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Question: A line having direction ratios 1, -4,2 intersects the lines $\frac{x-7}{3}=\frac{y-1}{-1}=\frac{z+2}...

A line having direction ratios 1, -4,2 intersects the lines x−73=y−1−1=z+21\frac{x-7}{3}=\frac{y-1}{-1}=\frac{z+2}{1} and x2=y−73=z1\frac{x}{2}=\frac{y-7}{3}=\frac{z}{1} at the points A and B resp., then co-ordinates of points A and B are

A

A(-8,6, -7) B(-6, -2, -3)

B

A(8,6,7) B(6,2,3)

C

A(8,6,7) B(6,-2,-3)

D

A(7,6,8)B(-3,-2,6)

Answer

A(-8,6, -7) B(-6, -2, -3)

Explanation

Solution

Let the line with direction ratios (1,−4,2)(1,-4,2) be LL. It meets:

  • Line 1: x−73=y−1−1=z+21=μ\frac{x-7}{3} = \frac{y-1}{-1} = \frac{z+2}{1} = \mu, so

    A=(7+3μ,  1−μ,  −2+μ).A = (7+3\mu,\; 1-\mu,\; -2+\mu).

  • Line 2: x2=y−73=z1=λ\frac{x}{2} = \frac{y-7}{3} = \frac{z}{1} = \lambda, so

    B=(2λ,  7+3λ,  λ).B = (2\lambda,\; 7+3\lambda,\; \lambda).

Since AA and BB lie on LL, the vector AB⃗\vec{AB} must be parallel to (1,−4,2)(1,-4,2). Hence, there exists kk such that:

B−A=k(1,−4,2).B-A = k(1,-4,2).

Calculate:

B−A=(2λ−(7+3μ),  (7+3λ)−(1−μ),  λ−(−2+μ)).B - A = \big(2\lambda - (7+3\mu),\; (7+3\lambda)-(1-\mu),\; \lambda -(-2+\mu)\big).

This gives:

{2λ−7−3μ=k(i)6+3λ+μ=−4k(ii)2+λ−μ=2k(iii)\begin{cases} 2\lambda - 7 - 3\mu = k \quad \text{(i)} \\ 6 + 3\lambda + \mu = -4k \quad \text{(ii)} \\ 2 + \lambda - \mu = 2k \quad \text{(iii)} \end{cases}

From (i): k=2λ−7−3μk = 2\lambda - 7 - 3\mu.

Substitute into (iii):

2+λ−μ=2(2λ−7−3μ)=4λ−14−6μ.2 + \lambda - \mu = 2(2\lambda - 7 - 3\mu) = 4\lambda - 14 - 6\mu.

Simplify:

2+λ−μ−4λ+14+6μ=0⟹−3λ+5μ+16=0,2 + \lambda - \mu - 4\lambda + 14 + 6\mu = 0 \quad \Longrightarrow \quad -3\lambda + 5\mu + 16 = 0,

or

3λ−5μ=16.(A)3\lambda - 5\mu = 16. \quad \text{(A)}

From (ii):

6+3λ+μ=−4(2λ−7−3μ)=−8λ+28+12μ.6 + 3\lambda + \mu = -4(2\lambda - 7 - 3\mu) = -8\lambda + 28 + 12\mu.

Simplify:

6+3λ+μ+8λ−28−12μ=0⟹11λ−11μ−22=0,6 + 3\lambda + \mu + 8\lambda - 28 - 12\mu = 0 \quad \Longrightarrow \quad 11\lambda - 11\mu - 22 = 0, λ−μ=2.(B)\lambda - \mu = 2. \quad \text{(B)}

From (B): λ=μ+2\lambda = \mu + 2. Substitute into (A):

3(μ+2)−5μ=16⟹3μ+6−5μ=16,3(\mu+2) - 5\mu = 16 \quad \Longrightarrow \quad 3\mu + 6 - 5\mu = 16, −2μ=10⟹μ=−5.-2\mu = 10 \quad \Longrightarrow \quad \mu = -5.

Thus, λ=−5+2=−3\lambda = -5 + 2 = -3.

Substitute back to find coordinates:

A=(7+3(−5),  1−(−5),  −2+(−5))=(−8,6,−7),A = (7+3(-5),\; 1-(-5),\; -2+(-5)) = (-8,6,-7), B=(2(−3),  7+3(−3),  −3)=(−6,−2,−3).B = (2(-3),\; 7+3(-3),\; -3) = (-6,-2,-3).