Find the direction cosines of a line segment whose direction ratios are:
(i) 2. – 6, 3
(ii) 2, - 1, - 2,
(iii) – 9, 6, -2
(i) direction ratios are:- (2, -6, 3)
So, the direction cosines are- (l, m, n), where, ,
So, l, m, and n are:-
The direction cosines are:-
(ii) direction ratios are:- (2, -1, -2)
So, the direction cosines are:- (l, m, n), where, ,
So, l, m, and n are:-
The direction cosines are:-
(iii) direction ratios are:- (-9, 6, -2)
So, the direction cosines are- (l, m, n), where, ,
So, l, m, and n are:-
The direction cosines are:-
Find the direction ratios and the direction cosines of the line segment joining the points:
(i) A (1, 0, 0) and B(0, 1, 1)
(ii) A(5, 6, -3) and B (1, -6, 3)
(iii) A (-5, 7, -9) and B (-3, 4, -6)
Given two line segments , we have the direction ratios,
Of the line joining these 2 points as,
AB = - + + k, (direction ratio)
The unit vector in this direction will be the direction cosines, i.e.,
Unit vector in this direction is:- (- + + k)/
The direction cosines are ()
(ii) Given two line segments , we have the direction ratios,
Of the line joining these 2 points as,
AB = -4 + (-12) + 6k
The direction ratio in the simplest form will be, (2, 6, -3)
The unit vector in this direction will be the direction cosines, i.e.,
Unit vector in this direction is:- (2 + 6 -3k)/
The direction cosines are ()
(iii) Given two line segments , we have the direction ratios,
Of the line joining these 2 points as,
AB = 2 - 3 + 3k, (direction ratio)
The unit vector in this direction will be the direction cosines, i.e.,
Unit vector in this direction is:- (2 -3 + 3k)/
The direction cosines are ()
Show that the line joining the points A(1,-1, 2) and B(3, 4, -2) is perpendicular to the line joining the points C(0, 3, 2) and D (3, 5, 6).
Given: A(1, -1, 2) and B(3, 4, -2)
The line joining these two points is given by,
AB = 2i + 5j -4k
C(0, 3, 2) and D(3, 5, 6),
The line joining these two points,
CD = 3i + 2j +4k
To prove that the two lines are perpendicular we need to show that the angle between these two lines is
So, AB.CD = 0 (dot product)
Thus, (2i + 5j -4k). (3i + 2j +4k) = 6 + 10 – 16 = 0.
Thus, the two lines are perpendicular.
Show that the line segment joining the origin to the point A(2, 1, 1) is perpendicular to the line segment joining the points B(3, 5, -1) and C(4,3, -1).
Given: O(0, 0, 0) and A(2, 1, 1)
The line joining these two points is given by,
OA = 2i + j +k
B(3, 5, -1) and D(4, 3, -1),
The line joining these two points,
BC = i - 2j +0k
To prove that the two lines are perpendicular we need to show that the angle between these two lines is
So, OA.BC = 0 (dot product)
Thus, (2i + j +k). (i - 2j +0k) = 2 - 2 + 0 = 0.
Thus, the two lines are perpendicular.
Find the value of p for which the line through the points A(3, 5, -1) and B(5, p, 0) 9 is perpendicular to the line through the points C(2, 1, 1) and D (3, 3, -1).
Given: A(3, 5, -1) and B(5, p, 0)
The line joining these two points is given by,
AB = 2i + (p-5)j + k
C(2, 1, 1) and D(3, 3, -1),
The line joining these two points,
CD = i + 2j - 2k
As the two lines are perpendicular, we know that the angle between these two lines is
So, AB.CD = 0 (dot product)
Thus, (2i + (p-5)j + k) .(i + 2j - 2k) = 0.
ð
ð p = 5
Thus, p = 5.
If O is the origin and P (2, 3,4) and Q (1, -2, 1) be any two points show that OPOQ.
Given O(0, 0, 0), P(2, 3, 4) So, OP = 2i + 3j + 4k
Q(1, -2, 1), So, OQ = i – 2j + k
To prove that OPOQ we have,
OP.OQ = 0, i.e. the angle between the line segments is
So, the dot product i.e. |OP||OQ|cos = 0,cos = 0,
OP.OQ = 0
Thus, (2i + 3j + 4k).( i – 2j + k) = 2 – 6 + 4 = 0
Hence, proved.
Show that the line segment joining the points A(1, 2, 3) and B (4, 5, 7) is parallel to the segment joining the points C(-4, 3, -6) and D (2, 9, 2).
Given A(1, 2, 3), B(4, 5, 7), the line joining these two points will be
AB = 3i + 3j + 4k
And the line segment joining, C(-4, 3, -6) and D(2, 9, 2) will be
CD = 6i + 6j + 8k
If CD = r(AB), where r is a scalar constant then,
The two lines are parallel.
Here, CD = 2(AB),
Thus, the two lines are parallel.
If the line segment joining the points A(7, p, 2) and B(q, -2, 5) be parallel to the line segment joining the points C(2, -3, 5) and D(-6, -15,11), find the values of p and q.
Given: A(7, p, 2) and B(q, -2, 5), line segment joining these two points will be, AB = (q-7)i + (-2-p)j + 3k
And the line segment joining C(2, -3, 5) and D(-6, -15, 11) will be, CD = -8i – 12j + 6k
Then, the angle between these two line segments will be 0 degree. So, the cross product will be 0.
ABCD = 0
ð ((q-7)i + (-2-p)j + 3k)( -8i – 12j + 6k) = 0
Thus, solving this we get,
p = 4 and q = 3
Show that the points A(2, 3, 4), B(-1, -2, 1) and C (5, 8, 7) are collinear.
We have to show that the three points are colinear , i.e. they all lie on the same line,
If we define a line which is having a parallel line to AB and the points A and B lie on it, if point C also satisfies the line then, the three points are colinear,
Given A(2, 3, 4) and B(-1, -2, 1), AB = -3i – 5j -3k
The points on the line AB with A on the line can be written as,
R = (2, 3, 4) +a(-3, -5, -3)
Let C = (2-3a, 3-5a, 4-3a)
ð (5, 8, 7) = (2-3a, 3-5a, 4-3a)
ð If a = -1, then L.H.S = R.H.S, thus
The point C lies on the line joining AB,
Hence, the three points are colinear.
Show that the points A(-2, 4.7), B(3, -6. -8) and C(1, -2, -2) are collinear.
We have to show that the three points are colinear , i.e. they all lie on the same line,
If we define a line which is having a parallel line to AB and the points A and B lie on it, if point C also satisfies the line then, the three points are colinear,
Given A(-2, 4, 7) and B(3, -6, -8), AB = 5i – 10j -15k
The points on the line AB with A on the line can be written as,
R = (-2, 4, 7) +a(5, -10, -15)
Let C = (-2+5a, 4-10a, 7-15a)
ð (1, -2, -2) = (-2+5a, 4-10a, 7-15a)
ð If a = 3/5, then L.H.S = R.H.S, thus
The point C lies on the line joining AB,
Hence, the three points are colinear.
Find the value of p for which the points A(-1, 3, 2), B(-4, 2, -2), and C(5, 5, p) are collinear.
We have to show that the three points are colinear , i.e. they all lie on the same line,
If we define a line which is having a parallel line to AB and the points A and B lie on it, as the points are colinear so C must satisfy the line,
Given A(-1, 3, 2) and B(-4, 2, -2), AB = -3i – j -4k
The points on the line AB with A on the line can be written as,
R = (-1, 3, 2) +a(-3, -1, -4)
Let C = (-1-3a, 3-1a, 2-4a)
ð (5, 5, p) = (-1-3a, 3-1a, 2-4a)
ð As L.H.S = R.H.S, thus
ð 5 = -1 – 3a, a = -2
Substituting a = -2 we get, p = 2-4(-2) = 10
Hence, p = 10.
Find the angle between the two lines whose direction cosines are:
and
Let
R1 =
And R2 =
R1.R2 = |R1||R2|cos
Here, as R1 and R2 are the unit vectors with a direction given by the direction cosines hence, |R1| and |R2| are 1.
So, cos = R1.R2 / 1
ð cos
ð =
The angle between the lines is
Find the angle between the two lines whose direction ratios are:
a, b, c and (b – c), (c – a), (a – b).
The angle between the two lines is given by
cos =
where R1 an R2 denote the vectors with the direction ratios,
So, here we have,
R1 = ai + bj + ck and R2 = (b-c)i +(c-a)j + (a-b)k
cos =
cos
Hence,
Find the angle between the lines whose direction ratios are:
2, -3, 4 and 1, 2, 1.
The angle between the two lines is given by
cos =
where R1 and R2 denote the vectors with the direction ratios,
So, here we have,
R1 = 2i - 3j + 4k and R2 = i +2j + k
cos =
cos
Hence,
Find the angle between the lines whose direction ratios are:
1, 1, 2 and ,4
The angle between the two lines is given by
cos =
where R1 and R2 denote the vectors with the direction ratios,
So, here we have,
R1 = i + j + 2k and R2 = ()i -()j + (4)k
cos =
cos
Hence,
Find the angle between the vectors and
The angle between the two lines is given by
cos =
where R1 and R2 denote the vectors with the direction ratios,
So, here we have,
R1 = 3i - 2j + k and R2 = 4i + 5j + 7k
cos =
cos
Hence,
Find the angles made by the following vectors with the coordinate axes:
(i)
(ii)
(iii)
(i) The angle between the two lines is given by
cos =
where R1 and R2 denote the vectors with the direction ratios,
So, here we have,
R1 = i - j + k and R2 = i for x- axis
cos =
cos
Hence,
With y- axis, i. e. R2 = j
cos =
cos
Hence,
With z- axis, i. e. R2 = k
cos =
cos
Hence,
(ii) The angle between the two lines is given by
cos =
where R1 and R2 denote the vectors with the direction ratios,
So, here we have,
R1 = j - k and R2 = i for x- axis
cos =
cos
Hence,
With y- axis, i. e. R2 = j
cos =
cos
Hence,
With z- axis, i. e. R2 = k
cos =
cos
Hence,
(iii) The angle between the two lines is given by
cos =
where R1 and R2 denote the vectors with the direction ratios,
So, here we have,
R1 = i - 4j + 8k and R2 = i for x- axis
cos =
cos
Hence,
With y- axis, i. e. R2 = j
cos =
cos
Hence,
With z- axis, i. e. R2 = k
cos =
cos
Hence,
Find the coordinates of the foot of the perpendicular drawn from the point A(1, 8, 4) to the line joining the points B(0, -1, 3) and C(2, -3, -1).
Given: A(1, 8, 4)
Line segment joining B(0, -1, 3) and C(2, -3, -1) is
BC = 2i – 2j – 4k
Let the foot of the perpendicular be R then,
As R lies on the line having point B and parallel to BC,
So, R = (0, -1, 3) + a(2, -2, -4)
R(2a, -1-2a, 3-4a)
The line segment AR is
AR = (2a-1)i + (-1-2a-8)j + (3-4x-4)k
As the lines AR and BC are perpendicular thus, (as R is the foot of the perpendicular on BC)
AR.BC = 0
ð 2(2a-1) + (-2)(-9-2a) + (-4)(-1-4a) = 0
ð 24a + 20 = 0
ð a =
Substituting a in R we get,
R()