2, -2) be given by
x+1/−4+1 = y−3/2−3 = z−2/−2−2
i.e., x+1/−3 = y−3/−1 = z−2/−4
The point C(5, 5, 10) lies on line (1) if (5, 5, 10) satisfies eqn. (1)
i.e., if 5+1/−3 = 5−3/−1 = 10−2/−4
if -2 = -2 = -2, which is true
Thus, the given points are collinear.
Que-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.
Sol: The eqn. of line passing through the points A(2, 1, 3) and B(5, 0, 5) is given by
x−2/5−2 = y−1/0−1 = 2−3/5−3
i.e., x−2/3 = y−1/−1 = 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).
∴ −4−2/3 = λ−1/−1 = −1−3/2
⇒ -2 = λ−1/−1 = -2
⇒ λ -1 = 2
⇒ λ = 3
Que-10: Find the equations of a line passing through the point P(1, 2, 3) and having direction cosines 2/3, −2/3, 1/3 Also find the coordinates of a point on the line at a distance of 6 units from P.
Sol: The required eqn. of line passing through the point P(1, 2, 3) and having direction cosines < 2/3, −2/3, 1/3 >
i.e., having direction ratios proportional to < 2, -2, 1 > is given by
x−1/2 = y−2/−2 = 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
∴ √(2t+1−1)²+(−2t+2−2)²+(t+3−3)² = 6
⇒ √4t²+4t²+t² = 6;
on squaring; we have
⇒ 9t² = 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)
Que-11: Find the values of p and q so that the points (p, q, 1), (-1, 4, -2) and (0, 2, -1) are collinear.
Sol: 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
x−p/−1−p = y−q/4−q = z−1/−2−1
Since points must A, B and C are collinear
∴ point C(0, 2, -1) lies on eqn. (1).
∴ 0−p/−1−p = 2−q/4−q
= −1−1−3
i.e., p/1+p = 2−q/4−q = 2/3
Now
p/1+p = 2/3
⇒ 3p = 2 + 2 p
⇒ p = 2
2−q/4−q = 2/3
⇒ 6 – 3 q = 8 – 2 q
⇒ q = -2
and
2−q/4−q = 2/3
⇒ 6 – 3q = 8 – 2 q ⇒ q = -2
Que-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 :
3−2x/4 = 2y−1/2 = 3+z/2
Sol: Equation of given line be, given by
3−2x/4 = 2y−1/2 = 3+z/2
⇒ −2(x−3/2)/4
= 2(y−(1/2))/2 = z+3/2
⇒ (x−(3/2))/−2 = (y−(1/2))/1
= z+3/2
⇒ (x−(3/2))/−2
=( y−(1/2))/1 = z−(−3)/2
which is the required line in standard form clearly line (1) passing through the point (3/2, 1/2, – 3) and having direction ratios < -2, 1, 2 >
∴ direction cosines of line (1) are
< −2/√(−2)²+1²+2², 1/√(−2)²+1²+2², 2/√(−2)²+1²+2² >
i.e., < −2/3, 1/3, 2/3 >
Que-13: Find the direction cosines of the line whose equations are x−2/2 = 2y−5/−3, z = -1.
Sol: Given line can be written in symmetrical form as
x−2/2 = 2y−5/−3 = x+1/0
⇒ x−2/2 =(y−(5/2))/−(3/2) = z+1/0
⇒ x−2/4 = (y−(5/2))/−3 = z+1/0
∴ Direction ratios of given line be < 4, -3, 0 >
∴ direction cosines of given line are; < 4/√4²+(−3)²+0², −3/√4²+(−3)²+0², 0 >
i.e., < 4/5, −3/5, 0 >
Que-14: Find the equations of a line through A(1 ,-1, 5) and parallel to the line
x−2/3 = y−5/−2, z = -1
Sol: Given eqn. of line can be written in symmetrical form as
x−2/3 = y−5/−2 = 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
x−1/3 = y+1/−2 = z−5/0
Que-15: The equation of a line is 2x−5/4 = y+4/3 = 6−z/6.
Find the direction cosines of a line parallel to the line.
Sol: Given eqn. of line be 2x−5/4 = y+4/3 = 6−z/6
eqn. (1) can be written in symmetrical form as :
2(x−(5/2))/4 = y+4/3 = −(z−6)/6
i.e., (x−(52))/2 = y+4/3 = z−6/−6
∴ direction ratios of line (1) are < 2, 3, -6 >
Thus the direction cosines of line (1) are ;
< 2/√2²+3²+(−6)², 3/√2²+3²+(−6)², −6/√2²+3²+(−6)² >
i.e., < 2/7, 3/7, −6/7 >
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