Answer :

Let first line be and second

Let A be the point (1, 1, 1) and B and C be points on first line and second line respectively

Consider first line:

⇒ x + 2 = λ and y – 3 = 2λ and z + 1 = 4λ

⇒ x = λ – 2 and y = 2λ + 3 and z = 4λ – 1

Hence the point B on line will have coordinates (λ – 2, 2λ + 3, 4λ – 1)

Let us write the equation of line AB we know a point on the line (1, 1, 1) we need the direction which we will get by writing the vector

Hence the equation of line AB will be

Consider second line

⇒ x – 1 = 2μ and y – 2 = 3μ and z – 3 = 4μ

⇒ x = 2μ + 1 and y = 3μ + 2 and z = 4μ + 3

Hence the point C on line will have coordinates (2μ + 1, 3μ + 2, 4μ + 3)

Let us write the equation of line AC we know a point on the line (1, 1, 1) we need the direction which we will get by writing the vector

Hence the equation of line AC will be

Now as we have to write equation of line which intersects first and second line and also passes through A

Which means we have to write equation of line which passes through A, B and C

AB intersects first line and AC intersects second line

AB and AC will represent the same line if the direction ratios of equation (i) and (ii) are in proportion

⇒ λ – 3 = 2μk …(p)

⇒ 2λ + 2 = k (3μ + 1)

⇒ 2λ + 2 = 3μk + k …(q)

⇒ 4λ – 2 = k (4μ + 2)

⇒ 2λ – 1 = k (2μ + 1)

⇒ 2λ – 1 = 2μk + k …(r)

Rewrite equation (q) as

⇒ 2λ + 2 = (2μk + k) + μk

Using (p) and (r)

⇒ 2λ + 2 = 2λ – 1 +

⇒ 6 = λ – 3

⇒ λ = 9

Put λ = 9 in (p)

⇒ 9 – 3 = 2μk

⇒ 2μk = 6

Put λ = 9 and 2μk = 6 in (r)

⇒ 2(9) – 1 = 6 + k

⇒ k = 11

Put k in 2μk = 6

⇒ 2μ (11) = 6

Put value of λ and μ in (i) and (ii) respectively which are representing the equation of same line

In (i)

In (ii)

Hence equation of line is

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