Chapter 11 Three Dimensional Geometry Ex 11.2 Solutions
Question - 11 : - Find the angle between the following pair of lines:
Answer - 11 : -
Question - 12 : - Find the values of p so that the lines
are at right angles.
Answer - 12 : -
So the directionratios of the lines are
-3, 2p/7, 2 and -3p/7,1, -5
Now, as both the linesare at right angles,
So, a1a2 +b1b2 + c1c2 = 0
(-3) (-3p/7) + (2p/7)(1) + 2 (-5) = 0
9p/7 + 2p/7 – 10 = 0
(9p+2p)/7 = 10
11p/7 = 10
11p = 70
p = 70/11
∴ The value of p is70/11
Show that the lines
are perpendicular to each other.
Answer - 13 : -
The equations of thegiven lines are
Two lines withdirection ratios is given as
a1a2 +b1b2 + c1c2 = 0
So the directionratios of the given lines are 7, -5, 1 and 1, 2, 3
i.e., a1 =7, b1 = -5, c1 = 1 and
a2 =1, b2 = 2, c2 = 3
Now, Considering
a1a2 +b1b2 + c1c2 = 7 × 1 +(-5) × 2 + 1 × 3
= 7 -10 + 3
= – 3 + 3
= 0
∴ The two linesare perpendicular to each other.
Question - 14 : - Find the shortest distance between the lines
Answer - 14 : -
Let us rationalizingthe fraction by multiplying the numerator and denominator by √2, we get
∴ The shortest distanceis 3√2/2
Answer - 15 : -
∴ The shortest distanceis 2√29
Answer - 16 : -
Here by comparing theequations we get,
∴ The shortest distanceis 3√19
Find the shortest distance between the lines whose vector equations are
Answer - 17 : -
And,
∴ The shortest distanceis 8√29