Chapter 11 Three Dimensional Geometry Ex 11.1 Solutions
Question - 1 : - If a line makes angles 90°, 135°, 45° with the x, y and z-axes respectively, find its direction cosines.
Answer - 1 : -
Let the direction cosines of the line be l, m and n.
Here let α = 90°, β = 135° and γ = 45°
So,
l = cos α, m = cos β and n = cos γ
So direction cosines are
l = cos 90° = 0
m = cos 135°= cos (180° – 45°) = -cos 45° = -1/√2
n = cos 45° = 1/√2
∴ The direction cosines of the line are 0, -1/√2, 1/√2
Question - 2 : - Find the direction cosines of a line which makes equal angles with the coordinate axes.
Answer - 2 : -
Given:
Angles are equal.
So let the angles be α, β, γ
Let the direction cosines of the line be l, m and n
l = cos α, m = cos β and n = cos γ
Here given α = β = γ (Since, line makes equal angles with thecoordinate axes) … (1)
The direction cosines are
l = cos α, m = cos β and n = cos γ
We have,
l2 + m 2 + n2 =1
cos2 α + cos2β + cos2γ= 1
From (1) we have,
cos2 α + cos2 α + cos2 α= 1
3 cos2 α = 1
Cos α = ± √(1/3)
∴ The directioncosines are
l = ± √(1/3), m= ± √(1/3), n = ± √(1/3)
Question - 3 : - If a line has the direction ratios –18, 12, –4, then what are its direction cosines?
Answer - 3 : -
Given
Direction ratios as -18, 12, -4
Where, a = -18, b = 12, c = -4
Let us consider the direction ratios of the line as a, b and c
Then the direction cosines are
∴ Thedirection cosines are
-18/22, 12/22, -4/22 => -9/11, 6/11, -2/11
Question - 4 : - Show that the points (2, 3, 4), (–1, –2, 1), (5, 8, 7) are collinear.
Answer - 4 : -
If the directionratios of two lines segments are proportional, then the lines are collinear.
Given:
A(2, 3, 4), B(−1, −2,1), C(5, 8, 7)
Direction ratio ofline joining A (2, 3, 4) and B (−1, −2, 1), are
(−1−2), (−2−3), (1−4)= (−3, −5, −3)
Where, a1 =-3, b1 = -5, c1 = -3
Direction ratio ofline joining B (−1, −2, 1) and C (5, 8, 7) are
(5− (−1)), (8− (−2)),(7−1) = (6, 10, 6)
Where, a2 =6, b2 = 10 and c2 =6
Now,
∴ A, B, C arecollinear.
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).
Answer - 5 : -
Given:
The vertices are (3,5, –4), (-1, 1, 2) and (–5, –5, –2).
The direction cosinesof the two points passing through A(x1, y1, z1)and B(x2, y2, z2) is given by (x2 –x1), (y2-y1), (z2-z1)
Firstly let us findthe direction ratios of AB
Where, A = (3, 5, -4)and B = (-1, 1, 2)
Ratio of AB = [(x2 –x1)2, (y2 – y1)2,(z2 – z1)2]
= (-1-3), (1-5),(2-(-4)) = -4, -4, 6
Then by using theformula,
√[(x2 –x1)2 + (y2 – y1)2 +(z2 – z1)2]
√[(-4)2 +(-4)2 + (6)2] = √(16+16+36)
= √68
= 2√17
Now let us find thedirection cosines of the line AB
By using the formula,
-4/2√17 , -4/2√17,6/2√17
Or -2/√17, -2/√17,3/√17
Similarly,
Let us find thedirection ratios of BC
Where, B = (-1, 1, 2)and C = (-5, -5, -2)
Ratio of AB = [(x2 –x1)2, (y2 – y1)2,(z2 – z1)2]
= (-5+1), (-5-1),(-2-2) = -4, -6, -4
Then by using theformula,
√[(x2 –x1)2 + (y2 – y1)2 +(z2 – z1)2]
√[(-4)2 +(-6)2 + (-4)2] = √(16+36+16)
= √68
= 2√17
Now let us find thedirection cosines of the line AB
By using the formula,
-4/2√17, -6/2√17,-4/2√17
Or -2/√17, -3/√17,-2/√17
Similarly,
Let us find thedirection ratios of CA
Where, C = (-5, -5,-2) and A = (3, 5, -4)
Ratio of AB = [(x2 –x1)2, (y2 – y1)2,(z2 – z1)2]
= (3+5), (5+5), (-4+2)= 8, 10, -2
Then by using theformula,
√[(x2 –x1)2 + (y2 – y1)2 +(z2 – z1)2]
√[(8)2 +(10)2 + (-2)2] = √(64+100+4)
= √168
= 2√42
Now let us find thedirection cosines of the line AB
By using the formula,
8/2√42, 10/2√42,-2/2√42
Or 4/√42, 5/√42,-1/√42