Find the equation of the plane passing through the point (-1,2,-3) and making intercepts on the coordinate axis in the ratio 2:1:3.
Answers
Answer:
If a line makes angles 90°, 135°, 45° with x, y and z-axes respectively, find its direction cosines.
Answer:
Let direction cosines of the line 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 line are 0, 1/√2 and 1/√2
Question 2:
Find the direction cosines of a line which makes equal angles with the coordinate axes.
Answer:
Let the direction cosines of the line make an angle α with each of the coordinate axes.
l = cos α, m = cos α, n = cos α
Now, l2 + m2 + n2 = 1
=> cos2 α + cos2 α + cos2 α = 1
=> 3 cos2 α = 1
=> cos2 α = 1/3
=> cos α = ±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?
Answer:
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/√(324 + 144 + 16), 12/√(324 + 144 + 16), -4/√(324 + 144 + 16)
= -18/√(484), 12/√(484), -4/√(484)
= -18/22, 12/22, -4/22
Thus, the direction cosines are: -18/22, 12/22 and -4/22
Question 4:
Show that the points (2, 3, 4), (−1, −2, 1), (5, 8, 7) are collinear.
Answer:
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 (x2, y2, 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, points 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).
Answer:
The vertices of ∆ABC are A (3, 5, −4), B (−1, 1, 2), and C (−5, −5, −2).
The direction ratios of side AB are (−1 − 3), (1 − 5), and (2 − (−4)) i.e., −4, −4, and 6.
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 ratios of BC are (−5 − (−1)), (−5 − 1), and (−2 − 2) i.e., −4, −6, and −4.
Therefore, the direction cosines of BC are
-4/√{(-4)2 + (-4)2 + (6)2}, -6/√{(-4)2 + (-4)2 + (6)2}, -4/√{(-4)2 + (-4)2 + (6)2}
= -4/2√(17), --6/2√(17), -4/2√(17)
= -2/√(17), -3/√(17), -2/√(17)
The direction ratios of CA are (−5 − 3), (−5 − 5), and (−2 − (−4)) i.e., −8, −10, and 2.
Therefore, the direction cosines of AC are
-8/√{(-8)2 + (10)2 + (2)2}, -10/√{(-8)2 + (10)2 + (2)2}, 2/√{(-8)2 + (10)2 + (2)2}
= -8/√(64 + 100 + 4), -10/√(64 + 100 + 4), 2/√(64 + 100 + 4)
= -8/√(168), -10/√(169), 2/√(168)
= -8/2√(42), -10/2√(42), 2/2√(42)
= -4/√(42), -5/√(42), 1/√(42)
Exercise 11.2
Question 1:
Show that the three lines with direction cosines
12/13, -3/13, -4/13, 4/13, 12/13, 3/13, 3/13, -4/13, 12/13
are mutually perpendicular.
Answer:
Two lines with direction cosines, l1, m1, n1 and l2, m2, n2, are perpendicular to each other, if
l1l2 + m1m2 + n1n2 = 0
(i) For the lines with direction cosines 12/13, -3/13, -4/13 and 4/13, 12/13, 3/13
l1l2 + m1m2 + n1n2 = (12/13) * (4/13) + (-3/13) * (12/13) + (-4/13) * (3/13)
= 48/169 – 36/169 – 12/169
= 48/169 – 48/169
= 0
Therefore, the lines are perpendicular.
(ii) For the lines with direction cosines 4/13, 12/13, 3/13 and 3/13, -4/13, 12/13
l1l2 + m1m2 + n1n2 = (4/13) * (3/13) + (12/13) * (-4/13) + (3/13) * (12/13)
= 12/169 - 48/169 + 36/169
= 48/169 – 48/169
= 0
Therefore, the lines are perpendicular.
(iii) For the lines with direction cosines 3/13, -4/13, 12/13 and 12/13, -3/13, -4/13
l1l2 + m1m2 + n1n2 = (3/13) * (12/13) + (-4/13) * (-3/13) + (12/13) * (-4/13)
= 36/169 + 12/169 – 48/169
= 48/169 – 48/169
= 0
Thus, all the lines are mutually perpendicular.
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