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QUESTION: 1

Two lines whose direction ratios are a_{1}, b_{1}, c_{1} and a_{2}, b_{2}, c_{2} are parallel, if

Solution:

QUESTION: 2

For which value of a lines and are perpendicular?

Solution:

(x-1)/(-3) = (y-2)/(2p/7) = (z-3)/2

(x-1)(-3p/7) = (y - 5)/1 = (z - 6)/(-5)

The direction ratio of the line are -3, 2p/7, -2 and (-3p)/7, 1, -5

Two lines with direction ratios, a1, b1, c1 and a2, b2, c2 are perpendicular to each other if a1a2 + b1b2 + c1c2 = 0

Therefore, (-3)(-3p/7) + (2p/7)(1) + 2(-5) = 0

(9p/7) + (2p/7) = 10

11p = 70

p = 70/1

QUESTION: 3

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

Solution:

QUESTION: 4

Two lines whose direction ratios are a_{1},b_{1},c_{1} and a_{2},b_{2},c_{2} are perpendicular, if

Solution:

QUESTION: 5

The shortest distance between the parallel lines whose equations are and

Solution:

QUESTION: 6

The angle between the pair of lines given byand is:

Solution:

QUESTION: 7

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

Solution:

x = 2y = -3z -4x = 6y = -z

x/1 = y/(½) = z(-⅓) x/(-¼) = y/(⅙) = z/(-1)

Cosθ = [(a1a2 + b1b2 + c1c2)/(a1 + b1 + c1)^{½} * (a2 + b2 + c2)^{½}]

Cosθ ={[(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}}

Cosθ = 0

θ = 90deg

QUESTION: 8

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

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}+1^{2}))*√((-2)^{2}+1^{2}+1^{2}))

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

The angle between the lines and is:

Solution:

QUESTION: 10

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

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)

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