Three-dimensional Geometry Test

Result

Three-dimensional Geometry Test - 7

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Weekly Quiz Competition

Question 1
1 / -0.25

Two lines whose direction ratios are a₁ , b₁ , c₁ and a₂ , b₂ , c₂ are parallel, if

A

B
a₁ a₂ = b₁ b₂ = c₁ c₂

C
a₁ a₂ + b₁ b₂ + c₁ c₂ = 0

D
Question 2
1 / -0.25

For which value of a lines and are perpendicular?

A
11/70

B
5

C
1

D
70/11

Solution
Question 3
1 / -0.25

The shortest distance between the lines whose equations are and is:

A

B

C

D
Question 4
1 / -0.25

Two lines whose direction ratios are a₁ ,b₁ ,c₁ and a₂ ,b₂ ,c₂ are perpendicular, if

A
a₁ a₂ + b₁ b₂ + c₁ c₂ = 0

B

C
a₁ a₂ = b₁ b₂ = c₁ c₂

D
Question 5
1 / -0.25

The shortest distance between the parallel lines whose equations are and

A

B

C

D
Question 6
1 / -0.25

The angle between the pair of lines given by and is:

A

B

C

D
Question 7
1 / -0.25

The angle between the lines x = 2y = –3z and –4x = 6y = –z is:

A
0 °

B
cos^-1 (1/√3)

C
90 °

D
180 °

Solution

x = 2y = -3z -4x = 6y = -z
x/1 = y/(½) = z(-⅓) x/(-¼) = y/(⅙) = z/(-1)
Cos θ= [(a1a2 + b1b2 + c1c2)/(a1 + b1 + c1)^½ * (a2 + b2 + c2)^½ ]
Cos θ={[(1*(-¼)) + (½)(⅙) + (-⅓)(-1)]/[(1)² + (½)² + (-⅓)² ]1/2 * [(-¼)² + (⅙)² + (-1)² ]1/2}
= {[(-¼+ 1/12 - ⅓)]/[2 + 1 - ⅔]^1/2 * [ -½+ ⅓-½]^½ }
Cos θ= 0
θ= 90deg
Question 8
1 / -0.25

The angle between the lines whose direction cosines are given by the equations 3l + m + 5n = 0, 6nm - 2nl + 5lm = 0 is:

A

B

C

D

Solution

3l + m + 5n = 0
m = - (3l + 5n) -----------(1)
6mn - 2nl + 5lm = 0 ----------(2)
Substitute m=-(3l+5n) in eq(2)
⇒6[- (3l + 5n)]n - 2nl + 5l[- (3l + 5n)] = 0
⇒( -18ln - 30n)n-2nl-15l^2+25ln=0
⇒l(l + 2n) + n(l + 2n) = 0
⇒(l + n) (l + 2n) = 0
∴l = - n and l = -2n
( l / -1 ) = ( n / 1) and ( l / -2) = ( n / 1) -------(3)
Substitute l in equation 1, we get
m = - (3l + 5n)
m = -2n and m = n
( m / -2) = ( n / 1) and ( m / 1) = ( n / 1 ) --------(4)
From ( 3) and (4) we get
( l / -1 ) = ( m / -2) = ( n / 1),
( l / -2) = ( m / 1) = ( n / 1 )
l : m : n = -1 : -2 : 1
l : m : n = -2 : 1 : 1
i.e D.r 's ( -1, -2, 1) and ( -2 , 1 , 1)
Angle between the lines whose direction cosines are
Cos θ= ( -1 ×-2 + -2 ×1 + 1 ×1) / √((-1)^2+(-2)² +1² ))*√((-2)² +1² +1² ))
Cos θ= 1 / √6 √6
Cos θ= 1 / 6
∴θ= cos inverse of (1/6)
∴Angle between the lines whose direction cosines is cos^-1 (1/6)
Question 9
1 / -0.25

The angle between the lines and is:

A

B

C

D
Question 10
1 / -0.25

The length of the shortest distance between the lines and is:

A
Zero

B
6 √3

C
144

D
none of these

Solution

let P and Q be the points on the given lines, respectively. then the general coordinates of P and Q are:
P(k+3, -2k+5, k+7) and Q (7m-1, -6m-1, m-1)
therefore the direction ratios of PQ are (7m-k-4,-6m+2k-6, m-k-8)
now PQ will be the shortest distance if it is perpendicular to both the given lines, therefore by the condition of perpendicularity,
1(7m-k-4) -2(-6m+2k-6) + 1(m-k-8) = 0 (1)
7(7m-k-4) -6(-6m+2k-6) + 1(m-k-8) = 0 (1)
now solving (1) and (2),
m=0 and k = 0
hence the points are P(3,5,7) and Q (-1,-1,-1), therefore the shortest distance between the lines
PQ = sqrt((3+1)² +(5+1)² +(7+1)² )
= sqrt(16+36+64) = sqrt(116)
= 2sqrt(29)