Question 1: If a line makes angles 90°, 135°, and 45° with x, y, and z-axes respectively, find its direction cosines.
Solution:
Let the direction cosines of the lines be l, m, and n.
l = cos 90° = 0
m = cos 135° = – 1/√2
n = cos 45° = 1/√2
Therefore , the direction cosines of the lines are 0, – 1/√2, 1/√2
Question 2: Find the direction cosines of a line which makes equal angles with the coordinate axes.
Solution:
Let the direction cosines of the line make an angle with each of the coordinate axes.
l = cos a, m = cos a, n = cos a
As we know, l2 + m2 + n2 = 1
⇒ cos2𝛂 + cos2𝛂 + cos2𝛂 = 1
⇒ 3cos2 𝛂 = 1/3
⇒ cos2𝛂 = 1/3
⇒ cos𝛂= 1/√3
Thus, the direction cosines of the line, which is equally inclined to the coordinate axes are 1/√3 , 1/√3 and 1/√3 .
Question 3: If a line has the direction ratios-18, 12, -4, then what are its direction cosines?
If a line has direction ratios of -18, 12, and -4, then its direction cosines are
-18/√(-18)2 + (12)2 + (-4)2 , 12/√(-18)2 + (12)2 + (-4)2 , -4/√(-18)2 + (12)2 + (-4)2
= -18/22 , 12/22 , -4/22
= -9/11 , 6/11 , -2/11
Thus, the direction cosines are -9/11 , 6/11 , -2/11
Question4: Show that the points (2, 3, 4), (-1, -2, 1), (5, 8, 7) are collinear.
The given points are a( 2, 3 , 4), B( – 1, – 2, 1 ), and C (5, 8, 7)
It is known that the direction ratios of line joining the points, (x1, y1, z1) and ( x1 , y1 ,z2 are given by x2 – x1 , y2 – y1 , and z2 -z1.
The direction ratios of AB are (-1 -2), (-2 -3), and (1 -4) I.e., -3, -5, and -3.
The direction ratios of BC are ( 5 -(-1 )), (8-(2)), and (7-1) i.e., 6, 10, and 6.
It can be seen that the direction ratios of BC are -2 times that of AB i.e., they are proportional.
Therefore, AB is parallel to BC. Since Point B is common to both AB and BC, point A,B and C are collinear.
Question 5: Find the direction cosines of the sides of the triangle whose vertices are (3,5,-4), (-1,1,2) and (-5,-5,-2).
Solution:
The vertices of △ABC are A(3, 5, -4), B(-1, 1, 2) and C(-5, -5, -2)
The direction ratios of the side AB are (-1 -3), (1,-5) and (2-(-4))
Then, √(-4)2+(-4)2+(6)2 = √16+16+36
= √68
= 2√17
Therefore, the direction cosines of AB are-4/√(-4)2+(-4)2+(6)2 , -4/√(-4)2+(-4)2+(6)2 , 6/√(-4)2+(-4)2+(6)2
⇒ -4/2√17 , -4/2√17 , 6/2√17
⇒ -2/√17 , -2/√17 , 3/√17
The direction ratio of BC are (-5 – (-1)) , (-5 -1), and (-2 -2)
Therefore, the direction cosines of BC are
-4/√(-4)2 + (-6)2 + (-4)2 , -6/√(-4)2+(-6)2+(-4)2 , -4/(-4)2+(-6)2+(-4)2
⇒ -4/2√17 , -6/2√17 , -4/2√17
The direction ratios of CA are (-5 -3) , (-5 -5) and (-2 – (-4))
Therefore the direction cosines of AC are
-8/√(-8)2+(10)2+(2)2 , -5/√(-8)2+(10)2+(2)2 , 2/√(-8)2+(10)2+(2)2
⇒ -8/2√42 , -10/2√42 , 2/2√42